Conjecture Every card that has an even number on one side is .docx

Conjecture: Every card that has an even number on one side is
red
on the other side.
Which cards does one have to turn over to find out whether the
conjecture is true?
PHIL 110; Spring 2020; Lecture 15 1
Every card has a colour on one side and a number on the other.
Is this a valid inference?
Premise: Every person at the party was a twentysomething.
Conclusion: Every person at the party who was wearing a jacket
was
a twentysomething.
Valid! Not valid!
13: Everything
PHIL 110; Spring 2020; Tom Donaldson

Things to be getting on with
• Take it easy – relax after the midterm.
• There will be an assignment next week.
1: Beyond Statement
Logic
Beyond Statement Logic
• There are certain inferences which cannot be adequately
evaluated using the tools we’ve discussed so far.
• Let’s look at some examples.
Tense Logic
Premise: Ashni will swim and Ben will swim, but Ashni won’t
swim while Ben swims.
Conclusion: Either Ashni will swim and then Ben will, or Ben
will

swim and then Ashni will.
Deontic Logic
Premise: You may have coffee.
Premise: You may have tea.
Conclusion: You may have coffee and tea.
Premise: C
Premise: T
Conclusion: (C & T)
The Logic of Quantification
Premise: Every dog is a mammal.
Premise: Fido is a dog.
Conclusion: Fido is a mammal.

We’ll focus on the logic of quantification …
• Tense isn’t relevant in (pure) mathematics.
• Deontic notions (such as obligation and permission) are also
not
relevant.
• But “every” is everywhere in mathematics!
• Every natural number has a unique prime factorization.
• Every polynomial of degree three has a real root.
• Every polynomial is differentiable.
• The negation of an “every” statement is equivalent to a
“some”
statement.
• So we’ll focus on “every” and “some”.
2: Introducing “Every”
Universal Generalizations
Universal generalizations in English often contain the word
“every”, or
“everything” or “everyone”, or “any”, or “all”:
• Every whale is a mammal.
• Everything is broken.

• All dogs are hairy.
But there are exceptions:
• Dogs have four legs.
• A bear is a mammal.
• Man is born free, but everywhere he is in chains.
The Need for Symbols
Compare:
• A bear is a mammal.
• A bear goes through my trash can every night.
As we said earlier in the term, English is extremely
complicated, so
in logic we need to use artificial symbols instead.
We won’t introduce any new symbols today, however.
Strict vs. Loose
• There are two sorts of universal generalization – strict and

loose.
• Strict: “Every single dog without exception is a mammal.”
• Loose: “Dogs have four legs.”
• A strict universal generalization can be refuted by just one
example, a “counterexample”.
• For example, if someone claims that all birds can fly, you can
prove him
wrong by showing him a single penguin.
Strict vs. Loose
• There are two sorts of “every” statement – strict and loose.
• Strict: “Every single dog without exception is a mammal.”
• Loose: “Dogs have four legs.”
• A strict universal generalization can be refuted by just one
example, a “counterexample”.
• Loose universal generalizations are not so easily refuted.
• It is sometimes unclear whether a universal generalization is
strict
or loose. Consider: “Abortions are immoral.”
• When doing philosophy, it is a good idea often to ask, “Is that
strict

or loose?”
Domains of Quantification
• When one says “everything”, it is rare that one means to
consider
every single thing in the whole universe without restriction.
• Example: “Every beer bottle is empty!”
• Example: “Every number is either odd or even.”
• Typically, one means to consider only the things within a
certain
“domain of quantification”.
Vacuous Generalizations
• The universal generalization “Every A is a B” is said to be
“vacuous” if there are no A’s. Consider:
• Every unicorn has a horn.
• Every witch wears a black hat.
• Logicians assume that all vacuous universal generalizations
are
true.
• This might seem a bit odd at first. (Think about “All the

kryptonite
in Vancouver is stored in my basement.”)
Premise: Every person at the party was a twentysomething.
Conclusion: Every person at the party who was wearing a jacket
was
a twentysomething.
Premise: Every A is C.
Conclusion: Every A that is B is C.
Existential Generalizations
Existential generalizations often contain “some” or “there is” or
“a”:
• A dog is barking in the garden.
• Some dog is barking in the garden.
• There is a dog barking the garden.
The negation of a universal generalization is equivalent to an
existential generalization:

• It is not true that everyone enjoyed the party.
• Someone didn’t enjoy the party.
The negation of an existential generalization is equivalent to a
universal generalization:
• It is not true that one of the men at the party was unmarried.
• All of the men at the party were married.
3: Venn Diagrams
Famous people
Famous people
Denzel
Washington

Famous people
Denzel
Washington
Kim
Kardashian
Famous people
Denzel
Washington
Kim
Kardashian
Tom
Donaldson
Famous people
Denzel
Washington
Kim
Kardashian
People who

should be
famous
Tom
Donaldson
Dogs
Black things
x
There is a dog that isn’t black.
Dogs
Black things
x
Some dog is black.
Dogs

Black things
x
Something is black.
x
No dog is black.
Every dog is black.
Canadians Singers
Talented People
x
Canadians Singers

Talented People
x
Canadians Singers
Talented People
x
x
3: Four Kinds of
Statements
Code Form Examples
A All A are B. Every zebra is a mammal.
All men are mortal.
Every single member of the club was at the party.

E No A are B. No person can hold their breath for thirty
minutes.
Not one person in this room is honest.
Expensive moisturizing creams are never worth buying.
I Some A is B. Some foxes live in Iceland.
There are people who can run a mile in four minutes.
At least one singer was off key.
O Some A is not B. Some logicians are not well groomed.
Some famous people do not deserve to be famous.
There are basketball players who aren’t tall.
All A are B.
No A are B.
A
B

Some A is B.
x
A
B
Some A is not B.
x
4: Carrol l Diagrams
Venn Diagrams With Four Categories …
… are rather hard to draw.
Venn Diagrams with Five Categories …
… are even harder!

This is where Carroll Diagrams come in
handy!
Carroll diagrams work just like Venn diagrams, except they use
rectangular grids rather than overlapping ellipses.
B
A
B
A
C
B
A
C
D

5 : Eva lua t i ng In f e r ences
Us ing Ve nn D ia g ra ms
Evaluating Inferences Using Venn Diagrams
I recommend the following procedure for evaluating inferences
using
Venn diagrams:
1. Write down a list all of the premises and the negation of the
conclusion.
2. Try to draw a Venn diagram depicting a situation in which all
the
statements on your list are true.
3. If you succeed, you have shown that the inference is invalid.
4. If you find that it is impossible to depict such a situation, this
is an
indication that the inference is valid.
A tip for step three: When drawing your diagram, deal with the
universal
generalizations first. Then think about the existential
generalizations.
To put it another way: Do your shading first!!
A B

C
All A are B.
No B are C.
Therefore:
No A are C.
A B
C
All A are B.
No B are C.
Therefore:
No A are C.
All A are B.
No B are C.
Some A is C.

A B
C
Some A is B.
All B are C.
Therefore:
Some A is C.
A B
C
Some A is B.
All B are C.
Therefore:
Some A is C.
Some A is B.
All B are C.
No A is C.

A B
C
Some A is B.
Some B is C.
Therefore:
Some A is C.
A B
C
Some A is B.
Some B is C.
Therefore:
Some A is C.
Some A is B.
Some B is C.
No A is C

Some for you to try
(1) No A is B. (2) All A are B.
No B is C. All B are C.
Therefore: Therefore:
No A is C. All B are C
Valid! Invalid!
Two More
(1) No A are B. (2) All A are B.
Some A is C. No C are A.
Some C is not B. No C are B.
Valid! Not valid!
One More

Every A is either B or C.
Everything that is not C is not B.
Something is A.
Therefore:
Something is C.
Valid! Not valid!
2 2 : “ L o v e ” a n d O t h e r
Tw o - P l a c e P r e d i c a t e s
P H I L 1 1 0 ; S p r i n g 2 0 2 0 ; To m D o n a l d s o n
1 : N a m e s a n d P r e d i c a t e s
Proper Names
A proper name is a word that represents an individual member
of the
domain of quantification. For example, if our domain is
philosophers, we
might use the following names:

• “Socrates”
• “Mary Wollstonecraft”
• “Mozi”
Similarly, if our domain is countries, we might use the
following names:
• “Canada”
• “Mexico”
• “India”
Predicates
If you take a statement and remove one or more proper names
from
it, the result is a predicate. (If you like, a predicate is a
sentence with
one or more proper-name-shaped holes in it.)
• A one-place predicate has one hole.
• A two-place predicate has two holes.
• A three-place predicate has three holes.
• (And so on!)

Predicates
One can make a sentence by taking a predicate, and then “filling
in”
the hole with a name (or filling in the holes with names):
“Ashni” + “____ likes muffins” = “Ashni likes muffins”.
Predicates
• So far, we’ve considered only one-place predicates like “____
likes
dancing.” and “____ is having fun.”
• Today, we’re going to look at two-place predicates.
• For simplicity, let’s restrict our attention to a single example:
x loves y. Lxy
Love
Lrj Romeo loves Juliet.
Ljr Juliet loves Romeo.

Lnn Narcissus loves himself.
Lqe Quasimodo loves Esmerelda
Notice that there’s a big difference between loving and being
loved,
so the order of the names matters.
Romeo
Juliet
Quasimodo
Esmerelda
Narcissus
2 : S y m b o l i z a t i o n
P r a c t i c e
An Example
In this section, let’s suppose that the domain of quantification is

people at Tom’s party, and that this includes Ashni, Ben,
Chiara, and
nobody else.
Symbolization Practice
English Sentence Symbolization
Ashni loves Ben. Lab
Ben loves Ashni. Lba
Ashni and Ben love each other. (Lab & Lba)
Ben loves himself. Lbb
Ashni and Ben both love Chiara. (Lac & Lbc)
Existential Quantification
• The existential quantifier works in just the same way as
before!
• ‘∃ x Lxa’ means someone loves Ashni, and has the same truth
value as
this disjunction:

• ‘∃ x Lax’ means Ashni loves someone, and has the same truth
value as
this disjunction:
Existential Quantification
• Now for a more complex example. This is a symbolization of
“There is someone whom Ashni and Ben both love”:
∃ x (Lax & Lbx)
• And here is a symbolization of “There is someone who loves
both
Ben and Chiara”:
∃ x (Lxb & Lxc)
Universal Quantification
• The universal quantifier works just as before!
• The sentence ‘∀ x Lxa’ means everyone loves Ashni, and has
the same

truth value as this conjunction:
((Laa & Lba) & Lca)
• This is very different, of course, to ‘∀ x Lax’, which means
Ashni loves
everyone:
((Laa & Lab) & Lac)
Universal Quantification
• Now for a more complex example. Here is a symbolization of
“Everyone loves either Ashni or Ben”:
∀
Symbolization Practice
There is someone who Ashni doesn’t love. ∃
∃ x Lax
∀ x Lxb
Nobody loves Ben. ∀

3 : N a t u r a l D e d u c t i o n
P r a c t i c e
Exercise
In each case, show that the inference is valid by constructing a
natural deduction proof
with the given premises and the given conclusion:
(1) Premise: Everyone loves Ashni.
Conclusion: Someone loves themself.
(2) Premise: Everyone loves Ashni.
Premise: Ashni loves Ben.
Conclusion: Someone loves both Ashni and Ben.
Exercise
1. ∀ x Lxa Premise (“Everyone loves Ashni”)
2. Laa 1, UI

3. ∃ x Lxx 2, EG (“Someone loves themself ”)
Exercise
1. ∀ x Lxa Premise (“Everyone loves Ashni”)
2. Lab Premise (“Ashni loves Ben”)
3. Laa 1, UI
4. (Laa & Lab) 2, 3 Conj
5. ∃ x (Lxa & Lxb) 4, EG (“Someone loves both Ashni and
Ben.”)
4 : S t a t e m e n t s t h a t c o n t a i n
b o t h a o n e - p l a c e a n d a
t w o - p l a c e p r e d i c a t e
Love and Dancing
• Let’s continue our discussion of the party, with only Ashni,
Ben
and Chiara in attendance.

• Let’s use the following predicates:
Dx x is a dancer.
Lxy x loves y.
Love and Dancing
A Every dancer loves Ashni. ∀ x(Dx → Lxa)
E No dancer loves Ashni. ∀
I Some dancer loves Ashni. ∃ x(Dx & Lxa)
O Some dancer doesn’t love Ashni. ∃
Love and Dancing
Everyone who Ashni loves is dancing. ∀ x(Lax → Dx)
Nobody who loves Ashni is dancing. ∀
Ashni loves a dancer. ∃ x(Dx & Lax)

There’s this dancer who Ashni doesn’t love. ∃
5 : Q u a n t i f i e r s
I n s i d e Q u a n t i f i e r s
First Example: ∀ x ∀ y Lxy
• Consider the English sentence “Everybody loves everybody.”
• This has two universal quantifiers in it!
• The correct symbolization is this: ∀ x ∀ y Lxy
• This shouldn’t be confused with this: ∀ x Lxx
• This latter statement means everybody loves themself.
Second Example: ∀ ∀ y Lxy
• This is trickier to interpret!
• Here’s one approach.
∃ x∀ y
Lxy.

• In English: It is not the case that there is some one person who
loves everybody.
Third Example: ∃ x ∃ y Lxy
• This is an easy one!
• It means, someone loves someone.
Fourth Example:
(1) ∃ x ∀ y Lxy
(2) ∀ x ∃ y Lxy
• Both of these statements contain “everyone”, “someone”, and
“loves”. So each must mean something like Everyone loves
someone, or Someone loves everyone.
• But can we get clear on the difference between them?
Fourth Example:
(1) ∃ x ∀ y Lxy
(2) ∀ x ∃ y Lxy
• (1) is an existential generalization. Its instances are:

• ∀ y Lay Ashni loves everyone.
• ∀ y Lby Ben loves everyone.
• ∀ y Lcy Chiara loves everyone.
• So (1) amounts to: Either Ashni or Ben or Chiara loves
everyone.
• In short, (1) means: There is a single (very amorous!) person
who
loves all.
Fourth Example:
(1) ∃ x ∀ y Lxy
(2) ∀ x ∃ y Lxy
• (2) is a universal quantification. Its instances are:
• ∃ y Lay Ashni loves someone.
• ∃ y Lby Ben loves someone.
• ∃ y Lcy Chiara loves someone.
• So (2) means something like this: Ashni loves someone, and
Ben
loves someone, and Chiara loves someone.
• In short, (2) means: Every person has someone that they love.

Fourth Example:
(1) ∃ x ∀ y Lxy
(2) ∀ x ∃ y Lxy
• In summary:
• (1) means: There is a single (very amorous!) person who loves
all.
• (2) means: Every person has someone that they love.
Fourth Example:
(1) ∃ x ∀ y Lxy
(2) ∀ x ∃ y Lxy
• In summary:
• (1) means: There is a single (very amorous!) person who loves
all.
• (2) means: Every person has someone that they love.
(1) (2)
a
a b c
b c

Fifth Example: ∃ x(Dx & ∀ y Lxy)
• Let’s start with this: ∀ y Lxy
• This means, x loves everybody.
• So the whole statement means: There is some person x, where
x is
a dancer and x loves everybody.
• To put it more succinctly: There is some (amorous!) dancer
who
loves everybody.
Mx: x is male. Pxy: x is a parent of y.
Fx: x is female. Lxy: x loves y.
Yx: x is young. Kxy: x kills y.
1. ∃ x ∃ y ((Yx & Mx) & (Yy & Fy) & (Lxy & Lyx) & (Kxx &
Kyy))
2. ∃ x ∃ y ∃ z ((Mx & My & Fz) & (Pyx & Pzx) & (Kxy &
Lxz))
Great Works of Literature, in Symbols

Continue to assume that the domain contains just three objects:
a, b
and c. For each of the following statements, express them in
natural
English, and draw an arrow diagram showing a situation in
which
the statement is true.
(1) ∀ x Lax
(2) ∀ x Lxa
(3) ∀ x ∀ y Lxy
(4) ∀ x ∀
(5) ∃ x ∀ y Lyx
(1) ∀ x Lax
Ashni loves everyone.
b a c
(2) ∀ x Lxa

Everyone loves Ashni.
b a c
(3) ∀ x ∀ y Lxy
Everyone loves everyone!
a
b c
(4) ∀ x ∀ Lxy
Nobody loves anyone.
a
b c
(5) ∃ x ∀ y Lyx
There is some single individual who is loved by all.

b c a
The Exam Has Been Scheduled
• Most of you will take the exam on WED 15-Apr, 1200-15:00,
C9001
• Some of you will take the exam at the CAL.
• Some of you may qualify for “hardship”:
• You have three exams within 24 hours.
• You have an examination at one location (e.g. the Burnaby
campus)
followed immediately by an exam at another location (e.g., the
Surrey
campus).
1 7 : M o r e o n t h e
U n i v e r s a l Q u a n t i f i e r
1 : R e c a p

Symbolizing A Statements
• To symbolize an A statement, you use the universal quantifier
“∀ ”
and the arrow “→”.
• For example:
All whales are mammals. ∀ x (Wx → Mx)
Every Canadian is polite. ∀ x (Cx → Px)
• If you find this hard to understand, don’t worry! You can
simply
memorize the fact that this is how A statements are symbolized.
Symbolizing E Statements
• To symbolize an E statement, you use the universal quantifier
“∀ ”,
the arrow “→”, and the negation operator
• For example:
No children play bridge. ∀
No mice understand calculus. ∀
• If you find this hard to understand, don’t worry! You can
simply

memorize the fact that this is how E statements are symbolized.
Instances
• Let’s write “M” for “____ is a mammal” and “W” for “____ is
a whale”.
• Suppose that the domain of quantification is animals.
• Suppose that “a” is a name for something in the domain.
• Here is a symbolization of Every whale is a mammal:
∀ x (Wx → Mx)
Instances
a whale”.
• Suppose that “a” is a name for something in the domain.
• Here is a symbolization of Every whale is a mammal:
∀ x (Wx → Mx) A universal generalization, and ...

(Wa → Ma) ... one of its instances.
Instances
• A universal generalization (i.e. a statement that starts with a
“∀ ”)
is true just in case all of its instances are true.
• A universal generalization is, in effect, a conjunction of all its
instances.1
1 I assume here that everything in the domain has a name.
O: ____ likes opera. Universe of discourse: people.
C: ____ is a child.
S: ____ is a snob.
(1) Everyone likes opera.
(2) Every snob likes opera. (Hint: This is an A statement!)
(3) No child likes opera. (Hint: This is an E statement!)
(4) Nobody likes opera.
(5) Only snobs like opera.

A Quick Symbolization Exercise
2 : T h e U n i v e r s a l
I n s t a n t i a t i o n R u l e
The UI Rule
• If you look at the inside of the back cover of your textbook,
you’ll
find a number of rules involving the universal quantifier …
• … some of them are rather complex! We’ll get to those later.
• For now, let’s focus on one rather simple rule – the UI rule:
From a universal generalization, you can infer any one of its
instances.
The UI Rule
For example, the following inferences are both valid …
Premise: ∀ x Dx (Everyone likes dancing.)
Conclusion: Da (Ashni likes dancing.)

Premise: ∀ x (Wx → Mx) (Every whale is a mammal.)
Conclusion: (Wd → Md) (If Moby Dick is a whale, he’s a
mammal.)
Example
Show that the following inference is valid, by giving a natural
deduction proof:
Pre
Example
1. ∀ x (Wx → Mx) Prem
3. (Wa → Ma) 1, UI

Exercise
Symbolize the following inference, and show that it is valid by
giving
a natural deduction proof:
Premise: Every SFU student is clever.
Premise: Dev is an SFU student.
Conclusion: Dev is clever.
3 : S o m e Wo r d s o f
C a u t i o n
1: The domain of quantification is
sometimes called the universe of discourse.
2: It’s bad practice to give one variable two
jobs in one statement.

• Suppose that you’re asked to symbolize “Everyone is dancing
and
everyone is smiling.”
• You could write:
(∀ x Dx & ∀ x Sx)
• This isn’t wrong, but it is potentially confusing, because
you’ve used the
variable “x” to do two different jobs in one statement.
• It would be much better to write:
(∀ x Dx & ∀ y Sy)
• In my lectures, I will assume that we adopt this convention!
3: The Universal Generalizations in Our
Symbolism are Strict …
Our Universal Generalizations are Strict.
• This means that a universal generalizations in our symbolism
can
be refuted by a single counterexample.

• For example, the following inference is valid:
an’t fly.)
∀ x(Bx → Fx) (It is not true that every bird can
fly.)
2. Bp 1, Simp
4. ∀ x(Bx → Fx) Supp/RA
5. (Bp → Fp) 4, UI
6. Fp 2, 5 MP
7. ⊥ 3, 6 Conj
∀ x(Bx → Fx) 4-7, RA

• It’s not possible to express loose universal generalizations in
our
symbolism …
• … but this is okay, since we’re trying to understand
mathematical
proof – and mathematicians don’t use loose generalizations in
their proofs!
4: Beware the following subtle error …
On a Subtle Error in Proofs
Suppose there are twenty people at the party (one of whom is
Ashni) and
only sixty bottles of beer.
Domain of quantification: People at the party.
D “____ drinks three bottles of beer.”
R We will run out of beer.
a Ashni
Premise: (∀ x Dx → R) (If everyone drinks three bottles of beer,
we will run out.)

Premise: Da (Ashni drinks three bottle of beer.)
Conclusion: R (We will run out.)
On a Subtle Error in Proofs
1. (∀ x Dx → R) Prem
2. Da Prem
3. (Da → R) 1, UI
4. R 2, 3 MP
5: Pay attention to the domain of
quantification!
Pay Attention to the Domain
• Suppose you’re asked to symbolize the statement “Everyone at
the
party who is dancing is happy.”
• If the domain of quantification for your symbolic sentences is

people,
you would write:
∀ x((Px & Dx) → Hx)
• If the domain of quantification for your symbolic sentences is
people at
the party, you would write:
∀ x(Dx → Hx)
4 : F o r e s h a d o w i n g t h e
U G R u l e
Square numbers: Rectangle numbers:

1

Hypothesis:
• If you add together two consecutive rectangle numbers, the
result
is always twice a square.
• For any n, the sum of the nth rectangle number and the (n+1)th
rectangle number is always twice a square.
Let n be any arbitrarily chosen natural number.
Then the nth rectangle number is: n(n+1)
Also, the (n+1)th rectangle number is: (n+1)(n+2)
So, the sum of the nth rectangle number and the (n+1)th
rectangle number is:
n(n+1) + (n+1)(n+2)
= (n2 + n) + (n2 + n + 2n + 2)
= 2n2 + 4n + 2
= 2(n+1)2
This is indeed twice a square!

Therefore:
For any n, the sum of the nth rectangle number and the (n + 1)th
rectangle number is twice a
square.
The UG Rule
• The statement we just proved is a universal generalization:
rectangle
number is twice a square.
• We proved it by proving that an “arbitrary instance” is true.
• This is an example of the UG rule at work.
• We’ll look at the rule in more detail next time …
Exercise
Symbolize the following inference, and show that it is valid by
giving
a natural deduction proof:
Premise: Everyone who is drinking beer is dancing.

Premise: Everyone who is dancing is having fun.
Premise: Ashni is drinking beer.
Conclusion: Ashni is having fun.
“If the inference from p to q is valid, and the inference from q
to r is
valid, then the inference from p to r must be valid as well.”
I agree!
I disagree!
“If two objects are indistinguishable, then it can’t be true that
one of
them is red and also true that the other is not red.”
I agree!
I disagree!

1 4 : B i v a l e n c e
Bivalence
• In this course, we’ve been assuming that every statement is
either
true or false.
• To put it another way, we’ve been assuming that given any
statement, either it or its negation is true.
• This is called the “principle of bivalence” (or even the “law of
bivalence”).
• In this lecture, we’ll look at some objections to the principle
of
bivalence, and discuss how to cope.
1 : Va g u e Te r m s
Vagueness
• Suppose we have a sequence of 1000 tiles. We can call them
“Tile 1”, “Tile 2”, “Tile 3”, … , “Tile 1000”.
• Tile 1 is the colour of the leaf on the Canadian flag.

• Each tile in the sequence is a little bit less red than its
predecessor – but the differences are imperceptibly small.
Adjacent tiles in the sequence are indistinguishable.
• Tile 1000 is the colour of a pumpkin.
Vagueness
• The principle of bivalence tells us that every tile in the
sequence is
either truly describable as “red” or truly describable as “not
red” –
like this:
• But, arguably, this is not plausible:
• There are tiles in the middle which we wouldn’t call “red” but
which we also
wouldn’t call “not red”.
• It isn’t credible that there are adjacent tiles, one of which is
truly describable
as “red” and one of which is truly describable as “not red”.
Vagueness
• Here, arguably, is a more attractive account. The tiles at the
beginning of the sequence are properly called “red”. The tiles at

the end of the sequence are properly called “not red”. Then
there
are some tiles in the middle which have a third, intermediate
status. These tiles can’t truly be described as “red”, but they
can’t
truly be described as “not red” either:
Some more vague terms
• To use the philosophical jargon, “red” is vague.
• Here are some other vague terms:
• “grownup”
• “cold day”
• “too much ice cream to eat in one sitting”
2 : S t a t e m e n t s A b o u t
t h e F u t u r e
The Correspondence Theory of Truth
• Some philosophers think that a statement is true just in case it
correctly depicts a fact – some chunk of reality.

Oscar is in the guitar case.
Oscar is next to the flowers.
The Correspondence Theory of Truth
• Some philosophers think that a statement is true just in case it
correctly depicts a fact – some chunk of reality.
• A sentence is false, on this view, if its negation correctly
depicts a
fact.
Oscar is in a red cupboard.
Statements About the Future

• Suppose we accept the correspondence theory of truth.
• Suppose we also accept the claim that the future doesn’t yet
exist.
• Now consider the statement “Canada will win an odd number
of medals in the
2020 Olympic Games.”
• Arguably, this statement isn’t true now (because there is
currently no fact to
which it corresponds).
• And arguably, this statement isn’t false now (because there is
currently no
fact to which its negation corresponds).
• If this is right, then the statement is a counterexample to the
principle of
bivalence. Our statement has some third status, “open,” perhaps,
or
“unsettled.”
• Many have attributed this view to Aristotle – although the
attribution is
contentious.
3 : T h e L i a r P a r a d o x

The Liar Paradox
Consider the following sentence:
The red sentence on slide 18 is false.
• If we say that this sentence is true, this implies that the
sentence is
false, which is a contradiction!
• If we say that the sentence is false, we’re saying that it’s false
that the
sentence is false – i.e. that the sentence is true. This is a
contradiction
again!
• Perhaps it’s best to refrain from saying that the sentence is
either true
nor false!
4 : R e f o r m i n g S t a t e m e n t
L o g i c
Bivalence
• As I said, the classical approach to logic assumes the principle
of
bivalence – we assume that every statement is either true or
false.

• However, there are apparent counterexamples to this thesis:
• Statements involving vague words.
• Statements about the future.
• Paradoxes
• Others?
• Perhaps we need to reform logic in order to accommodate such
cases …
Three-Valued Logic
• Suppose we accept the view that there are really three truth
values, not two. Some statements are true; some are false; some
are intermediate.
• Our truth tables will get bigger! For each binary connective,
we
now need a truth table with nine rows instead of just four.
• How will we fill in the rows?
• This is contentious – I will present one approach.
Conjunction

• A conjunction is true when both conjuncts are true.
• A conjunction is false when either one of its conjuncts is
false.
• Otherwise, the conjunction is intermediate.
Conjunction
p q (p & q)
T T T
T I I
T F F
I T I
I I I
I F F
F T F
F I F
F F F

Disjunction
• A disjunction is true when either one of the disjuncts is true.
• A disjunction is false when both of the disjuncts are false.
• Otherwise, the disjunction is intermediate.
Disjunction
T T T
T I T
T F T
I T T
I I I
I F I
F T T
F I I
F F F

Negation
• If a statement is true, its negation is false.
• If a statement is false, its negation is true.
• If a statement is intermediate, its negation is also
intermediate.
T F
I I
F T
Conditionals
• (p ↔ q) is equivalent to ((p → q) & (q → p)).
A New Connective

• We’ve seen that in our new logic we have to reform the truth
tables for the familiar connectives.
• We can also introduce some new connectives! For example,
we
could introduce a connective � meaning “It is neither true nor
false
that …” with the following truth table:
p �p
T F
I T
F F
The Definition of Validity
• If we assume the principle of bivalence, we will regard these
definitions as equivalent:
• An inference is valid just in case there is no possible situation
in which the
premises are true and the conclusion false.
in which the
premises are true and the conclusion is not true.
• But now we must regard these definitions as non-equivalent…

• Which should we choose?
• Well, let’s see what happens if we choose the first definition
…
Consider the following three statements:
A � �A)
T F T
I T F
F F F
Given our definition of validity, we have to say that from �A
you can
�A) …
but you
�A) from �A!!
This is absurd – so we have to reject this definition of validity.
• We have to choose between two definitions of validity:
in which the
premises are true and the conclusion false.

in which the
premises are true and the conclusion is not true.
• The first definition turns out to be ridiculous.
• So we have to choose the second.
Which of our rules are valid?
• It’s easy to check that many of our natural deduction rules are
still
valid in the new system. For example, Conj and Simp are still
valid!
• However, some of our natural deduction rules have to be
rejected –
or at least, reformed.
• Consider, for example, CP.
Conditional Proof
• Consider the following proof in our natural deduction system:
1. A Prem
│ 2. B Supp/CP
│ 3. (A & B) 1, 2 Conj

4. (B → (A & B)) 2-3,CP
• The inference from A to (B → (A & B)) is not valid in our
new system.
• So (within our new system) we must say that there is
something wrong
with the above proof.
• But Conj is valid in our new system, as I said.
• So we have to reject CP – or at least reform it somehow.
Reductio ad Absurdum
• Consider this proof in our natural deduction system:
1. A Prem
│3. ⊥ 2, R
-3, RA
system.
• So (within our new system) we must reject the above proof.
• So we must reject (or at least reform) the RA rule.

A Complaint About the New System
• Arguably, CP and RA are essential to mathematics.
• We can’t live without them.
• Thus, the new logic can’t be accepted.
5 : A n U n s o l v e d P r o b l e m
An Unsolved Problem
• We’ve seen that the principle of bivalence is problematic.
• However, our new logic has its own problems!
• Options:
• We could defend the principle of bivalence from the
objections.
• We could accept that the principle of bivalence is mistaken,
and then offer
some alternative defence of our natural deduction rules.
• We could learn to live with our three-valued logic.
• We could find an altogether new logic …
• All of these approaches have their defenders!

A Problem to Finish
Show that for any statement p one can construct a natural
deduction proof of the following statement:
Presumably, if we reject the law of bivalence we will also deny
that
wish to
reject at least one of the rules in your proof. Which one do you
think
we should reject?
1 9 : I n t r o d u c i n g t h e
E x i s t e n t i a l Q u a n t i f i e r
Domain: Animals at a particular zoo.

M ____ is a mammal.
T ____ has a tail.
B ____ is brown.
(a) Every mammal has a tail.
(b) Not every mammal has a tail.
(c) No mammal has a tail.
(d) Only the mammals are brown.
1 : I n t r o d u c i n g t h e
E x i s t e n t i a l Q u a n t i f i e r
Domain: People at a certain party
D: “____ likes dancing.”
M: “____ likes muffins.”
S: “____ likes swimming.”
• Suppose that there are five people at the party: a, b, c, d, and
e.
• Suppose we wish to symbolize the statement “Someone likes
dancing”.

• But what if the domain is very large?
• This is where the existential quantifier comes in!
The Existential Quantifier
∃ x
• There is at least one x such that …
• It’s true for some x that …
English sentence Sentence in our symbolism
Someone likes dancing. ∃ x Dx
Someone likes dancing but
not muffins.
∃
There’s someone who either
likes dancing, or likes both
swimming and muffins.
∃

Instances
• Like universal quantifications, existential quantifications have
“instances”.
∃ x (Mx & Dx) (An existential quantification…)
(Ma & Da) (… and one of its instances.)
• An existential generalization is true just in case one or more
of its
instances is true.1
1 I assume here that everything in the domain has a name.
Instances
For example, supposing that the people at the party are Ashni,
Ben,
Chiara, Deshaun, and Emma, the following two statements have
the
same truth value:
∃ x Mx
Someone likes muffins.
Either Ashni, Ben, Chiara, Deshaun, or Emma likes muffins.

Symbolizing I-Statements
It is straightforward to symbolise I statements, using the
existential quantifier:
1. There is at least one red fox. ∃ x(Rx & Fx)
2. Some foxes are red. ∃ x(Rx & Fx)
3. At least one bear lives in Vancouver. ∃ x(Bx & Vx)
(You might object that “Some foxes are red” doesn’t mean quite
the same thing
as “There is at least one red fox.” Don’t worry – we’ll deal with
this point later in
the term!)
Symbolizing O-Statements
It is straightforward to symbolise O statements, using the
existential quantifier:
1. There is at least one snake that is not poisonous. ∃ x(Sx &
2. Some snakes are not poisonous. ∃
3. Some philosophers are not atheists. ∃

A Symbolisation Exercise
Domain: The people at a certain party.
D: ____ is dancing.
B: ____ is drinking beer.
F: ____ is having fun.
(1) Someone is having fun.
(2) Some dancer is having fun.
(3) At least one dancer is not having fun.
(4) Someone is dancing and drinking beer, but not having fun.
∃
2 : T h e E G R u l e
The Existential Generalization Rule (EG)
• There are several inference rules associated with the
existential
quantifier.
• Today, we’ll just look at one of them.
• It’s very simple!

• The EG rule allows us to infer an existential generalization
from
any instance.
The Existential Generalization Rule (EG)
Here are some examples:
Premise: (Da & Fa) (Ashni is dancing and having fun.)
Conclusion: ∃ x(Fx & Fx) (Someone is dancing and having fun.)
Conclusion: ∃
having fun.)
Example
The following inference is valid. Establish this, by giving a
natural
deduction proof.
Premise: Da Ashni is dancing.
Premise: ∀ x(Dx → Fx) Every dancer is having fun.

Conclusion: ∃ x(Dx & Fx) Some dancer is having fun.
1. Da Prem
2. ∀ x(Dx → Fx) Prem
3. (Da → Fa) 2, UI
4. Fa 1, 3 MP.
5. (Da & Fa) 1, 4 Conj
6. ∃ x(Dx & Fx) 5, EG
Exercise
The following inference is valid. Establish this, by giving a
natural deduction proof.
Premise: Da Ashni is dancing.
Premise: (Da → Db) If Ashni is dancing, so is Ben.
Premise: Fb Ben is having fun.
Premise: ∀ x(Fx → Bx) Everyone who is having fun is drinking
beer.

Conclusion: ∃ x((Dx & Fx) & Bx) Someone is dancing, having
fun, and drinking beer.
2 1 : Q u a n t i f i e r N e g a t i o n
Announcements
• There is a tutorial handout for you to work on. I’ll upload
answers
tomorrow.
• If you have questions about the tutorial handout, you can post
them at
Sli.do, using the code S904. There’ll be a new code for next
week’s
handout.
• Using Sli.do, Duke asks: “I wanna ask what's going to be
covered on the
final exam? will it cover knowledge points before midterm 1
and
midterm 2?”
• The answer to Duke’s question is that the final is cumulative –
it will
include everything that we’ve covered this term.
• The final online assignment will also go up tomorrow. Good

luck with it!
Exercise
Show that the following argument is valid, by providing a
natural
deduction proof:
Premise: ∃ x Bx
Premise: ∀ x (Bx → Mx)
Conclusion: ∃ x Mx
1 : T h e Q N R u l e
These two statements are equivalent:
• Not everyone is having a good time.
• Someone’s not having a good time.
In symbols:
∀ x Gx

• ∃
Similarly, these two statements are equivalent:
• It’s not true that someone is married.
• Everyone is unmarried.
In symbols:
∃ x Mx
• ∀
The Quantifier Negation Rule
∀ x Φx, derive ∃
∃ x Φx, derive ∀
2 : T h e S q u a r e o f
O p p o s i t i o n

The Square of Opposition
• We say that two statements are “contradictories” if it’s not
possible for
both to be true, and not possible for both to be false.
from p,
and vice versa.
• It’s a useful fact to remember that every A-statement is
contradictory to
the corresponding O statement, and …
• … every E statement is contradictory to the corresponding I
statement.
• These points are traditionally represented on a diagram, the
“square of
opposition”.
• https://plato.stanford.edu/entries/square/
A
All A are B.
E
No A are B.
I
Some A are B.

O
Some A are not B.
Show that the following statements are equivalent, using natural
deduction proofs:
Not all A are B.
Some A are not B.
First Worked example
∀ x(Ax → Bx) Prem
2. ∃
3. ∃
4. ∃
5. ∃

1. ∃
2. ∃
3. ∃
4. ∃
∀ x(Ax → Bx) 4, QN
deduction proofs:
It is not true that some A are B.
No A are B.
Second Worked example
deduction proofs:
∃ x(Ax & Bx)
No A are B. ∀

∃ x(Ax & Bx) Premise
2. ∀
3. ∀
4. ∀
3 : E x e r c i s e s
An Exercise to Finish
Show that the following inference is valid, using a natural
deduction
proof:
Premise: ∃ x Bx
∃
Conclusion: ∃ x Mx

An Exercise to Finish
deduction
proof:
Premise: ∀ x (Bx → Mx)
∀ x Mx
∀ x Bx
1 8 : T h e U G R u l e
PHIL 110 and COVID 19
• There will be a final exam of some kind – I’m not sure yet
how this
will be done. I will do my very best to ensure that the
assessment
is fair to all of you.
• The TAs will deliver the graded midterm exams to me – I have

them in my office. If you want to see your midterm, let me
know.
• I will omit some of the more challenging material from this
iteration of the course. This is to ensure that you all have
adequate
time to prepare for the final, despite the unusual obstacles that
2020 has produced.
1 : T h e U G R u l e I n
A r i t h m e t i c
20
Hypothesis:
• If you add together two consecutive rectangle numbers, the
result
is always twice a square.

• For any n, the sum of the nth rectangle number and the (n+1)th
rectangle number is always twice a square.
Let n be any arbitrarily chosen natural number.
Then the nth rectangle number is: n(n+1)
Also, the (n+1)th rectangle number is: (n+1)(n+2)
So, the sum of the nth rectangle number and the (n+1)th
rectangle number is:
n(n+1) + (n+1)(n+2)
= (n2 + n) + (n2 + n + 2n + 2)
= 2n2 + 4n + 2
= 2(n+1)2
This is indeed twice a square!
Therefore:
rectangle number is twice a
square.

The UG Rule
• The statement we just proved is a universal generalization:
rectangle
number is twice a square.
• We proved it by proving that an “arbitrary instance” is true.
• This is an example of the UG rule at work.
2 : T h e U G R u l e i n
G e o m e t r y
Alternate Angles
Assuming that the red lines are parallel, the angles b and c are
equal!
a b
c
d

Alternate Angles
Assuming that the red lines are parallel, the angles b and c are
equal!
ab
cd
a
b
c
a
b
c
a

b
c
a
a
b
c
a
a
b
c
a c
a

b
c
a c
a
b
c
a c
We’ve shown that the angles inside any triangle add up to 180°.
We will acknowledge only those proofs in which one can appeal
step
by step to preceding propositions and definitions. If, for the
grasp of
a proof, the corresponding figure is indispensable, then the
proof
does not satisfy the requirements that we imposed on it. … in
any
complete proof the figure is dispensable. (Pasch, 1882)

[B]e careful, since [the use of diagrams] can easily be
misleading. A
theorem is only proved when the proof is completely
independent
of the diagram. The proof must call step by step on the
preceding
axioms. The making of figures is [equivalent to] the
experimentation
of the physicist … (Hilbert, 1894)
3 : T h e U G R u l e Wi t h i n
N a t u r a l D e d u c t i o n
The Exam Has Been Scheduled
• Most of you will take the exam on WED 15-Apr, 1200-15:00,
C9001
• Some of you will take the exam at the CAL.
• Some of you may qualify for “hardship”:
• You have three exams within 24 hours.
• You have an examination at one location (e.g. the Burnaby

campus)
followed immediately by an exam at another location (e.g., the
Surrey
campus).
1 6 : T h e U n i v e r s a l Q u a n t i f i e r
Contains Chocolate
Contains Garlic
Contains Cream
Domain: Traditional Dishes
Contains Chocolate
Contains Garlic
Contains Cream

Contains Chocolate
Contains Garlic
Contains Cream
Contains Chocolate
Contains Garlic
Contains Cream
Contains Chocolate
Contains Garlic
Contains Cream
x

Contains Chocolate
Contains Garlic
Contains Cream
x
Contains Chocolate
Contains Garlic
Contains Cream
x
x
Valid or not?
(1) No A are B. (2) All A are B.
Some A is C. No C are A.
Some C is not B. No C are B.

Valid! Not valid!
Valid or not?
(1) No A are B.
Some A is C.
Therefore:
Some C is not B.
Valid or not?
(2) All A are B.
No C are A.
Therefore:
No C are B.
1 : T h e N e e d f o r S y m b o l s

The Complexities of English
Sometimes when one uses the phrase “a bear”, you mean to talk
about some particular bear; sometimes you mean to generalise
about all bears.
A universal generalization: A bear is a mammal.
An existential generalization: A bear is in my garden.
Ambiguities in English Quantification
• “Everyone at the party isn’t dancing.”
• It is not true that everyone at the party is dancing.
• Nobody at the party is dancing.
• “Everything is caused by something.”
• There is some particular thing (God perhaps?) which causes
everything.
• Everything has a cause (though perhaps different things have
different
causes).

Goals for this lecture …
• Introduce the symbol which we use to express universal
generalizations – what we call the “universal quantifier”.
• Introduce one of the inference rules associated with this
symbol.
2 : N a m e s a n d P r e d i c a t e s
Proper Names
A proper name is a word that represents an individual member
of the
domain of quantification. For example, if our domain is singers,
we might
use the following names:
• “Celine Dion”
• “Beyoncé”
• “Pavarotti”
Similarly, if our domain is countries, we might use the
following names:
• “Canada”
• “Germany”

• “India”
Predicates
If you take a declarative sentence and remove one or more
proper
names from it, the result is a predicate. (If you like, a predicate
is a
name with one or more proper-name-shaped holes in it.)
• A one-place predicate has one hole.
• A two-place predicate has two holes.
• A three-place predicate has three holes.
• (And so on!)
Predicates
Some one-place predicates:
• “____ likes dancing.”
• “____ likes muffins”
• “____ likes swimming.”

Some two-place predicates:
• “____ and ____ are friends.”
• “____ is taller than ____.”
A three-place predicate:
• “____ and ____ together ate more food than ____.”
Predicates
One can make a sentence by taking a predicate, and then “filling
in”
the hole with a name (or filling in the holes with names):
“Ashni” + “____ likes muffins” = “Ashni likes muffins”.
For the moment, we’ll restrict our attention to one-place
predicates.
Names and Predicates in Our Symbolism
• We will use lower-case letters (usually from the beginning of
the alphabet) as
names. We will use capital letters as predicates:
Proper Names Predicates

a: Ashni D: “____ likes dancing.”
b: Ben M: “____ likes muffins.”
c: Chiara S: “____ likes swimming.”
• One can form a sentence in our symbolism by writing a
predicate, and then
the appropriate number of names.
• For example, “Ma” means Ashni likes muffins.
• We can form longer statements using the now-familiar
statement operators.
• For example, “(Ma & Mb)” means Ashni and Ben both like
muffins.
Examples
Ashni likes dancing. Da

Ben likes swimming. Sb
Ashni likes dancing and Ben likes
swimming.
(Da & Sb)
Either Ashni or Ben likes dancing.
Chiara and Ben like muffins, but
Ashni doesn’t.
If Chiara likes swimming, so do Ben
and Ashni.

(Da ↔ Db)
Q u a n t i f i e r
The Universal Quantifier
• Let’s suppose that my domain of quantification is people at
my party.
• There are five people in this domain: Ashni, Ben, Chiara,
Deshaun, and
Emma.
• I want to symbolise this statement: everyone likes dancing.
• How do I do it?
• I could write this: ((((Da & Db) & Dc) & Dd) & De)
• But what if the domain is, say, people who attended the most
recent
Canucks game?

• What if the domain is, numbers?
• What we need is a symbol that will allow us to ascribe some
property to
ALL the things in the domain, even if the domain is very large.
This is
what the universal quantifier is for!
∀ x
• For any x …
• Whatever x may be …
• It’s true for every x that …
Everyone likes dancing. ∀ x Dx
Everyone likes dancing and likes
muffins.
∀ x (Dx & Mx)
Everyone either likes swimming or
likes muffins.
∀

∀ x
• For any x …
Universal Quantifier
muffins.
∀ x (Dx & Mx)
likes muffins.
∀
∀ x
• For any x …

Variable
muffins.
∀ x (Dx & Mx)
likes muffins.
∀
∀ y
• For any y …
• Whatever y may be …
• It’s true for every y that …

Either everyone likes muffins or
everyone likes swimming.
(∀ ∀ y Sy)
If Ashni likes dancing, everyone
likes dancing.
(Da → ∀ y Dy)
∀ y
• For any y …
• Whatever y may be …
• It’s true for every y that …
Either everyone likes muffins or
everyone likes swimming.
(∀ ∀ y Sy)
If Ashni likes dancing, everyone
likes dancing.
(Da → ∀ y Dy)

What do variables represent?
• Note that when you use a universal quantifier, the variable
doesn’t
represent any particular entity in the domain:
∀ x Dx
• Rather, one might say, it represents “any object chosen freely
from
the domain”.
• This is very common in mathematics …
Claim: For any whole number k, if k is an odd number so is k2.
Proof
Suppose that k is an odd number.
Then for some number j: k = 2j + 1
Then, k2 = (2j + 1)(2j + 1)
So: k2 = 4j2 + 4j + 1
So: k2 = 2(2j2 + 2j) + 1
So k2 is odd!

So, in conclusion,
What do variables represent?
• Note that when you use a universal quantifier, the variable
doesn’t
represent any particular entity in the domain:
∀ x Dx
• Rather, one might say, it represents “any object chosen freely
from
the domain”.
• This is very common in mathematics …
• Something similar happens in English too: “When a dog is hot,
he
pants.
4 : I n s t a n c e s
Instances
• Suppose you take a universal generalization. That is …

• … a statement that begins with “∀ x”.
• You remove the initial “∀ x” …
• … and replace every occurrence of “x” in the statement with a
name for something in the domain (the same name each time!).
• The result is an instance of the universal generalization with
which you started.
Instances
Universal Generalization Instance
∀ x Dx Da
∀ y (Dy & My) (Db & Mb)
∀
A universal generalization is true just in case all of its instances
are
true.1
1 I assume here that everything in the domain has a name …
Instances

• To repeat, a universal generalization is true just in case all of
its
instances are true.
• Indeed, you might think of a universal generalization as a
conjunction of
all of its instances.
• Suppose that the domain of quantification is people at my
party, and
suppose that the people at my party are Ashni, Ben, Chiara,
Deshaun,
and Emma. Then the following two statements have the same
truth
value:
((((Da & Db) & Dc) & Dd) & De)
∀ x Dx
5 : S y m b o l i z i n g A
a n d E s t a t e m e n t s
Symbolizing A statements
a whale”.

• How should we symbolize “Every whale is a mammal”?
• The standard symbolization is this:
∀ x (Wx → Mx)
• This often puzzles students. (Where did the arrow come
from?!)
• So let’s take a closer look ...
Whales Mammals
Domain: Animals
• To symbolize the claim that an
animal x is inside the shaded
region, we would write
• To symbolize the claim that an
animal x is outside the shaded
region, we would write
• We can symbolize “All whales
are mammals” as
∀

• This is equivalent to
∀ x (Wx → Mx).
Every whale is a mammal.
Symbolizing A Statements
• More generally, we symbolize A statements with a universal
quantifier
and an arrow:
Every whale is a mammal. ∀ x (Wx → Mx)
Every dancer is happy. ∀ x (Dx → Hx)
Every SFU student is clever. ∀ x (Sx → Cx)
• I hope the last slide helped you to understand this!
• If not – that’s okay! You can just remember that this is how A
statements are symbolized.
Symbolizing E Statements
• Let’s write “R” for “____ is a reptile” and “W” for “____ is a
whale”.
• How should we symbolize “No whale is a reptile”?

• Here’s one way of doing it. Note that “No whale is a reptile”
is
equivalent to “Every whale is a non-reptile”.
• This can be formalized: ∀
I n s t a n t i a t i o n R u l e
The UI Rule
• If you look at the inside of the back cover of your textbook,
you’ll
find a number of rules involving the universal quantifier …
• … some of them are rather complex! We’ll get to those later.
• For now, let’s focus on one rather simple rule – the UI rule:
From a universal quantification, you can infer any one of its
instances.
The UI Rule
For example, the following inferences are both valid …

Premise: ∀ x Dx (Everyone likes dancing.)
Conclusion: Da (Ashni likes dancing.)
Conclusion: (Wd → Md) (If Moby Dick is a whale, he’s a
mammal.)
Example
deduction proof:
is not a whale.)
Example
1. ∀ x (Wx → Mx) Prem
3. (Wa → Ma) 1, UI

7 : S o m e P r a c t i c e
O: ____ likes opera. Universe of discourse: people.
C: ____ is a child.
S: ____ is a snob.
(1) Everyone likes opera.
(2) Every snob likes opera.
(3) Nobody likes opera.
(4) No child likes opera.
(5) Only snobs like opera.
(6) If a child likes opera, they’re a snob.
Show that this inference is valid, using a natural deduction

proof:
Premise: Every snob likes opera.
Premise: Sam is a snob.
Conclusion: Sam likes opera.
20: The EI Rule
PHIL 110; Spring 2020; Tom Donaldson
Domain: Movies
C ____ is a comedy.
A ____ was directed by Ang Lee.
J ____ stars Jake Gyllenhaal.
(a) Ang Lee has directed a comedy.
(b) Every movie Ang Lee has directed is a comedy.
(c) No movie directed by Ang Lee is a comedy.
(d) Ang Lee directed a movie, not a comedy, which stars Jake
Gyllenhaal.

Natural Deduction Practice
deduction proof:
Premise: Every SFU student lives in Vancouver or Burnaby.
Premise: Ahmed is an SFU student who doesn’t live in Burnaby.
Conclusion: There is an SFU student who lives in Vancouver.
1: The EI Rule
The Problem of Shared Names
• Suppose there are two people called “Ashni”. Now consider:
Ashni is currently in Toronto.
Ashni is currently in Vancouver.
Therefore:
Ashni is currently in both Toronto and Vancouver.

• This inference presents a challenge for us. We can suppose
that both premises are true:
• One of the Ashnis is in Toronto, so the first premise is true.
• One of the Ashnis is in Vancouver, so the second premise is
true.
• The inference appears to be an instance of the conjunction
rule.
• And yet the conclusion is false!
•
Solution
: In our symbolism, each name is only used once!
Witnesses
• We’ve said that a universal generalization can be refuted by
just
one example - a “counterexample”.
• For example, if someone claims that every bird can fly, we can
refute them

by telling them about Pingu, the TV star.
• An existential generalization can be shown to be true using
just
one example – a “witness”.
• Suppose someone asks whether the following statement is
true: “Some
bird knows how to ice skate.”
• We can show that it is true by presenting Pingu as an example.
• Pingu is a witness to the statement “Some bird knows how to
ice skate.”
The EI Rule
• When we use the EI rule, we start with an existential
generalization. We then introduce a name for an arbitrary
witness.

• For example:
• There are residents of Burnaby who play the piano. Let’s call
one of them
“Smith” …
• Some SFU students are champion wrestlers. Let’s call one of
them “Diana”.
Then …
• We know that there are prime numbers greater than one
million. Let N be
one of them. Then …
We know that someone broke in to the palace on Friday wearing
muddy shoes. Let’s call the guy “Smith”. Now Smith must have
come
in through the garden, and we know from his footprints that he
was
wearing shoes with a heel, and …

Claim: For all x and y, if x is rational and y is rational, then
x+y is rational.
Proof: Suppose that u and v are arbitrary rational numbers.
Since u is rational, there are integers x and y such that � =
�
�
.
Suppose that a and b are integers with � =
�
�
.
Since v is rational, there are integers x and y such that v =
�

�
.
Suppose that c and d are integers with � =
�
�
.
Then � + � =
�
�
+
�
�
=
��
��
+

��
��
=
��+��
��
Premise: ∃ x Dx Someone is dancing.
Conclusion: Di i is dancing.
Premise: ∃ x(Sx & Rx) Some singer is rich.
Conclusion: (Sj & Rj) j is a rich singer.

Premise: ∃ x(Sx & Rx)
Conclusion: ∃ x Sx
(1) ∃ x(Sx & Rx) Prem
(2) (Si & Ri) 1, EI
(3) Si 2, Simp
(4) ∃ x Sx 3, EG
From ∃ x Φx, infer Φi, where i is an arbitrary individual name
(one
that has not occurred either in the symbolization of the
argument or
on any previous line of the proof.)

Premise: ∃ x Tx
Premise: ∃ x Vx
Conclusion: ∃ x(Tx & Vx)
(1) ∃ x Tx Prem
(2) ∃ x Vx Prem
(3) Ti 1, EI
(4) Vi 2, EI
(5) (Ti & Vi) 3, 4 Conj
(6) ∃ x(Tx & Vx) 5, EG
Premise: ∃ x Tx
Premise: ∃ x Vx
Conclusion: ∃ x(Tx & Vx)
(1) ∃ x Tx Prem
(2) ∃ x Vx Prem
(3) Ti 1, EI

(4) Vi 2, EI
(5) (Ti & Vi) 3, 4 Conj
(6) ∃ x(Tx & Vx) 5, EG
5: An Exercise To Finish
deduction
proof:
Premise: ∀ x(Bx → Fx) (Every bird can fly.)
∃
fly.)
Hint: Use reductio ad absurdum.

2 3 : I d e n t i t y
Exercise
Consider the following two inferences. You should assume that
the
domain of quantification is people. In each case:
• If the inference is valid, show that it is valid by giving a
natural
deduction proof.
• If the inference is not valid, that it is not valid by drawing an
arrow
diagram depicting a situation in which the premise is true and
the
conclusion false.
First Inference Second Inference

Premise: ∀ x ∃ y Lxy Premise: ∃ y ∀ x Lxy
Conclusion: ∃ y ∀ x Lxy Conclusion: ∀ x ∃ y Lxy
Exercise
First Inference
Premise: ∀ x ∃ y Lxy
Conclusion: ∃ y ∀ x Lxy
Exercise
First Inference

Premise: ∀ x ∃ y Lxy
Conclusion: ∃ y ∀ x Lxy
Exercise
Second Inference
Premise: ∃ y ∀ x Lxy
Conclusion: ∀ x ∃ y Lxy
Exercise
1. ∃ y ∀ x Lxy Prem
2. ∀ x Lxi 1, EI

∀ x ∃ y Lxy Supp/RA
4. ∃ ∃ y Lxy 3, QN
5. ∃ x ∀
6. ∀
8. Lji 2, UI
9. ⊥ 7, 8 Conj
10. ∀ x ∃ y Lxy 3-9, RA, DN
1 : I n t r o d u c i n g I d e n t i t y
Qualitative And Numerical Sameness
• Suppose I say that Ashni and Ben’s partners are “the same”.
There
are two things I might mean:
• I might mean that Ashni and Ben are dating two people who
are very

similar. (Perhaps their partners are twins.)
• I might mean that Ashni and Ben are dating the very same
person – i.e.
there is one person who is dating both of them.
• There is a similar ambiguity in the English word “identical”.
• Philosophers avoid confusion by distinguishing between
“qualitative” and “numerical” identity.
• To say that x and y are qualitatively identical is to say that x
and y
are exactly alike (or at least very similar).
• To say that x and y are numerically identical is to say that x
and y
are not two things, but one.

• If x and y are not numerically identical, they are said to be
“distinct”.
For example:
• Adrian Brody is qualitatively identical to John Locke (but they
are
not numerically identical).
For example:

are
For example:
are
• Eric Blair is numerically identical to George Orwell.
The Identity Symbol: =
a) In mathematics, we use the symbol “=” to express
propositions

about numerical identity. For example:
c) Eric Blair = George Orwell
d) 7 + 5 = 12
e) 12 = 7 + 5
g) 12 ≠ 7 + 7
a
d
c

b
e
a
d
c
b
e
Everything is what it is,
and not another thing.
2 : U s e s o f t h e

I d e n t i t y S y m b o l
First Example
• How should we symbolize “Ashni loves Ben and someone
else”?
• The statement means: Ashni loves Ben, and Ashni also loves
someone who isn’t Ben.
• In symbols: (Lab & ∃ x (Lax & x ≠ b))
Second Example
• How about “Ashni loves Ben and nobody else”?
• This means: Ashni loves Ben, and it is not the case that there
is
somebody other than Ben who Ashni loves.

∃ x(Lax & x≠b))
• Equivalently: (Lab & ∀ x(Lax → x = b))
Third Example
• How would we symbolize “There are (at least) two people
dancing”?
• We might try: ∃ x ∃ y(Dx & Dy)
• But this doesn’t quite work!
• The correct approach: ∃ x ∃ y ((Dx & Dy) & x≠y)
Fourth Example

• What about “There are (at least) three people dancing”?
• This is no good: ∃ x ∃ y ∃ z ((Dx & Dy) & Dz)
• The correct symbolization is this:
∃ x ∃ y ∃ z (Dx & Dy & Dz & x≠y & y≠z & z≠x)
(I’ve omitted some brackets to make the statement easier to
read…)
Domain of Quantification: People
b: Bob Dylan P: ____ is a poet.
r: Robert Zimmerman
d: Dylan Thomas
• Robert Zimmerman and Bob Dylan are the same person.

• Bob Dylan and Dylan Thomas are not the same person.
• There are at least two poets.
• Dylan Thomas is a poet, and there are no other poets.
• Robert Zimmerman and Bob Dylan are the same person.
r = b
• Bob Dylan and Dylan Thomas are not the same person.
b ≠ d
• There are at least two poets.
∃ x ∃ y ((Dx & Dy) & x≠y)
• Dylan Thomas is a poet, and there are no other poets.

(Pd & ∀ x(Px → x = d)
3 : I n f e r e n c e R u l e s
f o r I d e n t i t y
The Reflexivity Of Identity
• Here is an obvious fact about identity: ∀ x x=x
• Richard Arthur allows you to assume this whenever you want.
It
should be labelled “Implicit Premise” or “Impl. Prem.” for
short.
• Suppose for example that we wish to prove the following
inference:
Premise: Da

Conclusion: ∃ x(Dx & x=a)
The Reflexivity Of Identity
1. Da Prem
2. ∀ x x=x Impl. Prem
3. a=a 2, UI
4. (Da & a=a) 1, 3 Conj
5. ∃ x(Dx & x=a) 4, EG
The Substitution of Identicals
• Suppose you know that Ashni is in Vancouver. And suppose

you
know that Ashni is Mrs. Anand. Then obviously you can infer
that
Mrs. Anand is in Vancouver.
• Here is a mathematical example. Suppose you know that N is a
prime number, and you know that N=K. Then you can infer that
K
is a prime number.
• These are examples of the “SI” rule.
• For example, suppose you are asked to prove the following
inference:
Premise: a=b
Premise: Da

Premise: Sb
Conclusion: ∃ x(Dx & Sx)
1. a=b Prem
2. Da Prem
3. Sb Prem
4. Db 1, 2 SI
5. (Sb & Db) 3, 4 Conj
6. ∃ x(Dx & Sx) 5, EG

4 : S o m e P r a c t i c e
(1) Premise: ∀ x(Lax → b = x)
Premise: b ≠ c
(2) Premise: a = b
Premise: b = c
Premise: Da
Premise: Sc
(3) Premise: Da
Premise: Sa
Premise: Sb
Conclusion: ∃ x ∃ y((Sx & Sy) & x ≠ y)

(1) Premise: ∀ x(Lax → b = x)
Premise: b ≠ c
1. ∀ x(Lax → b = x) Premise
2. b ≠ c Premise
3. (Lac → b = c) 1, UI
(2) Premise: a = b
Premise: b = c
Premise: Da
Premise: Sc
1. a = b Premise

2. b = c Premise
3. Da Premise
4. Sc Premise
5. Db 1, 3 SI
6. Dc 2, 5 SI
7. (Dc & Sc) 4, 6 Conj
8. ∃ x(Dx & Sx) 7, EG
(3) Premise: Da
Premise: Sa
Premise: Sb
Conclusion: ∃ x ∃ y((Sx & Sy) & x ≠ y)
1. Da Premise
3. Sa Premise
4. Sb Premise
5. a = b Supp/RA
6. Db 1, 5 SI

7. ⊥ 2, 6 Conj
8. a ≠ b 5-7, RA
9. (Sa & Sb) 3, 4 Conj
10. ((Sa & Sb) & a ≠ b) 8, 9 Conj
11. ∃ y((Sa & Sy) & a ≠ y) 10, EG
12. ∃ x ∃ y((Sx & Sy) & x ≠ y) 11, EG
PHIL 110; Spring 2020
Final Exam
This is an “open book” exam, in the sense that as you complete
the exam you may look at your notes, the
textbook, my lecture slides … or any other resource that you
find helpful.

The exam is designed to be taken in three hours, but you may
complete it more quickly or more slowly if
you prefer.
If you find one of the questions unclear, you may email me (at
[email protected]) to ask for clarification.
Otherwise, you should complete this exam without help from
anyone, and without collaborating.
You should upload your answers to Canvas as a pdf:
• You could write your answers by hand, and then scan your
answers to pdf.
• Alternatively, you could type your answers. For help with this,
please refer to the document “How
to Type Your Answers to the PHIL 110 Final”, on Canvas.

mailto:[email protected]
Question One (5 points)
Which of the following three inferences are valid? (There is no
need to explain your answers.)
(a) Premise: If Jon forgot to turn the oven on, then the dinner
has been ruined.
Premise: The dinner has been ruined.
Conclusion: Jon forgot to turn the oven on. NOT VALID
(b) Premise: When he retired, Jon moved from Vancouver to
Jamaica.
Conclusion: Jon likes sunny weather. NOT VALID
(c) Premise: Jon is either in Vancouver or in Jamaica.

Premise: Jon is not in Vancouver.
Conclusion: Jon is in Jamaica. VALID
Question Two (5 points)
Consider the following argument:
If moral laws are made by people, then moral relativism is true.
But moral relativism is
not true. Therefore, moral laws are not made by people. Moral
laws are either made by
people, or by God. Therefore, moral laws are made by God.
(a) What is the conclusion of this argument?
MORAL LAWS ARE MADE BY GOD.

(b) Identify two premises of this argument.
FOR FULL MARKS, STATE ANY OF TWO OF THE
FOLLOWING:
• IF MORAL LAWS ARE MADE BY PEOPLE, THEN MORAL
RELATIVISM IS TRUE.
• MORAL RELATIVISM IS NOT TRUE.
• MORAL LAWS ARE EITHER MADE BY PEOPLE, OR BY
GOD.
(c) Identify two inference rules that are used in the argument.
MT, DS
Question Three (5 points)
Briefly explain the distinction between strict and loose

generalizations, giving examples. What is a
counterexample? Can a loose universal generalization be refuted
by a single counterexample?
When you assert a strict universal generalization, you say that
something is true in every single case
without exception. When you assert a loose universal
generalization, you say that something is normally
true, or usually true, or typically true, or true in most cases.
For example:
STRICT: Every single triangle without exception has three
sides.
LOOSE: Dogs typically have four legs.
A strict universal generalization can be refuted a single example

– a “counterexample”. For example, if
someone says that every bird can fly, and intends this as a strict
generalization, you can refute them by
showing them a single penguin.
A loose universal generalization cannot be refuted by just one
counterexample.
Question Four (9 points)
Premise One: It’s not true that Ashni and Ben were both
Premise Two: Ashni was cooking. A
Premise Three: If Ashni was cooking and Ben wasn’t, then the

Conclusion: The meal was delicious. D
Symbolize the argument, using the following abbreviations:
A: Ashni was cooking.
B: Ben was cooking.
D: The meal was delicious.
Is the argument valid? Justify your answer in detail.
The argument is valid. You can justify this either by giving a
proof or by using a truth table. I’ll do it both
ways:

2. A Prem
6. D 3, 5, MP
T T T F T F F T
T T F F T F F T
T F T T F T T T
T F F T F T T F
F T T F F T F T
F T F F F T F T
F F T T F T F T
F F F T F T F T

Question Five (2 points)
Krishna has been asked to prove the following statement:
For any natural number n, (n2 + 3n + 10) is even.
Krishna responds by using a computer to check that there are no
counterexamples to the statement
below 1,000,000,000.
Has Krishna proved the statement? Briefly explain your answer.
No, Krishna has not proved the statement. Krishna has shown
that there are no counterexamples to the
statement below one billion, but he has not shown that there are
no counterexamples above one billion.

Question Six (9 points)
Exactly one of these two inferences is valid. Give a natural
deduction proof of the valid inference.
(1) Premise: ∀ x(Ax → Bx)
Premise: ∃ x Ax
Conclusion: ∃ x Bx
(2) Premise: ∀ x(Ax → Bx)
Premise: ∃ x Bx
Conclusion: ∃ x Ax

(1) is valid:
1. ∀ x(Ax → Bx) Premise
2. ∃ x Ax Premise
3. Ai 2, EI
4. (Ai → Bi) 1, UI
5. Bi 3, 4 MP
6. ∃ x Bx 5, EG
Question Seven (2 points)
Consider the following mathematical proof:
Claim There do not exist whole numbers a and b such that (

�
�
)
3
= 2.
Proof Suppose for contradiction that there do exist whole
numbers a and b,
where (
�
�
)
3
= 2.
Then by “cancelling down” the fraction, we can find numbers c
and d which are not both

even, where (
�
�
)
3
= 2.
Then c 3 = 2d 3, so c 3 is even. Now we know that the cube of
an odd number must always
be odd. So c is even. So for some number k, c = 2k.
Now since c 3 = 2d 3 and c = 2k, we have (2k)3 = 2d 3. Thus 8k
3 = 2d 3, and so 4k 3 = d 3.
This implies that d 3 is even, and so d is even.
But now we’ve shown that both c and d are even – which
contradicts our initial
statement that c and d are not both even. Thus we have arrived
at a contradiction and the

proof is complete.
Identify one inference rule that is used in this proof.
The RA rule is used – note the giveaway phrase “suppose for
contradiction”.
Question Eight (5 points)
Using the following symbols, symbolize the statements listed
below.
Universe of Discourse: the people at a party
a Ashni
b Ben
c Chiara

Lxy x loves y.
Sx x is a singer.
(1) Ashni loves Ben, but Ben doesn’t love her back.
(2) Ben loves Chiara and nobody else.
(Lbc & ∀ x(Lbx → x = c)
(3) There are at least two people who love Ben.
∃ x ∃ y((Lxb & Lyb) & x ≠ y)
(4) Everyone who Ashni loves, Ben loves too.
∀ x(Lax → Lbx)

(5) Ashni loves someone, and that person loves everyone who
loves Ben.
∃ x(Lax & ∀ y(Lyb → Lxy))
Question Nine (6 points)
In each part of this question, there are two statements. You
should choose one statement to be the
premise, and the other to be the conclusion, in such a way that
the resulting argument is valid. (You do
not need to explain your answers). I use the same symbols as in
Question Eight.
(a) ((Sa → Sb) & (Sb → Sc)) PREMISE
(Sa → Sc) CONCLUSION

(b) ∃
∃ x (Sx & Lxc) PREMISE
(c) ∃ x ∀ y Lxy PREMISE
∀ y ∃ x Lxy CONCLUSION
Question Ten (10 points)
“Every statement is either true or false.” Do you agree or
disagree? Explain your answer.
SEE LECTURE 14 FOR MY ANSWER.
Question One (3 points)
Which of the following three inferences are valid? (There is no

need to explain your answers.)
(a) Premise: Liu Yang spends two hours every day playing the
violin.
Conclusion: Liu Yang wants to be a good violinist.
(b) Premise: If Liu Yang is in class, she is on campus.
Premise: Liu Yang is on campus.
Conclusion: Liu Yang is in class.
(c) Premise: If Liu Yang is in class, she is on campus.
Premise: Liu Yang is in class.
Conclusion: Liu Yang is on campus.
Question Two (2 points)

If God is both omnipotent and loving, then His creatures never
suffer. But it just isn’t true
that God’s creatures never suffer (just look around!) so it is not
true that God is both
omnipotent and loving. But we know for sure that God is
loving. That is certain.
Therefore, God is not omnipotent. The conventional wisdom is
wrong on this point.
Identify two inference rules that are used in the argument.
Question Three (2 points)
Juan has shown that a certain inference is valid, using a truth
table. Ella is trying to show that the
inference is valid by giving a natural deduction proof. Do you

think it’s possible for Ella to find a proof of
the inference? Briefly explain your answer.
Question Four (2 points)
Zeynep is asked to prove the following statement:
The sum of the internal angles in a pentagon is 540°.
Zeynep responds by carefully drawing a number of pentagons,
and measuring their internal angles. She
confirms that, in each case, the sum of the angles is 540°.
Has Zeynep proved the statement? Briefly explain your answer.
Question Five (10 points)

Premise One: Either Ashni or Ben attended the party.
Premise Two: If Ashni attended the party, it was a great
success.
Premise Three: Ben didn’t attend the party, if it wasn’t a great
success.
Conclusion: The party was a great success.
Symbolize the argument, using the following abbreviations:
A: Ashni attended the party.
B: Ben attended the party.
S: The party was a great success.

Is the argument valid? Justify your answer in detail.
Question Six (7 points)
Exactly one of these two inferences is valid. Give a natural
deduction proof of the valid inference.
(1) Premise: ∀
Premise: ∃
Conclusion: ∃ x Bx
(2) Premise: ∀
Premise: ∃ x Bx
Conclusion: ∃
Question Seven (2 points)

Identify an inference rule that is used in the mathematical proof
written in this box:
Theorem For any whole numbers x and y, x2 – 4y ≠ 2.
Proof Suppose for contradiction that it is false that for any
whole numbers x
and y, x2 – 4y ≠ 2.
Then there exist whole numbers x and y, where x2 – 4y = 2.
Let’s say that a and b are whole numbers, where a2 – 4b = 2
Then a2 = 2 + 4b, so a2 = 2(1 + 2b).
So a2 is even.
So a is even.
So for some whole number c, a = 2c.
Thus, (2c)2 – 4b =2, and so 4c2 – 4b = 2.

So 2c2 – 2b = 1, and so 2(c2 – b) = 1.
Now clearly 2(c2 – b) is even, so 1 is even.
But this is absurd: 1 is not an even number! Thus the proof is
complete.
Question Eight (5 points)
Using the following symbols, symbolize the statements listed
below.
Universe of Discourse: the people at a certain party
b Ben
c Chiara
Axy x admires y.

Mx x is a mathematician.
(1) Ben doesn’t admire Chiara, even though Chiara admires
Ben.
(2) Ben and Chiara admire each other.
(3) Every mathematician admires Chiara.
(4) There are at least two mathematicians who admire Ben.
(5) There’s a mathematician who admires everyone.
Question Nine (2 points)
For this question, we will use the same symbols as in Question
Eight. We will continue to assume that the
universe of discourse is the class of people at a certain party.
Here are three statements. Choose two of them to be premises,
and one of them to be the conclusion, in

such a way that the resulting argument is valid:
(1) ∀ x(Mx → ∃ y Ayx)
(3) ∀
There is no need to explain your answer.
Question Ten (10 points)
“Every statement is either true or false”.
Do you agree or disagree? Explain your answer.

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