This document provides an introduction to predicate logic. It discusses how predicate logic builds on propositional logic using quantifiers, predicates, and logical connectives. It presents the basic approach of stripping and reinserting quantifiers when proving arguments. It then introduces four new inference rules for quantifier stripping and insertion. Several examples are provided demonstrating how to construct proofs of arguments using these rules. The document also discusses using temporary hypotheses and proving verbal arguments in predicate logic.

1. Formal Logic
Mathematical Structures
for Computer Science
Chapter 1
Copyright © 2006 W.H. Freeman & Co. MSCS Slides Formal Logic

2. Predicate Logic
● Similar to propositional logic for solving arguments, build from
quantifiers, predicates and logical connectives.
● A valid argument for predicate logic need not be a tautology.
● The meaning and the structure of the quantifiers and predicates
determines the interpretation and the validity of the arguments
● Basic approach to prove arguments:
■ Strip of quantifiers
■ Manipulate the unquantified wffs
■ Reinsert the quantifiers
Four new inference rules
Two rules to strip the qualifiers
Two rules to reinsert the qualifiers
● Note to remember: P(x) could be (∀y) (∀z) Q(x,y,z)
Section 1.4 Predicate Logic 1

3. Inference Rules
From
Can Derive
Name / Restrictions on Use
Abbreviation
(∀x)P(x)
P(t) where t is a Universal If t is a variable, it must
variable or constant Instantiation- ui
not fall within the scope
symbol
of a quantifier for t
(∃x)P(x)
P(t) where t is a Existential Must be the first rule
variable or constant Instantiation- ei
used that introduces t
symbol not previously
used in a proof
sequence
P(x)
(∀x)P(x)
Universal P(x) has not been
Generalization- ug
deduced from any
hypotheses in which x is
a free variable nor has P
(x) been deduced by ei
from any wff in which x
is a free variable
P(x) or P(a)
(∃x)P(x)
Existential To go from P(a) to (∃x)P
Generalization- eg
(x), x must not appear in
P(a)
Section 1.4 Predicate Logic 2

4. Examples: Proofs using Predicate Logic
● Prove the following argument:
■ All flowers are plants. Sunflower is a flower. Therefore, sunflower
is a plant.
■ P(x) is “ x is a plant”
■ a is a constant symbol (Sunflower)
■ Q(x) is “x is a flower”
● The argument is (∀x)[Q(x) → P(x)] Λ Q(a) → P(a)
● The proof sequence is as follows:
1. (∀x)[Q(x) → P(x)]
hyp
2. Q(a)
hyp
3. Q(a) → P(a)
1, ui
4. P(a)
2, 3, mp
Section 1.4 Predicate Logic 3

5. Examples: Proofs using Predicate Logic
● Prove the argument (∀x)[P(x) → Q(x)] Λ [Q(y)]ʹ′ → [P(y)]ʹ′
● Proof sequence:
1. (∀x)[P(x) → Q(x)]
hyp
2. [Q(y)]ʹ′
hyp
3. P(y) → Q(y)
1, ui
4. [P(y)]ʹ′
2, 3, mt
● Prove the argument (∀x)P(x) → (∃x)P(x)
● Proof sequence:
1. (∀x)P(x)
hyp
2. P(x)
1, ui
3. (∃x)P(x)
2, eg
Section 1.4 Predicate Logic 4

6. Examples: Proofs using Predicate Logic
● Prove the argument (∀x)[P(x) → Q(x)] Λ (∀x)P(x) → (∀x)Q(x)
● Proof sequence:
1. (∀x)[P(x) → Q(x)]
hyp
2. (∀x)P(x)
hyp
3. P(x) → Q(x)
1, ui
4. P(x)
2, ui : no restriction on ui about reusing a name
5. Q(x)
3, 4, mp
6. (∀x)Q(x)
5, ug
● Note: step 6 is legitimate since x is not a free variable in any
hypothesis nor was ei used before
Section 1.4 Predicate Logic 5

7. Examples: Proofs using Predicate Logic
● Prove the argument
(∀x)[P(x) Λ Q(x)] → (∀x)P(x) Λ (∀x)Q(x)
● Proof sequence:
1. (∀x)[P(x) Λ Q(x)]
hyp
2. P(x) Λ Q(x)
1, ui
3. P(x)
2, sim
4. Q(x)
2, sim
5. (∀x)P(x)
3, ug
6. (∀x)Q(x)
4, ug
7. (∀x)P(x) Λ (∀x)Q(x)
5, 6, con
Section 1.4 Predicate Logic 6

8. Examples: Proofs using Predicate Logic
● Prove the argument
(∀y)[P(x) → Q(x,y)] → [P(x) → (∀y)Q(x,y)]
● Using the deduction method, we can derive
(∀y)[P(x) → Q(x,y)] Λ P(x) → (∀y)Q(x,y)
● Proof sequence:
1. (∀y)[P(x) → Q(x,y)]
hyp
2. P(x)
hyp
3. P(x) → Q(x,y)
1, ui
4. Q(x,y)
2, 3, mp
5. (∀y)Q(x,y)
4, ug
Section 1.4 Predicate Logic 7

9. Temporary hypotheses
● A temporary hypothesis can be inserted into a proof sequence.
If T is inserted as a temporary hypothesis and eventually W is
deduced from T and other hypotheses, then the wff T → W has
been deduced from other hypotheses and can be reinserted into
the proof sequence
● Prove the argument
[P(x) → (∀y)Q(x,y)] → (∀y)[P(x) → Q(x,y)]
● Proof sequence:
1. P(x) → (∀y)Q(x,y)
hyp
2. P(x)
temporary hypothesis
3. (∀y)Q(x,y)
1, 2, mp
4. Q(x,y)
3, ui
5. P(x) → Q(x,y)
temp. hyp discharged
6. (∀y)[P(x) → Q(x,y)]
5, ug
Section 1.4 Predicate Logic 8

10. More Examples
● Prove the sequence [(∃x)A(x)]ʹ′ ⇔ (∀x)[A(x)]ʹ′
■ To prove equivalence, implication in each direction should be proved
● Proof sequence for [(∃x)A(x)]ʹ′ → (∀x)[A(x)]ʹ′
1. [(∃x)A(x)]ʹ′
hyp
2. A(x)
temp. hyp
3. (∃x)A(x)
2, eg
4. A(x) → (∃x)A(x)
temp. hyp discharged
5. [A(x)]ʹ′
1, 4, mt
6. (∀x)[A(x)]ʹ′
5, ug
● Proof sequence for (∀x)[A(x)]ʹ′ → [(∃x)A(x)]ʹ′
1. (∀x)[A(x)]ʹ′
hyp
2. (∃x)A(x)
temp. hyp
3. A(a)
2, ei
4. [A(a)]ʹ′
1, ui
5. [(∀x)[A(x)]ʹ′]ʹ′
3, 4, inc
6. (∃x)A(x) → [(∀x)[A(x)]ʹ′]ʹ′ temp. discharged
7. [((∀x)[A(x)]ʹ′)ʹ′]ʹ′ 1, dn
8. [(∃x)A(x)]ʹ′ 6, 7, mt
Section 1.4 Predicate Logic 9

11. Proving Verbal arguments
● Every crocodile is bigger than every alligator. Sam is a crocodile. But
there is a snake, and Sam isn’t bigger than that snake. Therefore,
something is not an alligator.
■ Use C(x): x is a crocodile; A(x): x is an alligator, B(x,y): x is bigger than y,
s is a constant (Sam), S(x): x is a Snake
● Hence prove argument
(∀x) (∀y)[C(x) Λ A(y) → B(x,y)] Λ C(s) Λ (∃x)(S(x) Λ [B(s,x)]ʹ′) → (∃x)[A(x)]
1. (∀x) (∀y)[C(x) Λ A(x) → B(x,y)] hyp
2. C(s) hyp
3. (∃x)(S(x) Λ [B(s,x)]ʹ′) hyp
4. (∀y)[C(s) Λ A(y) → B(s,y)] 1, ui
5. S(a) Λ [B(s,a)]ʹ′) 3, ei
6. C(s) Λ A(a) → B(s,a) 4, ui
7. [B(s,a)]ʹ′ 5, sim
8. [C(s) Λ A(a)]ʹ′ 6, 7, mt
9. [C(s)]ʹ′ V [A(a)]ʹ′ 9, De Morgan
10. [[C(s)]ʹ′]ʹ′ 2, dn
11. [A(a)]ʹ′ 9, 10, ds
12. (∃x)[A(x)]ʹ′ 11, eg
Section 1.4 Predicate Logic 10

12. Class Exercise
● Prove the argument
(∀x)[(B(x) V C(x)) → A(x)] → (∀x)[B(x) → A(x)]
Proof sequence:
1. (∀x)[(B(x) V C(x)) → A(x)] hyp
2. (B(x) V C(x)) → A(x) 1, ui
3. B(x) temp. hyp
4. B(x) V C(x) 3, add
5. A(x) 2, 4, mp
6. B(x) → A(x) temp. hyp discharged
7. (∀x)[B(x) → A(x)] 6, ug
Section 1.4 Predicate Logic 11

13. Class exercise
● Every ambassador speaks only to diplomats, and some ambassadors speak
to someone. Therefore, there is a diplomat.
● Use A(x): x is an ambassador; S(x,y): x speaks to y; D(x): x is a diplomat
Prove the argument
(∀x) (∀y)[(A(x) Λ S(x,y)) → D(y)] Λ (∃x)(∃y)(A(x) Λ S(x,y)) → (∃x)D(x)
Proof Sequence:
1. (∀x) (∀y)[(A(x) Λ S(x,y)) → D(y)] hyp
2. (∃x)(∃y)(A(x) Λ S(x,y)) hyp
3. (∀y)((A(a) Λ S(a,y)) → D(y)) 1, ui
4. (∃y)(A(a) Λ S(a,y)) 2, ei
5. A(a) Λ S(a,b) 4, ei
6. (A(a) Λ S(a,b)) → D(b) 3, ui
7. D(b) 5, 6, mp
8. (∃x)D(x) 7, eg
Section 1.4 Predicate Logic 12