Dokumen ini membahas tentang logika matematika, termasuk definisi pernyataan, nilai kebenaran, kalimat terbuka, negasi, serta pembentukan pernyataan komposit yang meliputi konjungsi, disjungsi, implikasi, dan biimplikasi. Juga dibahas tentang tautologi, kontradiksi, dan kontinjensi serta cara membuktikan kesimpulan melalui bukti langsung dan tidak langsung. Selain itu, dokumen ini memberikan latihan untuk memahami negasi, invers, konvers, dan kontrapositif dari pernyataan.

1. 
MATHEMATICAL LOGIC
By
Septi Ratnasari
(4101412082)
SEMARANG STATE UNIVERSITY
2013

2. 
Statement
Statement is sentences which may be true or
false, but noth both. True or false refers to the
actually of event.
In mathematics , statement is usually called
proposition.
Examples :
1. Jakarta is the capital city of the Republic of
Indonesia. (true)
2. Merapi is located in West Java. (false)
3. 2 + 3 = 6 (false)
A statement is usually written with a small
letter symbol such as p, q, r, s, and so on.

3. 
Truth Value of a Statement
A truth value is used to determine whether a
statement is true or false
The truth value of a statement can be denoted
as .
= tao, which is a Greek letter that chosen to
represent the word truth.
Examples
1. p : Human is breathing with lungs.
(p) = T ( read : the truth value of p is
true )
2. q : 12 + 3 = 5
(q) = F ( read : the truth value of q is
false )

4. 
Open Sentence
An open sentence is a sentence that still contain a
variable, hence its truth value cannot yet be
determined.
A variable is a symbol that is used to represent an
arbitrary element of a universal set.
Examples :
 2x – 3 < 9, x∈R
 x is prime number less than 20
A constant that replaces a variable which converts an
open sentence into a true statement is called an open
sentence solution.
A set that consists of all solutions is called a set of

5.  Negation of a Statement
A new statement that is constructed from a
previous statement such that it is true if the
previous statement is false, and it is false if
the previous statement is true.
Symbol of negation is ( ~ ).
p
~p
~(~p)
T
F
T
F
T
F

6. Truth Value of Compound Statements

Compound statement is statements which is
obtained by combining two or more statements.
Logical conjunctions
Symbol
Term
... and ...
... or ...
˄
˄
Conjunction
Disjunction
If ... then...
=>
Implication
... if and only if ...

Bi-implication
Single statements that are used to form
compound statement are called the components
of the resulting compound statement.

7. 
1. Conjunction
A compound statement is called
p
q
p ˄q
T
T
T
a conjunction if two statements
T
F
F
p and q are combined to form a
F
T
F
compound statement with a
F
F
F
conjunction “and”, denoted as
“p ˄ q”.
The truth value of p ˄ q is true
Examples :
(T)pif both + 3 =components
a)
: 2 of its 5 ( T )
b) p
: 12 is completely
areqtrue. 5 is a prime
:
divided by 3. ( T )
number. ( T )
q
: 15 is completely
p ˄ q : 2 + 3 = 5 and divided by 2. ( F )
5 is a prime
p ˄ q : 12 is completely

8. 
2. Disjunction
A compound statement is called
a disjunction if two statements
p and q can be combined by
using the logical conjunction
“or”, denoted by “p ˄ q” which
is read “p or q”.
The truth value of p ˄ q is only
Examples :
true if :at least= 8 ( Tp or q is
a)p
5 + 3 either )
b) p
true. : 8 is an even
q
number. ( T )
p˄q :5+3=8
or 8 is an even
p
T
T
F
F
q
T
F
T
F
pvq
T
T
T
F
:8>8(F)
q :8=8(T)
p ˄ q : 8 > 8 or 8 = 8 (
T ), can also be
stated
as 8 ≥ 8. ( T )

9. 
3. Implication
A compound statement is called an implication if two
statements p and q are combined to form a compound
statement by using the logical conjunction “if ... then ...”
, denoted by “p => q”, which can be read :
1. If p then q,
2. p implies q,
3. q only if p,
4. p is sufficient for q,
5. q is required for p.
p is called the antecedent (hypothesis) and q is called
the consequant.
The truth value of p => q is false (F) if p is true and q
is false. And for all the other composition of p => q is

10. 
p
T
T
F
F
q
T
F
T
F
p => q
T
F
T
T
Examples :
a) p : 5 + 3 = 8 ( T )
q : 8 is an even number. ( T )
p => q : If 5 + 3 = 8, then 8 is an even
number. ( T )
b) p : 5 > 3 ( T )
q : 5 is an even number. ( F )
p => q : If 5 > 3, then 5 is an even number.

11. 
4. Bi-implication
A compound statement is called a bi-implication if two
statements p and q are combined to form a compound
statement with a logical conjunction “... if and only if
...”, denoted by “p  q” that means “p if and only if q”,
i.e “if p then q and if q then p”. Hence p  q ≅ (p =>
q) ˄(q =>qp). p => q
(p => q) ˄ ( => p) ≅ p
q
p
q => p
q
T
T
T
T
T
T
F
F
T
F
F
T
T
F
F
F
F
T
T
T

12. Tautology

Tautology is a compound statement which is always
true for all possibilities from its components.
Example of Tautology :
p
q
~p
~q
p => q
(p => q) ˄ ~q
((p => q) ˄ ~q) =>
~p
T
T
F
F
T
F
T
T
F
F
T
F
F
T
F
T
T
F
T
F
T
F
F
T
T
T
T
T

13. 
Contradiction
Contradiction (the reverse of tautology) is a
compound statement which is false for all
possibilities from its components.
Example of Contradiction :
p
q
p˄q
~ (p ˄ q)
~ (p ˄ q) ˄ p
T
T
T
F
F
T
F
T
F
F
F
T
T
F
F
F
F
F
T
F

14. 
Contingency
Contingency is a compound statement
which is not a tautology and not a
contradiction for all possibilities from its
components.
p
q
p˄q
(p  q)  p
T
T
T
T
T
F
T
T
F
T
T
F
F
F
F
T

15. 
Negation of Compound Statements
a. Negation of Conjunction
p
q
~p
~q
p˄q
~(p ˄ q)
~p ˄ ~q
T
T
F
F
T
F
F
T
F
F
T
F
T
T
F
T
T
F
F
T
T
F
F
T
T
F
T
T
equivalent
From the truth table above we see that ~(p ˄q) ≅ ~p ˄~q. Therefore,
(~p ˄~q) is the negation of (p ˄q).
It can be concluded that :
~(p ˄q) ≅ ~p ˄~q

16. 
b. Negation of Disjunction
p
q
~p
~q
p˄q
~(p ˄ q)
~p ˄ ~q
T
T
F
F
T
F
F
T
F
F
T
T
F
F
F
T
T
F
T
F
F
F
F
T
T
F
T
T
equivalent
From the truth table above we see that ~(p ˄q) ≅ ~p ˄~q. Therefore,
(~p ˄~q) is the negation of (p ˄q).
It can be concluded that :
~(p ˄q) ≅ ~p ˄~q

17. 
c. Negation of Implication
p
q
~p
~q
p => q
~p ˄ q
~( p => q)
~(~p ˄ q) ≅ (p ˄ ~q)
T
T
F
F
T
T
F
F
T
F
F
T
F
F
T
T
F
T
T
F
T
T
F
F
F
F
T
T
T
T
F
F
equivalent
From the table, we can conclude that :
~( p => q) ≅ ~(~p ˄q) ≅ (p ˄~q)
equivalent

18. d. Negation of Bi-implication
p  q ≅ (p => q) ˄ ( => p)
q



Remember that
From this equivalence, we can derive :
~( p  q)
=
~((p => q) ˄ ( => p))
q
=
=
=
~(p => q) ˄ ~( => p)
q
~(~p ˄q) ˄~(~q ˄p)
(p ˄~q) ˄ ( ˄~p)
q
p
q
~p
~q
pq
~( p  q)
T
T
F
F
T
F
F
F
F
T
F
F
T
F
T
T
F
T
F
T
T
F
F
T
F
T
T
F
F
T
T
T
F
F
F
F
(p ˄ ~q) (q ˄ ~p)
equivalent
Hence,
~( p  q) ≅ (p ˄~q) ˄ ( ˄~p)
q
(p ˄ ~q) ˄ (q ˄
~p)

19. 
Converse, Inverse, and Contrapositive
Implication
=
p => q
Converse
=
q => p
Inverce
=
~p => ~q
Contrapositive =
~q => ~p

20. The relationships among the implications can be shown

by the following diagram.
p => q
Converse
q=>p
Inverse
Contrapositive
Inverse
~p =>~q
Converse
~q=>~p

21. 
implication
converse
inverse
contrapositive
p
q
~p
~q
p => q
q => p
~p => ~q
~q => ~p
T
T
F
F
T
T
T
T
T
F
F
T
F
T
T
F
F
T
T
F
T
F
F
T
F
F
T
T
T
T
T
T
equivalent
equivalent

22. 
Statements with Quantor
1. Universal Quantor

23. 
2. Exsistential Quantor

The solution set from an open sentence p(x) which
has at least one member of the whole set S, can be
written as follow
Symbol
is called existential quantor, read as '
There exists' or 'At least one of '.

24. 
Example
Given an open sentence 2x + 1 = 7. Expree it by using
exsistential quantor, and then determine its truth value,
whether the whole set is real number R.
Answer :

25. Negation of Quantor Statement


26. 

27. Making Conclusion

•
One of the important purposes of mathematical logic is to
have knowledge regarding testing argument or making
conclusion.
•
In logic, a premise is a claim that is a reason ( or element
of a set of reason ) for, or objection against, some other claim.
In other words, it is a statement presumed true within the
context of an argument toward a conclusion.
•
In the context of ordinary argumentation, the rational
acceptability of a disputed conclusion depends on both the

28. 1. Modus Ponens

Premise 1
Premise 2
Conclusion
: p => q
: p
: q
The argument above can be written in a form of
implication as follow
[(p => q) ˄p] => q

29. 
1. Modus Tollens
Premise 1
Premise 2
Conclusion
: p => q
: ~q
: ~p
The argument above can be written in a
form of implication as follow
[(p => q) ˄~q ] => ~p

30. 3. Syllogism

Premise 1
Premise 2
Conclusion
: p => q
: q => r
: p => r
The argument above can be written in a
form of implication as follow
[(p => q) ˄(q => r)] => (p => r)

31. 
Direct and Indirect Proof
1.
Direct Proof
In order to direcly prove a conditional statement of the
form ( p =>
q ), it is only necessary to consider
situations where the statement p is true to conclution q.
The common proof rules used are modus ponens,
modus tollens, and syllogism.

32. 
Example
Show that for all integer n, if n is an odd numbers, then n2 is
also an odd number !
Answer :
For example
p : n is an odd integer
q : n2 is an odd integer
It will be shown that p => q is true.
Because n is an odd number, thus n = 2k + 1, k ∊ C
Then, we have n2
= (2k + 1)2
= 4k2 + 4k + 1
= 2(2k2 + 2k) + 1
= 2m + 1
Where m = 2k2 + 2k, so n2 is an odd integer.
Thus, it is shown that p => q is true.

33. 
2.
a.
Indirect proof
Indirect Proof by Contradiction
To prove statement (p => q) is true, we can
proceed by assuming ~q is true. And show that it
leads to a logical contradiction.
Thus, according to the law of contradiction, ~q
must be true, and so, statement (p => q) is true.

34. 
Example
Show that ' If n2 is an odd number, then n will be also an
odd number ' by using indirect proof by contradiction.
Answer :
Suppose that n is an even number, that is n = 2k, k ∊ B.
Because n = 2k
Then n2 = (2k) 2 = 4k2 = 2(2k2 ) = 2m
Where m = 2k2
We get n2 is an even number, contradiction with n2 is an
odd number.
Thus, it is shown that if If n2 is an odd number, then n
will be also an odd number.

35.  b. Indirect Proof by Contraposition
To prove statement (p => q) is true, we can proceed
by assuming ~q is true. And show that it leads ~p is
true.
Thus ~q according to the law of contraposition ~p, is
true.
So statement (~q => ~p) is true.

36.  Example
Show that for all integer n2 is an odd number, then n will be also an odd
number.
Answer :
Use the indirect contraposition to prove the statement above.
Suppose that
p : n2 is an odd number
q : n is an odd number
We asssume that ~q is true, it is mean that n is an even number, n = 2k
We will get n2 = (2k2) = 4k2 = 2(2k2) = 2m, where m = 2k2
It is mean that n2 is an even number.
Thus, ~p : n2 is an even number
~q : n is an even number
Because (~q => ~p) is true and p => q ≅ (~q => ~p)
Then, p => q is true.
Thus, it is shown that if n2 is an odd number, then n will be also an odd

37.  Exercises
Give the negation of the statement “If all leaders put
forward the interest of their people, then all people will
live prosperously”.
2. Determine the inverse,converse, and contrapositive of
the statement form (p ˄q) =>r.
3. Write the valid conclusions of these premises.
p1 : Students do not like math or teachers like to teach.
p2 : If teachers like to teach, then math grades are good.
p3 : Students like math.
1.
∴ ...

38. 
4.
Write the valid conclusions of the following premises.