1. UNIVERSITI PENDIDKAN SULTAN IDRIS
PREPARED BY : MOHAMAD AL FAIZ BIN
SELAMAT

3. Tautology:
A proposition that is always true for all possible value of its propositional
variables.
Example of a Tautology
The compound proposition p ˅¬p is a tautology because it is always true.
P ¬p p ˅ ¬p
T F T
F T T

5. P ¬p p ˅ ¬p
T F F
F T F

7. A contingency table is a table of counts. A two-dimensional
contingency table is formed by classifying subjects by two
variables. One variable determines the row categories; the other
variable defines the column categories. The combinations of
row and column categories are called cells. Examples include
classifying subjects by sex (male/female) and smoking status
(current/former/never) or by "type of prenatal care" and
"whether the birth required a neonatal ICU" (yes/no). For the
1.
mathematician, a two-dimensional contingency table with r
rows and c columns is the set {xi j: i =1... r; j=1... c}.

9. :
,
Definition:
An argument is a sequence of propositions written
The symbol ∴ is read “therefore.” The propositions, ,… are
called the hypotheses (or premises) or the proposition q is
called the conclusion. The argument is valid provide that if the
proposition are all true, then q must also be true; otherwise,
the argument is invalid (or a fallacy).

11. p→q
Determine whether the argument
p
∴q
Is valid
[First solution] We construct a
truth table for all the propositions
involved.
P q p→q p q
T T T T T
T F F T F
F T T F T
F F T F F

13. Rule of inference Name
p→q Modus ponens
p
∴q
p→q Modus tollens
⌐q
∴ ⌐p
Addition
p
∴p˅q
Simplification
p˅q
∴p
p Conjunction
q
∴p˅q
p→q Hypothetical syllogism
q→r
∴p→r
p˅q Disjunctive syllogism
⌐p
∴q

15. Represent the argument.
The bug is either in module 17 or in module 81
The bug is a numerical error
Module 81 has no numerical error
___________________________________________
∴ the bug is in module 17.
Given the beginning of this section symbolically and
show that it is valid.
If we let
p : the bug is in module 17.
q : the bug is in module 81.
r : the bug is numerical error.

17. The argument maybe written
pVq
r
r → ⌐q
∴p
From r → ⌐q and r, we may use modus
ponens to conclude ⌐q. From r V q and
⌐q, we may use the disjunctive syllogism to
conclude p. Thus the conclusion p follows
from the hypotheses and the argument is
valid.

19. This method is based on Modus Ponens,
[(p ⇒ q) ˄ p ] ⇒ q
Virtually all mathematical theorems are
composed of implication of the type,
(
The are called the hypothesis or premise, and
q is called conclusion. To prove a theorem
means to show the implication is a tautology.
If all the are true, the q must be also true.

21. Solution:
Let p: x is odd, and q: x2 is odd. We want to prove
p → q.
Start: p: x is odd
→ x = 2n + 1 for some integer n
→ x2 = (2n + 1)2
→ x2 = 4n2 + 4n + 1
→ x2 = 2(2n2 + 2n) + 1
→ x2 = 2m + 1, where m = (2n2 + 2n) is an
integer
→ x2 is odd
→ q

23. Definition:
An indirect proof uses rules of inference on the negation of
the conclusion and on some of the premises to derive the
negation of a premise. This result is called a contradiction.
 Contradiction: to prove a conditional proposition
p ⇒ q by contradiction, we first assume that the
hypothesis p is true and the conclusion is false
(p˄ ~ q). We then use the steps from the proof of
~q ⇒ ~p to show that ~p is true. This leads to a
contradiction (p˄ ~ p), which complete the proof.

25. Proof: Assume that x is even (negation of
conclusion).
Say x = 2n (definition of even).
Then = (substitution)
= 2n · 2n (definition of exponentiation)
= 2 · 2n2 (commutatively of
multiplication.)
Which is an even number (definition of even)
This contradicts the premise that is odd.