ADVANCED MATHEMATICS FORM 5 – LOGIC

BI CONDITIONAL STATEMENT

Consider the truth table for (P → Q) ∧ (Q → P)                      

The statement (P → Q) ∧ (Q → P) is known as bi-conditional statement and is abbreviated as P  Q

Truth table for P      Q

Note: P  Q is read P if and only if Q

          P  Q is true when both P and Q are true or when P and Q are false

    Example. The truth value of 43 > 3 if and only if 5< 3 (F)

                                                43 < 3 if and only if 3 < 5 (F)

                                                43< 3 if and only if 5 < 3 (T)

                                                43> 3 if and only if 5 > 3 (T)

CONVERSE, CONTRA POSITIVE, INVERSE

Given a proposition: if a quadrilateral is a parallelogram then its opposite sides are parallel, P → Q

Converse: If the opposite sides are parallel, then the quadrilateral is a parallelogram i.e.   Q → P.

Contra positive: If the positive sides are not parallel, then the quadrilateral is not a parallelogram. i.e. ~ Q → ~ P

Inverse: if a quadrilateral is not a parallelogram, then the opposite sides are not parallel i.e. ~ P → ~ Q

Truth table for implication, converse, contra positive, inverse        

Column 3 has exactly truth value as column 7

            i. e P → Q  ~Q → ~ P

            Q → P  ~ P → ~Q

EQUIVALENT STATEMENTS

Two propositions are logically equivalent if they have exactly the same truth values

E.g. P V Q and Q V P are logically equivalent

Solution: Draw truth for P V Q and Q V P

                        1            2                        3                      4

Since column 3 has exactly the same truth values as column 4 then

 P V Q  Q V P

Questions

Show whether or not the following propositions are logically equivalent

i)P → Q, ~ P V Q

Since column 3 and 5 have exactly the same truth value therefore
P → Q  ~ P V Q         

  ii) P → (P V Q); P → Q

 

Since column 4 does not have exactly same truth value as column 5 then   p → (P V Q)  P → Q

           iii) P → Q: ~ P → Q

Since column 3 does not have exactly same truth values as column 5 therefore

            P → Q   ~ P → Q

iv) P → Q; Q → P

Since column 3 does not have exactly same truth values as column 4 therefore P → Q  Q → P

v) ~ (P → Q);PV ~ Q
(5)                          (6)

Since column 5 does not have exactly same truth value as column 6 therefore ~ (P → Q)  P V ~Q

          vi) ~ (P V Q); ~P ∧ ~Q

Since column 6 has exact same truth values as column 7 therefore ~ (P V Q) →( ~P ∧ ~Q)

COMPOUND STATEMENTS

Compound statements with three components P, Q, R.

Consider the following compound statement,

Triangles have all three sides and either the area of a circular region of radius r is  or it is false that the diagonals of a parallelogram do not meet.

Solution

 (To symbolize the above statement)

  Let P triangles have three sides

 Let Q  circular region of radius r is

Let R diagonals of parallelogram do not meet

            P ∧ (Q V ~R)

To find the truth values of the above statement

        The statement has a truth value true

TAUTOLOGY

A tautology is a proposition which is always true under all possible truth conditions of its component parts

Example

Show that whether or not
~ (P ∧ Q) V (~P → ~Q) is a tautology

(6)               (7)

Since column 8 has all the truth values True (T) therefore it is TAUTOLOGY

Since column 8 has truth value true throughout then, ~ (P ∧ Q) V (~ P → ~Q) is a tautology

Questions

1.  Show whether the given compound statements are tautology or not

            i) (P ∧ Q) → P

Since column 4 has truth value true throughout then (P ∧ Q) → P is a tautology.

ii) P → (P ∧ Q)

Since column 4 does not have truth value true throughout then P → (P ∧ Q) is not a tautology.

iii)    P → ~P

Since column 3 does not have the truth value true throughout then

P → ~ P is not a tautology.

iv) (P → Q) → (~ P → Q)

Since column 6 does not have the truth value true throughout then (P → Q) →(~ P → Q) is not a tautology

v) (P → Q) V (Q → P)

Since column 5 has all truth values true throughout then  (P → Q) V (Q → P) is a tautology.

2.  Express the following in symbolic form and then find its truth value

            i) 2 is a prime, and either 4 is even or it’s not true that 5 is even  

Solution

            Let P  2 is a prime

            Let Q  4 is even

            Let R  5 is even

            P ∧ (Q V ~R)

            P ∧ (Q V ~R) has a truth value true.

            ii) 7 is odd, or either London is in France and it is false that Paris is not in Denmark

            Let P 7 is odd

                  Q  London is in France

                   R  Paris is not in Denmark

                  P V (Q ∧ ~R)     

             P V (Q ∧ ~R) has a truth value True

3.   Find the truth values of P ∧ (Q V ~R) if

            i) P, Q, R all has truth value T

            ii) If P, Q, R all have truth value of F

            iii) If P is true, Q is false and R is false

            A complete truth table for general cases

1.    Only one compound P
Two rows

2.  Two components P and Q
Four rows

3.  Three components P, Q, R
       Eight Rows

4.   Four components P, Q, R, S
       Sixteen rows

            Example constructs a truth table for the compound statement

                        ((P → Q) Λ R)  Q