Conditional StatementsCS 2312, Discrete Structures II

Poorvi L. Vora, GW

Conditional StatementA implies B

A => B

If A then B

A is the hypothesis; B the conclusion

Example: If x is a prime larger than 2 then x is oddA: “x is a prime number larger than 2”B: “x is odd”

• Generally, the truth of the conditional statement is a function of one or more variables; in this case, x.

• This example statement is true for all values of x.

Truth value a function of variables

Generally speaking, the truth value of a statement depends on variables.

For example, consider the inequality: 0 ≤ n3 ≤ Bn2

– Whether it is true or not depends on the values of the variables: n and B

– There are quite likely value of n and B for which it is true, as well as values for which it is false

: for every

Consider a modification of the previous example: 0 ≤ n3 ≤ Bn2 n ≥ b

The truth value of this inequality is now a function of B and b

On fixing b and B, • If it is untrue for a single value of n ≥ b then the statement is false• If it is true n ≥ b then it is true• Thus the truth is not dependent on n

Counterexample

Sometimes you may have statement that is false: x is divisible by 4 even integers x

The way to show it is false is to show that the statement is not true “ even integers x”

one even integer for which it is false is enough.

However, this does not mean that x is not divisible by 4 even integers x

Counterexample, contd

The statement “x is divisible by 4” is sometimes true and sometimes false

Thus neither of the statements below is true, because there are counterexamples for both: • x is divisible by 4 even integers x• x is not divisible by 4 even integers x

: There exists

Sometimes you may be asked to show that there exist values of variables such that a statement is true.

Continuing the previous example: $ B, b > 0 such that 0 ≤ n3 ≤ Bn2 n ≥ b

One way of proving above is to • find one value of b, B and • show that the inequality is true for these values.

How and and change a statement

Consider the statement: “x is divisible by 6”Its truth depends on the value of x

Adding or can eliminate the dependence on x to allow you to say definitively whether the statement is true or false:

x such that x divisible by 6: True$ odd x such that x is divisible by 6: False

x is divisible by 6 integers x that are multiples of 12: Truex is divisible by 6 integers x: False

The truth table of A=>B

We want to know the truth value ofA => B

Note: statement makes no claims when A is false. – when A is false, the statement is true independent of the

value of B

Negation or Counterexample

Examining the statement A(x) => B(x) x (where both A and B are functions of x)

– If you think it is true, you need to provide a proof.• Begin with the assumption A(x) is true and show B(x) is true.

– If you think it is false, you could provide a counterexample. • Exactly one example of the variable which makes A true but B false.• Notice that providing an example where B is true but A false gives you nothing at

all– because the statement is making no claims when A is false.

don’t care

A Counterexample

Suppose you are presented with the conditional statement:

integers x, if 2 divides x then 4 divides x

What is a valid counterexample?

Are there values of x for which the statement is true?

Converse

Given: A => B

The converse isB => A

Does it follow?

Example: If x is a prime larger than 2 then x is odd

What is the converse?

Does it follow?

Proving the converse does not prove the main statement

Suppose you have to showA => B

And you begin with B to show A

You have not shown A => B

You have shown B => A

Examples of incorrect proofs

To show that If 2 divides x, then 4 divides x (this is not true, but examples of incorrect proofs will conclude it is)

Example 1: Suppose 4 divides x. Then x = 4q (for q the quotient when 4 divides x). Hence x is even and 2 divides x.

Incorrect! Why?

Examples of incorrect proofs, contd.

To show that If 2 divides x, then 4 divides x (this is not true, but examples of incorrect proofs will conclude it is)

Example 2: Suppose 2 divides x. Then x = 2q1

Suppose 4 divides x. Then x = 4q2

x = x => 2q1 = 4q2 => q1 = 2q2

LHS = RHS

Incorrect! Why?

ContrapositiveOne can show

A=> B by assuming A is true and showing that then B is true

OR

by assuming that B is not true and showing that then A is not true.

That is, by showing: not B => not A

Which is the contrapositive

Which is logically equivalent to A => B

Why is the contrapositive equivalent to the original statement?

A B A => B not B => not A

T T T T

T F F F

F T T T

F F T T

Consider the truth table

Inverse

not A => not B

Is this related to the original statement? The converse? The contrapositive?

It is the contrapositive of the converse

Summary

Conditional Statement: A => B

– Converse: B => A

– Contrapositive: not B => not A

– Inverse: not A => not B

– Counterexample: an example of A and not B

Necessary and Sufficient

A => BA sufficient for B

B => A A necessary for B

A <=> BA is necessary and sufficient for BB is necessary and sufficient for A

Bidirectional

If and only if (iff)

If A then B AND If B then A

A <=> BA iff B

A and B are equivalent statements

False Premise, Valid Argument, False Conclusion

Claim: If 0 + 1 = 0 then 2=0

Proof: Begin with correct statement: 2 = 0 + 1 + 1 => 2 = (0+1)+1 = 0+1=0

Valid Premise, Invalid Argument, Valid Conclusion gives zero credit

Assume: If 4 divides x then 2 divides xShow that 4 divides 16

Invalid proof:2 divides 16. Hence, applying assumption, 4 divides 16.

Proof by Contradiction

To show A=> B

Recall, if A is not true, you cannot determine anything.

Assume A is true. Suppose B is not. Then show there is a contradiction. That is, B has to be true.