EF Proving biconditionals

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Section 6.8 Proving biconditionals

We also often want to prove that two statements

P, Q

are equivalent; i.e. that

P \Leftrightarrow Q .

Fact 6.8.1.

The equivalence

P \leftrightarrow Q \Leftrightarrow (P \to Q) \land (Q \to P)

holds; i.e. a biconditional is equivalent to the conjunction of the corresponding conditional

P \to Q

and its converse.

Proof.

You are asked to prove this by truth table in Exercise 2.5.5.

Procedure 6.8.2. Proving a biconditional.

To prove

P \Leftrightarrow Q,

prove

P \Rightarrow Q

and

Q \Rightarrow P

separately.

As usual, this also works in the universal case since

\forall

distributes over

\land

(Proposition 4.2.6).

Worked Example 6.8.3.

Prove: A number is even if and only if its square is even.

We want to prove that the following quantified biconditional (“for all

n

” omitted, domain is nonnegative, whole numbers).

biconditional: $n$ is even if and only if $n^{2}$ is even.
conditional and converse: (if $n$ is even then $n^{2}$ is even) and (if $n^{2}$ is even then $n$ is even)
contrapositive and converse: (if $n^{2}$ is odd then $n$ is odd) and (if $n^{2}$ is even then $n$ is even)
conditional and inverse: (if $n$ is even then $n^{2}$ is even) and (if $n$ is odd then $n^{2}$ is odd)

These are all equivalent, so we could prove any one pair.

Original conditional.

This is proved as Worked Example 6.3.2.

Converse.

If

n^{2}

is even, then there exists an integer

m

such that

n^{2} = 2 m,

so that

n = \sqrt{2 m}

… ? We seem to be stuck.

Inverse.

If

n

is odd, then there exists an integer

m

such that

n = 2 m + 1 .

Then,

n^{2} = 4 m^{2} + 4 m + 1

is odd.

Checkpoint 6.8.4.

Attempt Exercise 6.12.10.

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