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Are p -> q and ~q -> ~p equivalent

Step by step, with the reason for every step · Equivalent statements · Discrete Math

Step by step

The question

Question: are p → q and ¬q → ¬p equivalent?

p → q and ¬q → ¬p

1 Rows for p and q

Each letter can be true or false.

Two letters give 2 × 2 = 4 rows: every way they can go.

= two letters: 2 × 2 = 4 rows

We make a table: a column for each letter, then a column for each part, from the inside out.

We write T for true and F for false.

T = true, F = false

p: the first half T, the second half F.

This way no row is missed.

q: T, F, T, F, down the column.

2 Column for p → q

We read the arrow as if, then.

→ = if ..., then ...

p → q is a promise: if p, then q.

It is broken only when p is true and q is false.

p → q: F only when p is T and q is F

So it is F only in row 2.

3 Column for ¬q and ¬p

We read this sign as not.

¬ = not

¬q is the opposite of q: T becomes F, F becomes T.

¬q: the opposite of q

So it is T only in rows 2 and 4.

¬p is the opposite of p too.

¬p: the opposite of p

4 Column for ¬q → ¬p

¬q → ¬p is a promise: if ¬q, then ¬p.

It is broken only when ¬q is true and ¬p is false.

¬q → ¬p: F only when ¬q is T and ¬p is F

So it is F only in row 2.

5 Compare the columns

Now compare the two last columns.

p → q in red, ¬q → ¬p in purple.

Row by row they are the same: T, F, T, T and T, F, T, T.

This is the contrapositive: swap the two parts and put not on both.

It always means the same.

the contrapositive

Careful!

To be equivalent they must match in every row.

One row that differs is enough to say no.

Careful: one row that differs is enough for "not equivalent"

So they are equivalent.

We write p → q ≡ ¬q → ¬p.

= p → q ≡ ¬q → ¬p

6 Check

Let's check row 2 by hand.

In row 2, p is true and q is false.

row 2: p = T, q = F

The first: T → F gives F.

T → F = F

And the second: ¬F → ¬T gives F.

The same, as the table says.

Right!

¬F → ¬T = T → F = F

Let's look back.

First: rows for p and q.

Then: column for p → q.

Then: column for ¬q and ¬p.

Then: column for ¬q → ¬p.

Then: compare the columns.

Last: check.

Answer
= p → q ≡ ¬q → ¬p

Your turn

Try this one yourself, then check:

Are q → r and ¬r → ¬q
Show the answer
q → r ≡ ¬r → ¬q

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