I saw in a writing: ~p -> p = false. This can't be true since 0 -> 1 = 1, so ~p -> p = p!

It can be true since 1 -> 0 = 0.

3 hours ago, Willem F Esterhuyse said:

I saw in a writing: ~p -> p = false. This can't be true since 0 -> 1 = 1, so ~p -> p = p!

p is a proposition, not a number. If p is "Love is blind", then ~p -> p would mean "If love is not blind, then love is blind." That can't be true, because either love is blind or it's not blind.

EDIT: Okay, this isn't technically a contradiction. It's considered true if p is true, and of course it's false if p is false.

"~p ↔ p" would be a contradiction.

Edited January 16, 20232 yr by Lorentz Jr

You have to be very careful with negating a compound proposition or statement.

For example the negation of "love is blind"   could be "something other than love is blind"  eg "A worm is blind" .

In this case not p is true.

2 hours ago, Lorentz Jr said:

p is a proposition, not a number.

I assume that 0 and 1 in OP don't stay for arithmetic numbers, but rather represent logical values, which by usual convention are: 0 for FALSE and 1 for TRUE.

6 minutes ago, Genady said:

I assume that 0 and 1 in OP don't stay for arithmetic numbers, but rather represent logical values, which by usual convention are: 0 for FALSE and 1 for TRUE.

I see. I guess I should have figured that out. Maybe time for some coffee.

5 hours ago, Willem F Esterhuyse said:

since 0 -> 1 = 1

Can you demonstrate that this is true?

18 hours ago, swansont said:

Can you demonstrate that this is true?

No, you learn it from the truth table for "therefore". There is a reason for it though, more of a convention.