I haven’t formally learnt logic so I’m not sure if my proof is what you’d call rigorous, but the result is pretty useful for splitting up conditionals in proofs like some of the number theory proofs I’ve been trying. E.g.
Show that if a is greater than 2 and a^m + 1 is prime, then a is even and m is a power of 2
In symbolic form this is:
∀a >= 2 ( a^m + 1 is prime -> a is even AND m is a power of 2 )
The contrapositive is:
∀a >= 2 ( a is odd OR m is NOT a power of 2 -> a^m + 1 is composite )
and due to the result above, this becomes
∀a >= 2 ( a is odd -> a^m + 1 is composite ) AND ( m is NOT a power of 2 -> a^m + 1 is composite )
so you can just prove two simpler conditionals instead of one more complicated one.
Maybe I'm dumb, but the first line reads to me as "if a or b then c" and the second one reads as "not a or b, otherwise c" and those two aren't equivalent. One is a condition for C, and the other is a condition in general. But it's been a while, so maybe that's an okay thing to do?
Afaik they are equivalent since using the truth table of a conditional A->B, it’s false when A is true but B is false (like how a philosophical argument is invalid if the premise A is true yet the conclusion B is false) so ~(A->B) = A and ~B and A->B = ~A or B. Were you asking about something else?
Yes, something else. you went from
(a OR b) -> c
to
~(a OR b) OR c.
The first one you are stating that C will be true if A or B is true
But in the second one, there is no ->, so how can they be equivalent?
The first line has a truth table of
A B
00: X
01: 1
10: 1
11: 1
where X 1 1 1 are the values of C
but the second line has a truth table of
A B C
000: 1
001: 1
010: 0
011: 1
100: 0
101: 1
110: 0
111: 1