this is called the distributivity of implication over disjunction in classical propositional logic
this post was submitted on 23 Jul 2023
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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
where 1 1 0 1 0 1 0 1 are values of truth
so I don't understand how you can compare them
I’m not sure but could it be because, in your first truth table, you assumed the truth value of (a OR b) -> c to be true and you are finding the truth values of c that correspond with pairs of values of a and b?
However, in the second table you are finding the truth values of ~(a OR b) OR c that correspond with truth values of c as well as a and b so just like you said, you cannot compare the two tables you present above.
To get the truth table for the proposition (a OR b) -> c, you would find the corresponding truth values to those of a, b and c (like you did in the first table). Something like this:
A B C A OR B (A OR B) -> C
000 0 1
001 0 1
010 1 0
011 1 1
100 1 0
101 1 1
110 1 0
111 1 1
since it’s possible for the conditional proposition to be false (i.e. if either A or B are true yet C is false)
Oh I think I'm starting to get it. You're converting whether or not the proposition is true into a conditional, not the proposition itself.
I don't think (a OR b) -> c = ~(a OR b) OR c is valid, I think it needs to be written such that it communicates The proposition (a OR b) -> c is true if the following conditional ~(a OR b) OR c is true. It would be more clear in the opposite order too, If the conditional ~(a OR b) OR c is true, then the proposition (a OR b) -> c is also true.
Without a set of data, you have my first truth table, so you can't actually say whether or not (a OR b) -> c is true. However, the conditional has an associated complete truth table. They're not equivalent. I can prove that they're not equivalent by giving you another conditional that satisfies (a OR b) -> c: ~(A or B or C) or C which simplifies to C. The truth tables are different, so c != ~(a OR b) OR c however in your notation, (a OR b) -> c = ~(a OR b) OR c, which means the way I solved it would be written (a OR b) -> c = c. I'm going to switch to double equals for clarity:
1) (a OR b) -> c == ~(a OR b) OR c.
2) (a OR b) -> c == c.
3) If 1 is true, and 2 is true, then ~(a OR b) OR c == c. (Transitive property: “things which are equal to the same thing are also equal to each other.”)
4) ~(a OR b) OR c != c, therefore 1 and 2 can't both be true
A B C C (A OR B) -> C
00 0 0 1 <- This line is where the difference is
00 1 1 1
01 0 0 0
01 1 1 1
10 0 0 0
10 1 1 1
11 0 0 0
11 1 1 1
So observing that this looks true because a OR b only fails when both a and b do, here's an alternative:
a OR b -> c
Conditional contraposition:
~c -> ~(a OR b)
De Morgan's law 1:
~c -> ~a AND ~b
I'm not actually sure what this kind of distribution is called, but it makes sense:
(~c -> ~a) AND (~c -> ~b)
Two more contrapositions:
(a -> c) AND (b -> c)
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