Thread: Conditional Connective in Logic

Hello.

Having a bit of trouble understanding the truth table for the statement:

.

Could somebody run me through the scenarios and explain how they affect the truth value of . In particular, the scenario where is false, but is true... how does this imply that is true?

So :

true, true: ?

true, false: ?

false, true: ?

false, false: ?

I know the answers are true, false, true, true. But can someone explain the logic behind it?

Originally Posted by Mush

Hello.

Having a bit of trouble understanding the truth table for the statement:

.

Could somebody run me through the scenarios and explain how they affect the truth value of . In particular, the scenario where is false, but is true... how does this imply that is true?

So :

true, true: ?

true, false: ?

false, true: ?

false, false: ?

I know the answers are true, false, true, true. But can someone explain the logic behind it?

The "logic" behind it is that it is DEFINED that way! You are asking a slightly different question, I think- you want to connect that with out "everyday" idea of "if... then...". And the difficulty with that is our usual concept of "if A then B" does NOT assign a truth value in the case that A is false, no matter what B is. You have to be careful to distinguish between "If A then B" and "A if and only if B". "If A then B" only talks about what happens if A is TRUE. It says nothing at all about what happens if A is false. But for symbolic logic purposes, we must have a value in all cases and it is simplest to assign "true". I like to think of it as "innocent until proven guilty". Suppose a teacher says to his class, "If you get "A" on every test, I will give you an "A" in the course." Okay, you get an "A" on every test and get an "A" in the course (the "T->T" case). His statement was obviously true. On the other hand, suppose you get a "B" on every test and get a B for the course (the "F->F" case). Again, his statement is true. But suppose you got an A on every test but one and a B on that one. If he gives you an "A" was his statement false (the F-> T case)? No, because he never said what would happen if you DIDN'T get an A on every test. The only way you could be SURE his statement was false was if you got an A on every test and did NOT get an A in the course (the "T->F" case).

Originally Posted by HallsofIvy

The "logic" behind it is that it is DEFINED that way! You are asking a slightly different question, I think- you want to connect that with out "everyday" idea of "if... then...". And the difficulty with that is our usual concept of "if A then B" does NOT assign a truth value in the case that A is false, no matter what B is. You have to be careful to distinguish between "If A then B" and "A if and only if B". "If A then B" only talks about what happens if A is TRUE. It says nothing at all about what happens if A is false. But for symbolic logic purposes, we must have a value in all cases and it is simplest to assign "true". I like to think of it as "innocent until proven guilty". Suppose a teacher says to his class, "If you get "A" on every test, I will give you an "A" in the course." Okay, you get an "A" on every test and get an "A" in the course (the "T->T" case). His statement was obviously true. On the other hand, suppose you get a "B" on every test and get a B for the course (the "F->F" case). Again, his statement is true. But suppose you got an A on every test but one and a B on that one. If he gives you an "A" was his statement false (the F-> T case)? No, because he never said what would happen if you DIDN'T get an A on every test. The only way you could be SURE his statement was false was if you got an A on every test and did NOT get an A in the course (the "T->F" case).

Indeed. I had the hunch that I was correct in saying that P being false gave no indication of the falsity of the statement P-> Q... but due to the nature of boolean algebra, the statement had to be either true or false.

So, to conclude, if the premises neither negate nor confirm the conditional statement, then we assume it is true by default for the purposes of accordance with boolean algebra?