Homework Help: Logical Implication - If p, then q

1. Aug 17, 2010

michonamona

If p, then q.

Suppose p is false but q is true. Why is it that the implication "If p, then q" is still true?

For example,

If x=2, then x + 3 = 5.

Suppose x is NOT 2 (i.e. p is false), but x+3=5 (q is still true). Why is the implication
"If x=2, then x + 3 = 5" still true?

Is the truth of the consequence all that matters?

Thanks,

M

2. Aug 17, 2010

Dickfore

How is this a homework problem?

3. Aug 17, 2010

loseyourname

Staff Emeritus
What you've set up isn't really a material conditional. It's true that "if x=2, then x+3=5," but it's also true that "if x+3=5, then x=2." The logical relationship between p and q as you've specified them is equivalence, not implication. As such, the truth value of p and q has to be the same.

4. Aug 17, 2010

Dickfore

Here's a problem in mathematical logic:

What binary logical function f satisfies the modus tolens rule, i.e. the function:

$$\left[(p f q) \wedge \neg q \right] f \neg p$$

is a tautology?

5. Aug 17, 2010

Staff: Mentor

The way I learned it was that there was only one set of truth values for p and q that make the logical implication false: p being true and q being false.

For every other possible combinations of truth values, the implication is considered true.

6. Aug 17, 2010

loseyourname

Staff Emeritus
It's probably easiest to see why if you think of a plain language proposition. For instance,

"If Dracula is mortal, Dracula will die."

This is true even though Dracula is not mortal. It would only be false if it was possible for Dracula to be mortal but not die.

7. Aug 17, 2010

Hurkyl

Staff Emeritus
This is time for a logic puzzle!

Your challenge is to write down a truth table for a binary operator I. That is, you need to decide the value "true" or "false" for each of:
• I(true, true)
• I(true, false)
• I(false, true)
• I(false, false)

But you need to do it in a way that
• If I(P,Q) and P are both true, then Q must also be true
• It is possible for I(P,Q) to be true, and P to be true
• It is possible for I(P,Q) to be true, yet P to be false.
• I(P,Q) is not a synonym for Q
• I(P,Q) is not a synonym for P=Q

8. Aug 18, 2010

HallsofIvy

Pretty much, yes. I always think of "p-> q with p false, q true" as "innocent until proven guilty". In order to assert that "p-> q" is false, we must find some instance in which p is true but q is false. Showing a case with p false and q true simply isn't enough.

Here's an example I have used with classes: I tell you "If you get an 'A' on every test, then you will get an 'A' in the course". That is a "p-> q" implication with "p" being "you get an A on every test" and "q" being "you get an A in the course".

Now consider the 4 cases:
1) p= T, q= T. You did, in fact, get an A on every test, and you got an A for the course- exactly what I said.

2) p= F, q= F. You did NOT get an A on every test (perhaps, in fact, you failed every test!) and did not get an A for the course. No surprise there!

3) p= T, q= F. You get an A on every test but do NOT get an A in the course. Okay, now you have a right to complain! You met the conditions, I did not live up to my promise.

4) p= F, q= T. You did NOT get an A on every test (perhaps you got an A in all but one and got a high B in that test) but you got an A for the course anyway. Obviously, you are not going to complain but did I fail to keep my promise? Was my original statement false? No, because I did NOT say what would happen if you did NOT get an A in every test. I said what would happen if you got an A in every test but did NOT say anything about any other possibility.

Thanks,

M[/QUOTE]

9. Aug 18, 2010

michonamona

Thank you everyone for your replies!

But isn't my getting an A in the course conditional upon whether or not I got an A on every test? Can I say that you broke your promise because you said that I will get an A in the course only if I get an A on every test?

My confusion with this implication arises from the proof of the empty set being a set of every set.

Just to remind everyone, I've included the proof below.

$$\oslash \subset A$$ for every set A.

Proof:

In order for $$\oslash \subset A$$ to be true, the implication

If $$x\in \oslash$$, then $$x\in A$$

must be true. Since x is never in the empty set, then the implication above is always true. Therefore, the empty set is the subset of every set.

10. Aug 18, 2010

Staff: Mentor

No, the condition (hypothesis) was "if you get an A on every test" and the conclusion was "you get an A for the course". This is the difference between p ==> q and p <==> q. The implication doesn't say what happens if you don't get As on all the tests.

Relative to HallsofIvy's 4th case, the hypothesis is false, but the conclusion is true.

The only statement that says exactly the same thing as this implication is what's called the contrapositive: ~q ==> ~p. In words, "if you don't get an A for the course, you didn't get A's on each of the tests".

No. There is no "only if" in this implication. It does not exhaustively state the circumstances under which you get an A. The promise was: if you get all A's on the tests, you will get an A for the course. The only way the promise could be broken is when you actually get all A's on the tests, but don't get an A for the course.

11. Aug 19, 2010

HallsofIvy

NO, I did NOT say that! Specifically, I did NOT use the word "only". You are still confusing the conditional, "If p then q", with "p if and only if q", the bi-conditional.

I said what would happen if you got an A on every test. I did not say what would happen in any other case. I did NOT say that was the only circumstance under which I would give an A. Strictly speaking, if I went ahead and gave an "A" to everyone in the class, it would still be true that "if you get an A on every test you will get an A in the course."