# If P then Q is true and so is P. Then Q is not necessarily true, right?

Say, "$$A$$ or $$B$$" (this is the inclusive 'or') and $$A$$ are both true propositions. Then we still don't know whether proposition $$B$$ is also true.

What if you are instead told that "If $$P,$$ then $$Q$$" and $$P$$ are both true propositions. Then you can't necessarily say that proposition $$Q$$ is true, can you? Because "If $$P,$$ then $$Q$$" could be true in the following three cases:

1. $$P$$ is true and $$Q$$ is true
2. $$P$$ is false and $$Q$$ is false
3. $$P$$ is false and $$Q$$ is true.

So, there is this unsaid assumption then in mathematics, right? That it was shown that $$P$$ is true and $$Q$$ is true before the proposition "If $$P,$$ then $$Q$$" was presented to us as a true proposition.

EDIT

I think I'll change my question if I can:

Say, you were proving other things true by using "If $$P,$$ then $$Q$$". And you were told that "If $$P,$$ then $$Q$$" is true, but told the truth value of neither $$P$$ nor $$Q.$$ You would then make the assumption that it was proven that $$P$$ is true and $$Q$$ is true, right? Otherwise the statement "If $$P,$$ then $$Q$$" is kinda useless in the cases where $$P$$ is false, right? (as in case 2 and case 3 that I wrote)

• $P$ and $P\Rightarrow Q$ is sufficient to deduce $Q$. When we do proofs, we generate truth by showing $P\Rightarrow Q$. For the purposes of the proof, we assume $P$. Commented Sep 30, 2021 at 6:49
• See Modus Ponens. A typical math proof works this way: we prove the proposition "if A, then T", where A is an Axiom or previously proven Theorem. Thus, by MP, we have proved the new theorem T. Commented Sep 30, 2021 at 7:07
• There are several ways here to motivate why $Q$ is true. (1) Your Options 2. and 3. are not possible, since you already assume that $P$ is true. (2) The implication $P \implies Q$ is logically equivalent to $\neg P \lor Q$, so you can again think of it in terms of "or". (3) There is a logical principle which allows to deduce $Q$ from the truth of $P \implies Q$ and $P$, called modus ponens. Commented Sep 30, 2021 at 7:10
• If Proposition P is true, why are you considering cases 2 and 3?
– A.J.
Commented Sep 30, 2021 at 7:17
• @user6750519 The typical way this comes up in practice is you have a conjecture $Q$ that you think is true and want to prove. Then you try to prove $Q$ and discover you can prove it if you assume $P$, which you also think (but don't know) is true. Thus you have given a proof of $P\to Q.$ Although it would be useless if you're wrong and $P$ is false, if your instincts are right, then it could be progress, because you have reduced proving $Q$ to proving $P.$ Then you can set to work proving $P$ and can let everyone else know that if they want to prove $Q,$ they can do so by proving $P.$ Commented Sep 30, 2021 at 22:11

Not sure what the question is. But $$P\implies Q$$ means we have one of the $$3$$ possibilities, as stated in the question:

1. P is true and Q is true
2. P is false and Q is false
3. P is false and Q is true

And if $$P$$ is true then we must have $$1$$, so $$Q$$ is true.

'P implies Q' literally means 'if P is true, then Q is true'. So if we know that P implies Q and P is true, then Q is also true.

• Let $$P$$ and $$Q$$ be statements—possibly compound statements—in a specified context. The truth function $$P\to Q$$ is false when $$P$$ is true and $$Q$$ false, and is true otherwise, in which case we say “$$P$$ being true implies that $$Q$$ is true” or “if $$P$$ is true, then $$Q$$ is true” and write $$P\Rightarrow Q.\tag1$$

Say, you were told that "If $$P,$$ then $$Q$$" is true, but told the truth value of neither $$P$$ nor $$Q.$$ You would then make the assumption that it was proven that $$P$$ is true and $$Q$$ is true, right?

No, no, statement $$(1)$$ does not assume, nor ask you to assume, that statement $$P$$ is true; if we have no information about $$P$$'s truth, then statement $$(1)$$ does not help us infer whether statement $$Q$$ is true and is useless.

After all, statement $$(1)$$ does not assert, “$$P$$ is true; consequently $$Q$$ is true.”

• In the context of mathematics, the statement $$P(x) \Rightarrow Q(x)\tag2$$ typically means “if $$P(x)$$ is true for some value of $$x,$$ then $$Q(x)$$ is true for the same value of $$x$$”, that is, that $$P(x)\to Q(x)$$ is universally true.

Many mathematics theorems are of this form. Applying such a theorem is basically asserting that $$P(c)$$ is true, and that since $$P(c)\Rightarrow Q(c),$$ by Modus Ponens, $$Q(c)$$ must therefore be true.

On the other hand, the process of proving statement $$(1)$$ or $$(2)$$ does involve assuming that $$P$$ or $$P(x)$$ is true.

Say you were proving other things true by using"If P, then Q". And you were told "If P, then Q" is true but NOT told anything about P or Q (you weren't told the truth values of P or Q). You would then make the assumption that it was proven that P is true and Q is true, right? Otherwise the statement "If P, then Q" is kinda useless in the cases where P is false, right? (as in case 2 and case 3 that I wrote)

No, you wouldn't make that assumption, because $$P\to Q$$ doesn't imply that either $$P$$ or $$Q$$ is true. Yes, generally speaking, showing $$P\to Q$$ is useless (and trivial) if you know that $$P$$ is false, but that doesn't absolve us of the responsibility of proving $$P$$ in order to make use of the implication.

In practice, when a mathematician proves $$P\to Q$$, it's usually when $$P$$ and $$Q$$ are both unproven conjectures believed to be true. This can be great progress, even though we can't conclude to have proven $$P$$ or $$Q,$$ since now we know that if we manage to prove $$P$$ then we will have proven $$Q$$ too.

For instance, if it were 1990 and we really wanted to be the person who finally proved Fermat's last theorem, we might find it very interesting that about five years before then, it was proved that $$P\to Q,$$ where $$P$$ is (a special case of) the Taniyama–Shimura conjecture and $$Q$$ is Fermat's last theorem. This it tells us that if we prove Fermat's last theorem, one way of doing so is to prove the Taniyama–Shimura conjecture. So if we're particularly sharp on Elliptic curves, modular forms, and the like, this might be an attractive strategy. And indeed, that's what Andrew Wiles tried and succeeded at.

That's just one example... this happens all the time and is in a large part how math progresses.

Another example is often we prove things of the form $$P\to Q$$ where $$P$$ and $$Q$$ have parameters. For instance $$P$$ might say "$$R$$ is a finite integral domain" and $$Q$$ might say "$$R$$ is a field". It is a theorem that $$P\to Q,$$ and this theorem is indeed useless when $$R$$ is not a finite integral domain,and useful when $$R$$ is one. But again, this doesn't absolve us from having to establish that $$R$$ is, in fact, a finite integral domain in order to use this theorem to establish that $$R$$ is a field.

You couldn't argue that from a programming perspective( which uses mostly identical logic) . Conditionals imply, you will run that block of code, if the condition is true. In other words the proposition that the block of code will run, is true. If the conditional is true. If $$P$$ then $$Q$$ is the same as $$Q$$ if $$P$$ .