[Math] If P then Q is true and so is P. Then Q is not necessarily true, right

logicproof-writingpropositional-calculus

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)

Best Answer

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.

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