# Why is modus ponens a q-implication?

Modus ponens is stated $$((P \implies Q) \land P) \implies Q$$ But isn't $(P \implies Q) = (\lnot P \lor Q)$?

Then we get $((P \implies Q) \land P) \implies Q$ $= (\lnot P \lor Q) \land P$ $= (\lnot P \land P) \lor (P \land Q)$ $= \text{False} \lor (P \land Q)$ $= P \land Q$ So $P\land Q$, not just $Q$. Why do we say that modus ponens then just implies $Q$?

• It also implies $P\wedge Q\wedge True$, but we don't say that either. – vadim123 Mar 11 '18 at 5:11
• We say that it implies $Q$, not that it implies only $Q$. – Alexander Burstein Mar 11 '18 at 5:11
• $P\implies (P\land Q)$ is equivalent to $P\implies Q$. “If this is a ruby, then it is red” is equivalent to “If this is a ruby, then it is red and it is a ruby”. – MJD Mar 11 '18 at 5:13

Yes, indeed, $P$ and $Q$ are both implied by $P$ and $P\to Q$.   In fact, $P$ and $Q$ and $P\to Q$ are all implied by $P$ and $P\to Q$.
However, we're most interested in learning new things.   $Q$ is new thing we learn from knowing $P$ and $P\to Q$.   It is not the only thing, but it is the new thing.