Why is modus ponens a q-implication?

Question

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 05:11
$P\implies (P\land Q)$ 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 05:13

Graham Kemp · Accepted Answer · 2018-03-11T05:29:01.650

3

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.

It is useful to know this, so we give the rule of inference a special name: "modus ponens".

edited Mar 11 '18 at 05:29

answered Mar 11 '18 at 05:17

Graham Kemp

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Why is modus ponens a q-implication?

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