I am reading Rosser's Logic for Mathematicians, here ($P \supset Q$ means $P \rightarrow Q$):
If we have $P \supset Q$ and $P$ proved, how come we have $Q$ without having to use modus ponens? It's as If for such occasion, there are two ways. It's not clear what are these ways. I know what using modus ponens is, but this alternate way is completely mysterious to me.
On
Suppose we proved $(P⊃Q)$. Also, suppose that in proving $(P⊃Q)$, we knew that $P$ held true, and then deduced $Q$. Thus, $Q$ is true. We do NOT need to assume $(P⊃Q)$ and $P$ and use modus ponens to deduce $Q$. Instead, we can just assume $P$, since we know that $P$ is true, and thus deduce $Q$.
In symbols, if we assume $P$ and deduce $Q$ we can write,
$$P\vdash Q$$
That differs from modus ponens,
$$\{P, (P⊃Q)\} \vdash Q$$
He is saying if you have a deduction of $P$ and you also have a deduction of $Q$ from $P,$ then you can just concatenate those deductions into one long deduction of $Q.$ This won't use modus ponens. In fact, you never even deduce $P\to Q.$