What do we mean when we say 'From $p$ and $p\implies q$, infer $q$'?

Question

There's a question: From $p$ and $p\implies q$, infer $q$. In my book, this is solved algebraically like this:
Given premises: $$P1: p$$ $$P2: p\implies q$$ Conclusion to be drawn: $q$ $$\therefore P1 \bullet P2 \implies C$$ $$= [p.(p\implies q)] \implies q$$ $$= [p.(\bar p + q)] \implies q$$ $$= (p \bullet \bar p + p \bullet q) \implies q$$ $$= (0 + p \bullet q) \implies q$$ Carrying out conditional elimination again, $$(\bar{p+q})+q$$ $$=\bar p + \bar q + q$$ $$=\bar p +1$$ $$=1$$ Hence the result is established.
Now my question is what do we mean when we are here trying to draw conclusion $q$? And what does $1$ as the final result signify?

Answer 1

The letters $p, q$ stand for propositions. So, for example, take

• $p$: "I study hard"
• $q$: "I get a good grade"

Then your derivation entails

• $p \Rightarrow q$ and $p$: "If I study hard I get a good grade, and I studied hard". Therefore:
• $q$: "I get a good grade."

The calculation as shown reduces \begin{align*} (p \text{ and } p \Rightarrow q) \Rightarrow q \end{align*} to $1$, i.e. to true so that above rule always holds.