I have this proof
I'm confused because from what I understand the $\vdash$ means that whenever the left-hand side is true, then the right hand side also is.
If I assume p and q are true and work my logic from there, doesn't that exclude other possibilities? For example what if p and r are true but q is false? Maybe I'm reading this wrong. Is the left hand side equivalent to $p\land (q \implies r)$?
My second question is, I have a different way of solving this but I don't know how to write it like a proof. Basically if the left hand side is true then $p$ and $q \implies r$ have to be both true. By modus ponens, this makes the right hand side true (if p is true then $p \implies (q \implies r)$ is equivalent to $q \implies r$ , and since p is also true then $1\implies 1$ is true, q.e.d). How do I write this like in the example?
Best Answer
That is the definition for semantic entailment, or models, $\models$.
Syntactic entailment, or derives, $\vdash$, means that the the right-hand-side can be infered from the left-hand-side, using the syntactic proof system you are using (the diagram indicates that is a natural deduction system).
The other possibilities are irrelevant to the prof. You just want to show a derivation of $p\to(q\to r)$ from $(p\land q)\to r$ -- that the premise implies that an assumption of $p$ then of $q$ will let you infer $r$. To do that, you need not care what happens when you assume other things, such as $p$ and $\lnot q$.
(The map says if I turn left here and there I will get to where I want to go. Well the map does not say what happens if I turn right at either place, but really, I just want to know if following the map will get me where I want to go.)