How to prove this theorem $(v\lor s)\land(v\rightarrow p)\land(s\rightarrow a)\land\lnot a\vdash p$

Question

I'm noob to Discrete Math, but I need to prove this $(v\lor s)\land(v\rightarrow p)\land(s\rightarrow a)\land\lnot a\vdash p$

If you can explain what shall I do to prove it.

I can create truth table for left part of formula $(v\lor s)\land(v\rightarrow p)\land(s\rightarrow a)\land\lnot a$, but

I don't completely understand what shall I do with right part of formula I mean $\vdash p$

Please explain what I need to prove: maybe equivalence or tautology or something other or maybe I need to compare left side of formula $(v\lor s)\land(v\rightarrow p)\land(s\rightarrow a)\land\lnot a$ and right side $\vdash p$

Answer 1

Suppose

$$(v\lor s)\land(v\rightarrow p)\land(s\rightarrow a)\land\lnot a.\tag{1}$$

Then from $(1)$,

$$\begin{align} v\lor s,\tag{2}\\ v\rightarrow p,\tag{3}\\ s\rightarrow a,\tag{4}\\ \lnot a\tag{5} \end{align}$$

From $(4)$, we have $\lnot s$ or $a$. The latter contradicts $(5)$, so suppose the former. Now $(2)$ gives $v$ or $s$, but $\lnot s$ holds from before, so we have $v$. But now $v$ implies $p$ from $(3)$.

Hence

$$\vdash((v\lor s)\land(v\rightarrow p)\land(s\rightarrow a)\land\lnot a)\to p,$$

but this holds if and only if

$$(v\lor s)\land(v\rightarrow p)\land(s\rightarrow a)\land\lnot a\vdash p.\,\square $$