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

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 $$\vdash p$$

• Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. Commented Apr 26, 2023 at 11:08
• I've never seen the notation $\vdash p$, is that the same as $\rightarrow p$? Commented Apr 26, 2023 at 11:20
• $\Gamma \vdash p$ in the context of proof theory means that there exists a proof of $p$ from the set of assumptions $\Gamma$, where a proof is meant in the syntactic way, loosely: a sequence of valid rules of deduction. Commented Apr 26, 2023 at 11:44
• Which proof system are you allowed to use?
– hff1
Commented Apr 26, 2023 at 12:52
• Commented Apr 26, 2023 at 13:01

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

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

• Thank you so much for help. I'll be trying to sort all of this. I don't know how to like you or say BIG THANK YOU to increase your status. Commented Apr 26, 2023 at 13:10
• You're welcome, @NadineSk. I'm glad I could help. Commented Apr 26, 2023 at 13:30