# Prove the logical equivalence $(p \land (q \lor (r ∧ \lnot s) \to v)) ∨ p \equiv p$

Can someone help me solve this?

I'm trying to solve this using the identities only, not truth tables.

In my tries, I came up that I use negation for $$(r \land \lnot s)$$, use distributive (but backwards) law for $$Q \lor (r \land \lnot s)$$ and use commutative afterwards?

Use the absorption identity: $$(A\wedge B)\vee A\equiv A$$
To proof a biconditional statement $$(A\leftrightarrow B)$$ it is sufficient to proof two conditional statements $$(A\rightarrow B)$$ and $$(B\rightarrow A)$$. In our case we need to prove: $$p\rightarrow((p\land(q\lor(r\land\neg s)\rightarrow v))\lor p)\\ ((p\land(q\lor(r\land\neg s)\rightarrow v))\lor p)\rightarrow p\\$$ For the first one start with assuming $$p$$. From $$p$$ you can infer $$(p\lor q)$$, where $$q$$ can be any compound formula, hence we can put $$(p\land(q\lor(r\land\neg s)\rightarrow v))$$ instead of it. We have $$((p\land(q\lor(r\land\neg s)\rightarrow v))\lor p)$$ now. After that use conditional proof on $$p$$ and you get to $$p\rightarrow((p\land(q\lor(r\land\neg s)\rightarrow v))\lor p)\\$$ For the second conditional assume $$((p\land(q\lor(r\land\neg s)\rightarrow v))\lor p)$$ then use distribution law to obtain $$((p\lor(q\lor(r\land\neg s)\rightarrow v))\land (p\lor p))$$ and since $$(p\lor p)\leftrightarrow p$$ you can rewrite it as$$((p\lor(q\lor(r\land\neg s)\rightarrow v))\land p)$$. Using AND elimination infer $$p$$, and by using conditional proof once again obtain $$((p\land(q\lor(r\land\neg s)\rightarrow v))\lor p)\rightarrow p\\$$