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? 
 A: Use the absorption identity: $(A\wedge B)\vee A\equiv A$
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
$$
((p\land(q\lor(r\land\neg s)\rightarrow v))\lor p)\rightarrow p\\
$$
