Convert to Conjunctive Normal Form exercise

Question

I got confused in some exercises I need to convert the following to CNF step by step(I need to prove it with logical equivalence)

$1.¬(((a→b)→a)→a)$
$2.¬((p→(q→r)))→((p→q)→(p→r))$

Answer 1

Use Logical equivalences we have: \begin{align} &¬(((a→b)→a)→a)\\ &\equiv\neg(\neg(\neg(\neg a \lor b)\lor a)\lor a)\tag*{Conditional equivalence}\\ &\equiv((a \land\neg b)\lor a)\land\neg a\tag*{De Morgan's law}\\ &\equiv((a\lor a) \land(\neg b\lor a))\land\neg a\tag*{Distributive law}\\ &\equiv(a \land(\neg b\lor a))\land\neg a\tag*{Idempotent law}\\ &\equiv((\neg b\lor a)\land a)\land\neg a\tag*{Commutative law}\\ &\equiv(\neg b\lor a)\land(a\land\neg a)\tag*{Associative law}\\ &\equiv(\neg b\lor a)\land\bot\tag*{Negation law}\\ &\equiv\bot\tag*{Identity law}\\ \\ &¬((p→(q→r)))→((p→q)→(p→r))\\ &\equiv(\neg p\lor(\neg q\lor r))\tag*{Conditional equivalence}\\ &\lor(\neg(\neg p\lor q)\lor(\neg p\lor r))\\ &\equiv(\neg p\lor(\neg q\lor r))\tag*{De Morgan's law}\\ &\lor((p\land\neg q)\lor(\neg p\lor r))\\ &\equiv(\neg p\lor(\neg q\lor r))\tag*{Distributive law}\\ &\lor((p\lor(\neg p\lor r))\land(\neg q\lor(\neg p\lor r)))\\ &\equiv(\neg p\lor(\neg q\lor r))\tag*{Associative law}\\ &\lor(((p\lor\neg p)\lor r)\land(\neg q\lor(\neg p\lor r)))\\ &\equiv(\neg p\lor(\neg q\lor r))\tag*{ Negation law}\\ &\lor((\top\lor r)\land(\neg q\lor(\neg p\lor r)))\\ &\equiv(\neg p\lor(\neg q\lor r))\tag*{Domination law}\\ &\lor(\top\land(\neg q\lor(\neg p\lor r)))\\ &\equiv(\neg p\lor(\neg q\lor r))\lor(\neg q\lor(\neg p\lor r))\tag*{Identity law}\\ &\equiv((\neg p\lor\neg q)\lor r)\lor((\neg q\lor\neg p)\lor r)\tag*{Associative law}\\ &\equiv((\neg p\lor\neg q)\lor r)\lor((\neg p\lor\neg q)\lor r)\tag*{Commutative law}\\ &\equiv(\neg p\lor\neg q)\lor r\tag*{Idempotent law}\\ \end{align}

Hence $(1)$ has $\bot$ as its minimal CNF & DNF, and $(2)$ has $(\neg p\lor\neg q)\lor r$ as its minimal CNF & DNF