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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)

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

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  • $\begingroup$ What rules you can use, and what you had tried, by the way it's a tautology $\dots$ you can use truth table method, or you want to prove it's a tautology use logical equivalence or natural deduction or any other valid methods $?$ $\endgroup$ – Manx Nov 22 '19 at 0:40
  • $\begingroup$ I just learned them and I tried so many times but I can't convert them because i am getting confused with so many negation out of parenthesis in demorgan step I don't need truth tables for example (¬p → q) → (q → ¬r) Propositional Logic Solution. ≡ ¬(¬p → q) ∨ (q → ¬r) ≡ ¬(p ∨ q) ∨ (¬q ∨ ¬r) ≡ (¬p ∧ ¬q) ∨ (¬q ∨ ¬r) ≡ (¬p ∨ ¬q ∨ ¬r) ∧ (¬q ∨ ¬r) $\endgroup$ – HellfireWorld Nov 22 '19 at 0:53
  • $\begingroup$ I see $\dots$ so you want to prove it with logical equivalence, did you missed an 'and' in your expression:$$ ¬(((a→b)→a)→a)\color{orange}\land¬((p→(q→r)))→((p→q)→(p→r))$$ $\endgroup$ – Manx Nov 22 '19 at 0:58
  • $\begingroup$ Sorry for my bad syntax I am new here... they are 2 different exercises $\endgroup$ – HellfireWorld Nov 22 '19 at 0:59
  • $\begingroup$ Should I wait? I would be really thankful if you could help $\endgroup$ – HellfireWorld Nov 22 '19 at 1:26
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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

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  • $\begingroup$ really thank you, I will try to understand it now because I really messed up with the negations and parenthesis $\endgroup$ – HellfireWorld Nov 22 '19 at 1:43
  • $\begingroup$ Manx correct me if I am wrong in the second exercise the 4rth equals its Associative law, after distributive? also I haven't done the negation law and identity law $\endgroup$ – HellfireWorld Nov 22 '19 at 11:17
  • $\begingroup$ @HellfireWorld you are right, the distributive law is a typo, was trying to copy the previous line, but forgot to change the tag, thanks $\endgroup$ – Manx Nov 22 '19 at 11:20
  • $\begingroup$ nice they haven't learned me the identity law and negation law with ⊤ ⊥ is it correct to skip those steps in the second ? and in the first one should i let it ≡(¬b∨a)∧(a∧¬a) $\endgroup$ – HellfireWorld Nov 22 '19 at 11:23
  • $\begingroup$ @HellfireWorld If you have rules like $(p\lor \neg p)\land q\equiv q$, you can use this instead of the Identity laws, but $\top$ simply stand for true and $\bot$ for false, and $(¬b∨a)∧(a∧¬a)$ is in CNF form, but not in minimal CNF form, since it can still be simplify further. $\endgroup$ – Manx Nov 22 '19 at 11:28

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