Mar 7, 2016 · So far we have shown that $ (P \to Q) ~\vdash (\neg P \lor Q)$. To finish proving the equivalency $ P \to Q \equiv \neg P \lor Q ~$ we also need to show $ (\neg P \lor Q) \vdash (\neg P \lor Q) $. I don't see an obvious inference rule at this point, but we could show it …

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Jan 14, 2016 · Start with the given statement, $$ [p \land (p \rightarrow q)] \rightarrow q.$$ As you noticed, from the first logical equivalence in Table 7, you can replace the part in the round brackets to get the equivalent statement

Aug 14, 2018 · Prove ~P ˅ Q entails P → Q, by assuming P and demonstrating that eliminating the disjunction will derive Q by means of explosion (P,~P ├ Q) and reiteration (P, Q ├ Q). Prove the converse, that P → Q entails ~P ˅ Q, either by (1) excluding the middle and introducing an appropriate disjunctive in each case, or (2) reducing to absurdity ...

Jan 19, 2020 · Given the premises p→q and ¬p→¬q, prove that p is logically equivalent to q. I understand why this works, but I do not know how to construct a complete formal proof. So far, I have this: premises: p→q ¬p→¬q (p→q) ∧ (¬p→¬q) ⇔ T apply contrapositive law (p→q) ∧ (q→p) ⇔ T ∴ (p↔q) ⇔ T

Oct 16, 2020 · And the idea is that I am trying to prove ~(P ^ Q) -> (~P v ~Q) so that I can use the biconditional rule to end up with (¬P ∨ ¬Q) ↔ ¬(P ∧ Q). However, I am stuck on this and I do not know how to start this other than maybe starting with another subproof startinga s ~(P ^ Q) but then I am stuck there again. Or maybe I started wrong.

Jun 15, 2019 · (p ∨ ¬q) ∧ (¬p ∨ ¬q) Looking at an equivalency table I did, it seems p ∨ ¬q gives the same results as (p ∨ ¬q) ∧ (¬p ∨ ¬q). However I'm not sure how you would deduce this without the table, as in, if I was outrightly asked to write the above in simpler terms I wouldn't know where to begin. Am I understanding this correctly?

Oct 20, 2016 · ∼(p ∨∼q) ∨ (∼p ^ ~ q) ≡ ~p. Please help I don't know where to start. These are the laws I need to list in each step when simplifying. Commutative laws: p ∧ q ≡ q ∧ p p ∨ q ≡ q ∨ p. Associative laws: (p ∧ q) ∧ r ≡ p ∧ (q ∧ r) (p ∨ q) ∨ r ≡ p ∨ (q …

Feb 12, 2017 · Once again, we see that 'P only if Q' has the same meaning as 'If P then Q'. And (once again) because we translate 'If P then Q' as 'P-→ Q', it follows that we translate 'P only if Q' as 'P-→ Q'. If you're still unconvinced then it's time to beat a dead horse.... Notice first that the sentence 'If P then Q' amounts to the following two claims:

Apr 16, 2019 · $(p→q)∧(p→r) $ is the same as $(\overline{p} \vee q)\wedge (\overline{p} \vee r)$ which is the same as $(\overline{p}\vee(\overline{p}\wedge r)\vee(q\wedge\overline{p})\vee(q\wedge r))$ From here, it is clear that if both $\overline{p}$ and $(q\wedge r)$ is false, the complete statement is false. If either is true, then the full statement ...