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Verify is tautology by using logical equivalence:

$$((p ∨ q) ∧ (p → r) ∧ (∼ r)) → q$$

by showing what laws are using, step by step explanation needed.

Logical equivalence: de morgan's laws, commutativity, associativity, double negative, distribution, idempotent, absorption, identity, universal bound, negation

Rules of inference: modus ponens, modus tollens, generalisation, specialisation, conjunction, ellimination, transivity, proof by division into cases, contradiction

What I got so far:

I need to prove

$$((p ∨ q) ∧ (p → r) ∧ (∼ r)) → q ≡ \text{tautology}$$ Since $p → r$ same as $∼ p ∨ r$: $$((p ∨ q) ∧ (p → r) ∧ (∼ r)) → q ≡ ((p ∨ q) ∧ (∼ p ∨ r ) ∧ (∼ r)) → q$$

I am stuck here

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  • $\begingroup$ Please include your work or at the very least what does one have to do to verify something is a tautology? For instance, I can tell that the premise tautologically implies $q$ without making use of the equivalences. $\endgroup$ – Alvin Lepik Aug 12 '18 at 15:29
  • $\begingroup$ @AlvinLepik please check my solution I got so far. $\endgroup$ – icefish Aug 12 '18 at 17:52
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Proceed by $$(p\lor q) \land (\neg p\lor r)\land\neg r \rightarrow q \equiv \neg (p\lor q) \lor \neg (\neg p\lor r)\lor r \lor q\\ \equiv (\neg p\land\neg q) \lor (p\land\neg r)\lor r\lor q $$ View the latter formula grouped as follows $$\left [(\neg p\land \neg q) \lor q\right ] \lor \left [(p\land\neg r)\lor r\right ]\equiv $$ Use distributivity $$\equiv \left [(\neg p\lor q) \land (\neg q\lor q)\right ] \lor \left [(p\lor r)\land (\neg r\lor r)\right ] \equiv \neg p\lor q \lor p\lor r \equiv t $$

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Next use absorption, since $~ (\neg p\vee r)\land \lnot r \equiv \neg p$

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