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By Mark V. Lawson

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Another method is to use logical equivalences. Let A be a wff. First convert A to NNF and then if necessary use the distributive laws to convert to a wff which is in DNF. 5. We show how to convert ¬(p → (p ∧ q)) into DNF using a sequence of logical equivalences. The first step is to replace →. We use the fact that x → y ≡ ¬x∨y. This gives us ¬(¬p∨(p∧q)). Now use de Morgan’s laws to push negation inside the brackets. This yields ¬¬p ∧ ¬(p ∧ q) and then ¬¬p ∧ (¬p ∨ ¬q). We now apply double negation to get p ∧ (¬p ∨ ¬q).

This means that have have exactly one wff on the LHS of the semantic turnstile. Suppose that A1 , . . , An B is a valid argument and that it is not the case that A1 ∧. ∧An B is a valid argument. Then there is some assignment of truth values to the atoms that makes A1 ∧ . . ∧ An true and B false. But this means that each of A1 , . . , An is true and B is false that contradicts that we are given that A1 , . . , An B is a valid argument. Suppose that A1 ∧. ∧An B is a valid argument but that A1 , .

Observe that (p → q) → r ≡ p → (q → r) since the truth assignment p q r F F F makes the LHS equal to F but the RHS equal to T . Our next example is an application of some of our results. 9. We have defined a binary propositional connective ⊕ such that p ⊕ q is true when exactly one of p or q is true. Our goal now is to extend this to three atoms. Define xor(p, q, r) to be true when exactly one of p, q or r is true, and false in all other cases. We can describe this connective in terms of the ones already defined.

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An introduction to logic [Lecture notes] by Mark V. Lawson


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