# Show that the conditional statement is a Tautology without using truth tables...

by VinnyCee
Tags: conditional, statement, tables, tautology, truth
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 P: 492 1. The problem statement, all variables and given/known data Show that $$\left[\neg\,p\,\wedge\,\left(p\,\vee\,q\right)\right]\,\longrightarrow\,q$$ is a tautology without using truth tables. 2. Relevant equations DeMorgan's Laws, etc. 3. The attempt at a solution $$\left[\neg\,p\,\wedge\,\left(p\,\vee\,q\right)\right]\,\longrightarrow\,q$$ by. EX 3 (see EX 8) $$\left[\neg\,p\,\wedge\,\left(p\,\vee\,q\right)\right]\,\vee\,q$$ $$\left[p\,\wedge\,\neg\,\left(p\,\vee\,q\right)\right]\,\vee\,q$$ $$\left[p\,\wedge\,\left(\neg\,p\,\wedge\,\neg\,q\right)\right]\,\vee\,q$$ $$\left[\left(p\,\wedge\,\neg\,p\right)\,\wedge\,\left(p\,\wedge\,\neg\,q\right )\right]\,\vee\,q$$ $$\left[F\,\wedge\,\left(p\,\wedge\,\neg\,q\right)\right]\,\vee\,q$$ Now what?
 P: 617 there is an error in your first line a -> b is logicaly equivlent to ~a or b

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