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

[VinnyCee]

Homework Statement

Show that \left[\neg\,p\,\wedge\,\left(p\,\vee\,q\right)\right]\,\longrightarrow\,q is a tautology without using truth tables.

Homework Equations

DeMorgan's Laws, etc.

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?

 

[JonF]

there is an error in your first line

a -> b is logicaly equivlent to ~a or b