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

1. Sep 10, 2007

### VinnyCee

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?

2. Sep 10, 2007

### JonF

there is an error in your first line

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