# Homework Help: Pv (Qv~P)

1. Oct 1, 2012

### lionely

1. The problem statement, all variables and given/known data
Show that Pv(Qv~P) is always true whatever the values of p and q.

Attempt

Pv(Qv~P)
(PvQ) v (Pv~P)
(PvQ) v T

2. Oct 2, 2012

### Staff: Mentor

Re: Pv(Qv~P)

The step above is incorrect. The logical disjunction operator (V) is associative, with P V (Q V R) <==> (P V Q) V R

3. Oct 2, 2012

### lionely

Re: Pv(Qv~P)

So.. it should it be P V (Q V ~P) = (~P V P) V Q?

4. Oct 2, 2012

### lionely

Re: Pv(Qv~P)

So is it T V Q?

5. Oct 2, 2012

### Staff: Mentor

Re: Pv(Qv~P)

Yes. Can you give reasons for each step? (There are a couple.)

That's not the final, simplified expression. What does that simplify to?

6. Oct 2, 2012

### lionely

Re: Pv(Qv~P)

Oh Umm because (~P V P) = T

hence T V Q

I think there is an indentity for P = T.. so p V q?

7. Oct 2, 2012

### Staff: Mentor

Re: Pv(Qv~P)

You're not done.

The part in red should have been a clue as to what you should conclude.

T V <whatever> ā T

Last edited by a moderator: Oct 3, 2012
8. Oct 3, 2012

### lionely

Re: Pv(Qv~P)

I don't see what's in red. I'm sorry :(

9. Oct 3, 2012

### lionely

Re: Pv(Qv~P)

but isn't it P V Q = T ? Since it's "OR" it's either one or the other or both?

10. Oct 3, 2012

### vela

Staff Emeritus
Re: Pv(Qv~P)

I fixed the tag in Mark's post, so you can see now what he intended.

No, look at the truth table for ā. If both P and Q are false, then PāQ is false.