Show that Pv(Qv~P) is always true whatever the values of p and q

[lionely]

Homework Statement

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

 

[Mark44]

Homework Statement

Show that Pv(Qv~P) is always true whatever the values of p and q.


Attempt

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

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

(PvQ) v T

 

[lionely]

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

 

[lionely]

So is it T V Q?

 

[Mark44]

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

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

So is it T V Q?

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

 

[lionely]

Oh Umm because (~P V P) = T

hence T V Q

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

 

[Mark44]

Oh Umm because (~P V P) = T

hence T V Q

You're not done.

Show that Pv(Qv~P) is always true[/color] whatever the values of p and q.

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


T V <whatever> ⇔ T

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

 

[lionely]

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

 

[lionely]

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

 

[vela]

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

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

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

No, look at the truth table for ⋁. If both P and Q are false, then P⋁Q is false.