- #1
lionely
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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
The step above is incorrect. The logical disjunction operator (V) is associative, with P V (Q V R) <==> (P V Q) V Rlionely said: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)
lionely said:(PvQ) v T
Yes. Can you give reasons for each step? (There are a couple.)lionely said:So.. it should it be P V (Q V ~P) = (~P V P) V Q?
That's not the final, simplified expression. What does that simplify to?lionely said:So is it T V Q?
You're not done.lionely said:Oh Umm because (~P V P) = T
hence T V Q
Show that Pv(Qv~P) is always true whatever the values of p and q.
lionely said:I think there is an indentity for P = T.. so p V q?
I fixed the tag in Mark's post, so you can see now what he intended.lionely said:I don't see what's in red. I'm sorry :(
No, look at the truth table for ⋁. If both P and Q are false, then P⋁Q is false.lionely said:but isn't it P V Q = T ? Since it's "OR" it's either one or the other or both?
The statement "Pv(Qv~P)" means "P or (Q or not P)". This is a logical statement that can be read as "P or Q or not P".
One way to prove that "Pv(Qv~P)" is always true is by using a truth table. This table will show that the statement is true for all possible combinations of the truth values of P and Q.
Yes, for example, if P is true and Q is false, the statement would be "true or (false or not true)", which would simplify to "true or (false or false)", and finally to "true". This shows that the statement is always true, regardless of the values of P and Q.
The significance of proving that "Pv(Qv~P)" is always true is that it demonstrates a fundamental logical principle, known as the law of excluded middle. This law states that a statement is either true or false, and there is no in-between. This proof also shows that the statement "Pv(Qv~P)" is a tautology, meaning it is always true.
The truth value of "Pv(Qv~P)" does not change regardless of the values of P and Q. This is because the statement is always true, as proven by the truth table. Even if the values of P and Q are switched or both changed, the statement will still evaluate to true.