Prove REPLACEMENT Theorem in Propositional Logic

[RyozKidz]

The book which i read for improving my logic sense~
There is a theorem called REPLACEMENT ..

( P \rightarrow Q ) \vee \neg ( P \rightarrow Q)
where (P\rightarrow Q) is the second occurence of ( P \rightarrow Q)

But what if the replace the second occurrence with \neg P\vee Q!
And i try to check with the truth table it does not gv me the values !
Help~~

 

[Elucidus]

The book which i read for improving my logic sense~
There is a theorem called REPLACEMENT ..

( P \rightarrow Q ) \vee \neg ( P \rightarrow Q)
where (P\rightarrow Q) is the second occurence of ( P \rightarrow Q)

But what if the replace the second occurrence with \neg P\vee Q!
And i try to check with the truth table it does not gv me the values !
Help~~

I am not absolutely sure what you mean by the truth table not giving you the values. Are you saying that

(P \rightarrow Q) \vee \neg (P \rightarrow Q) \text{ and } (P \rightarrow Q) \vee \neg (\neg P \vee Q)

are not appearing to be logically equivalent?

--Elucidus

 

[RyozKidz]

I am not absolutely sure what you mean by the truth table not giving you the values. Are you saying that

(P \rightarrow Q) \vee \neg (P \rightarrow Q) \text{ and } (P \rightarrow Q) \vee \neg (\neg P \vee Q)

are not appearing to be logically equivalent?

--Elucidus

yup yup~~coz i can't prove this is tvalidity by using truth table..~~

 

[Elucidus]

Here is the side-by-side truth table of the two expressions. The final values of each is in boldface.

\begin{array}{c|c|cccc|cccc}<br /> P &amp; Q &amp; (P \rightarrow Q) &amp; \vee &amp; \neg &amp; (P \rightarrow Q) &amp; (P \rightarrow Q) &amp; \vee &amp; \neg &amp; (\neg P \vee Q) \\<br /> \hline<br /> T &amp; T &amp; T &amp; \bold{T} &amp; F &amp; T &amp; T &amp; \bold{T} &amp; F &amp; T \\<br /> T &amp; F &amp; F &amp; \bold{T} &amp; T &amp; F &amp; F &amp; \bold{T} &amp; T &amp; F \\<br /> F &amp; T &amp; T &amp; \bold{T} &amp; F &amp; T &amp; T &amp; \bold{T} &amp; F &amp; T \\<br /> F &amp; F &amp; T &amp; \bold{T} &amp; F &amp; T &amp; T &amp; \bold{T} &amp; F &amp; T<br /> \end{array}

Additionally, any expression of the form (A \vee \neg A) (like the one on the left) is a tautology. Also the implication (P \rightarrow Q) is reducible to (\neg P \vee Q).

--Elucidus