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 occurrence 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 occurrence 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.

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

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