First Let us simplify :
P˄(P->Q)
=P˄(~P˅Q)
=(P˄~P)˅(P˄Q)
=P˄Q
To check whether an expression A logically implies B , we make A → B a tautology , because logic implication means my conclusion follows from my arguments, so arguments → conclusion is a tautology
Now , let us check whether (P˄Q) → option(x) is Tautology or not .
To check whether an implication is Tautology or not we will use Two Approaches :
Approach 1 : Make Premise or LHS of implication True and then try to make Conclusion or RHS of implication False .
“ If we are able to do so then it is not Tautology “
“ I we are not able to do so , then it is Tautology “ .
Approach 2 :
Make Conclusion or RHS of implication False and then try to make Premise or LHS of implication True .
“ If we are able to do so then it is not Tautology “
“ I we are not able to do so , then it is Tautology “ .
Option (i) : P˄Q → False
Using Approach 2 ,
RHS = False , Now let us try to make LHS = True
P˄Q = True , Taking P = T and Q = T we are able to to make P˄Q = True .
Since , LHS =True and RHS = False simultaneously , it is not Tautology.
Option (ii) : P˄Q → Q
Using Approach 2 ,
RHS = Q = False . Now Lets us try to make LHS = True .
LHS = P˄Q . Since Q is False we will not be able to make LHS True.
Hence , Tautology.
Option (iii) : P˄Q → TRUE
We know that , Anything → True = True .
So this is Tautology .
Option (iv) : P˄Q → P˅Q
Using Approach 2 ,
RHS = P˅Q = False . This will only happen when P = False and Q = False . Now let us make LHS = True .
LHS = P˄Q . Since P = F and Q =F , we cannot make LHS True .
Hence , Tautology.
Option (v) : P˄Q→ ~Q˅P
Using Approach 1 , LHS = P˄Q= True . This will only happen when P =T and Q =T . Now let us make RHS = False .
RHS = ~Q˅P . Since P = True we cannot make RHS false.
Hence , Tautology .
So , 4 is the answer .