GATE | GATE-CS-2016 (Set 2) | Question 11 - GeeksforGeeks

11 Oct, 2021
Consider the following expressions:(i) false(ii) Q(iii) true(iv) P ∨ Q(v) ¬Q ∨ PThe number of expressions given above that are logically implied by P ∧ (P ⇒ Q) is ______________
[This Question was originally a Fill-in-the-blanks Question](A) 2(B) 3(C) 4(D) 5Answer: (C)Explanation:  
 
This solution is contributed by Anil Saikrishna Devarasetty.
 
Alternate Explanation :Answer is 4. Here is the solution
If say X is ‘Logically Implied’ by [ P ∧ (P ⇒ Q) ] then[ P ∧ (P ⇒ Q) ] ⇒ X is always true i.e it is a tautologyso if the above expression is a tautologythen we can say that X is logically implied by P ∧ (P ⇒ Q)
So we need to find X for which [ P ∧ (P ⇒ Q) ] ⇒ X will be always true for all values of P, Q and X.Look at the below table
P....Q...(P ⇒ Q)...[P ∧ (P ⇒ Q)].......X.......[ P ∧ (P ⇒ Q) ] ⇒ X
0....0.....1............0.............1/0............1......
0....1.....1............0.............1/0....  ......1......
1....0.....0..... ......0.............1/0............1......
1....1.....1............1..............1.............1.......

notice that value of X doesn’t matter if premise of expression i.ePremise of [ P ∧ (P ⇒ Q) ] ⇒ X i.e [ P ∧ (P ⇒ Q) ] is 0meaning the final expression would be a tautology for all values of X if [ P ∧ (P ⇒ Q) ] is 0
but if premise is 1 (as in last row) then X must be 1 so that the final implication i.e., [ P ∧ (P ⇒ Q) ] ⇒ X is true for all values.
if you replace X by all 5 options then you will find thatfor X = Q, True, P ∨ Q, ¬Q ∨ P the said expression would always be truefor X = False the expression would not be a tautologyHence # of expression is 4——————————————————————
Note: 
An important inference rule called "modus ponenes" 
says this [ P ∧ (P ⇒ Q) ] ⇒ Q is a tautology
we noted that if we replace X by Q then it is 
indeed a tautology meaning Q is implied by 
[ P ∧ (P ⇒ Q) ] 
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