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

Consider the following expressions:
(i) false
(ii) Q
(iii) true
(iv) P ∨ Q
(v) ¬Q ∨ P
The 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) 5


Answer: (C)

Explanation:  

anil_16_11_2



 

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 tautology
so if the above expression is a tautology
then 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.e
Premise of [ P ∧ (P ⇒ Q) ] ⇒ X i.e [ P ∧ (P ⇒ Q) ] is 0
meaning 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 that
for X = Q, True, P ∨ Q, ¬Q ∨ P the said expression would always be true
for X = False the expression would not be a tautology
Hence # 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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