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3 Answers

2 votes
2 votes
I feel Option D) is the answer as
           A) fails when x = {} (empty set) , though ∀x [P(x)] is TRUE ,  ∃x [P(x)] is FALSE .
           B) is a trivial case and it doesnot hold
2 votes
2 votes
A) ∀x [P(x)] ---> ∃x [P(x)]
    It Convert to digital logic expression

P1.P2 -------> P1 + P2

You can not find 1--------->0

I mean If P1=1, P2=1 then P1.P2 =1 (true) otherwise 0 (false),So P1 + P2 =1 +1 =1

So Option (A) is True always

But

B) ∃x [P(x)] ---> ∀x [P(x)]

It Convert to digital logic expression

P1 + P2 ------> P1.P2

You can easily find 1------------>0

I mean If P1=1 ,P2=0 or P1=0 ,P2=1 then P1 + P2 =1(true) otherwise 0(false) So P1.P2 = 1.0 = 0.1 = 0

So this is Not true always
0 votes
0 votes

I think the answer should be A..however do correct me if I m wrong..

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