∃!xP(x) ->∃xP(x) and
∃!x`P(x) ->`∀xP(x),
(where the notation ∃!xP(x)
denotes the proposition “There
exists a unique x such that P(x) is true”) are :
(A) True and False
(B) False and True
(C) False and False
(D) True and True
Answer D
Statement 1.There exists a
unique x,where p(x) is true. Implies that there exists a x where p(x) is true.
Statement 2. There exists a unique x where p(x) is false.implies that ‘not
for all x. p(x) is true.’
Statement 2 is true. Where as statement 1 is also true.
Answer D.
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