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3 votes
3 votes
Which of the following are true?

∃x(P(x)->Q(x)) ->(∀xP(x)->∀xQ(x))

∃xP(x)->∀x Q(x) ->∀x(P(x)->Q(x))

2 Answers

7 votes
7 votes

HERE IS THE ANSWER:

1.∃x(P(x)->Q(x)) ->(∀xP(x)->∀xQ(x))
LHS:∃x(P(x)->Q(x))== ∃x(~P(x)vQ(x))

                                            ∃x~P(x)v∃xQ(x)

                                            ~∀xP(x)v∃xQ(x)which is ∀xP(x)->∃xQ(x); HENCE FALSE
2.∃xP(x)->∀x Q(x) ->∀x(P(x)->Q(x))

LHS:∃xP(x)->∀x Q(x)==~∃xP(x)v∀x Q(x) 

                                            ∀x(~P(x)v∀x Q(x) 

                                            ∀x(~P(x)v Q(x) ) TAKING ∀x common

                                           ∀x(P(x)->Q(x)):HENCE TRUE

0 votes
0 votes

1.) ∃x(P(x)->Q(x)) ->(∀xP(x)->∀xQ(x))

let P(x)= x is natural no

let Q(x)= x is odd no

let x is the set of natural no

to make above statement false

we should make RHS false and and LHS true

RHS is  (∀xP(x)->∀xQ(x)) is equivalent to (T--> F) is false

LHS is ∃x(P(x)->Q(x))  is equivalent to  true

so ∃x(P(x)->Q(x)) ->(∀xP(x)->∀xQ(x)) is false

2.) ∃xP(x)->∀x Q(x) ->∀x(P(x)->Q(x))

let P(x)= x is even no

let Q(x)= x is odd no

let x is set of natural no

RHS  is  ∀x(P(x)->Q(x)) equivalent to false

LHS is  ∃xP(x)->∀x Q(x) equivalent to (T-->F) is false

so ∃xP(x)->∀x Q(x) ->∀x(P(x)->Q(x)) is true

 not any case to make it false

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