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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))
| 261 views
+2
1. False

2. true
0
How did u evaluated it ?

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

by Active (2.1k points)
0
You cannot distribute for all quantifier over disjunction so then how can u take for all quantifier common ?
+1
Did u  really think before making such statement.....

http://nokyotsu.com/qscripts/2014/07/distribution-of-quantifiers-over-logic-connectives.html

The above solution is perfect .nothing wrong

And one more thing .Gate is near
0

∃xP(x)->∀x Q(x) ->∀x(P(x)->Q(x))

Anyone , Please give any example for the above logic in language.

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

by Active (2.3k points)

+1 vote
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+1 vote