First Order Logic

Question

Can Anyone confirm this FOL : (∃xp(x) ∨ ∃xq(x)) ⟹ ∃x(p(x) ∨ q(x)) is valid or not ?

I think it is NOT.

Comments

∃ is distributive over ∨  as ∀ is distributive over ∧

so for you question , arrow will go bidirectional means LHS and RHS are completely equivalent.

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about such type of questions , we can directly check by taking some value

like take x=2 for P2 is true so now even don't care about Q

LHS : P2=True and Q=false for every x  T+F=T

RHS : we have such at least one x available ans that is x=2 for this P is T so RHS=true

simple so LHS=>RHS

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yes valid

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Not valid. it should be ∃x(p(x) ∨ q(x)) ⟹  (∃xp(x) ∨ ∃xq(x)).

 

 

Answer 1

Yes, it is valid.

∃ is distributive over V, means

(∃xp(x) ∨ ∃xq(x)) is equivalent to ∃x(p(x) ∨ q(x))

So, We have to check (A ⟹ A) is valid or not?

A ⟹ A 

~A ∨ A which is always 1 (Valid)

Answer 2

let x be the set of boys.

p(x)=set of fat boys.

q(x)=set of tall boys.

Now LHS: there are some boys which are fat and there are some boys which are tall

RHS: there exist some boys which are fat or tall.

so it is valid statement . because when LHS is true RHS will be always true. and when LHS is false  p->q generates True.

Answer 3

it is a valid argument because ∃ is distributive over U.