3 votes 3 votes Can Anyone confirm this FOL : (∃xp(x) ∨ ∃xq(x)) ⟹ ∃x(p(x) ∨ q(x)) is valid or not ? I think it is NOT. Mathematical Logic mathematical-logic + – junaid ahmad asked Nov 22, 2017 junaid ahmad 607 views answer comment Share Follow See all 4 Comments See all 4 4 Comments reply Rupendra Choudhary commented Nov 22, 2017 reply Follow Share ∃ is distributive over ∨ as ∀ is distributive over ∧ so for you question , arrow will go bidirectional means LHS and RHS are completely equivalent. 2 votes 2 votes Rupendra Choudhary commented Nov 22, 2017 reply Follow Share 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 1 votes 1 votes srestha commented Nov 22, 2017 reply Follow Share yes valid 0 votes 0 votes chandan sahu commented Jan 14, 2019 reply Follow Share Not valid. it should be ∃x(p(x) ∨ q(x)) ⟹ (∃xp(x) ∨ ∃xq(x)). 0 votes 0 votes Please log in or register to add a comment.
Best answer 2 votes 2 votes 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) Rachit Agarwal answered Dec 3, 2017 • selected Dec 3, 2017 by joshi_nitish Rachit Agarwal comment Share Follow See all 0 reply Please log in or register to add a comment.
2 votes 2 votes 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. Red_devil answered Nov 22, 2017 Red_devil comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes it is a valid argument because ∃ is distributive over U. raviyogi answered Nov 22, 2017 raviyogi comment Share Follow See all 0 reply Please log in or register to add a comment.