Register

propositional logic

327 views

2 Answers

0 votes
0 votes
"Not everybody in the school is perfect"

A) NOT( ∀x(S(x)--->P(x) )
 ~∀x ~[S(x)------>P(x)]
   ∃x ~[~S(x) V P(x) ]
    ∃x [S(x) ^ ~P(x) ]
This is Equal to Option (B)
  
B) ∃x( S(x) ^ NOT (P(x)) )
So Answer is (C)
User Avatar
Voting & Accepting an answer helps improve the community
← PreviousNext →
← Previous in categoryNext in category →

Related questions

2 votes
2 votes
2 answers
2
kd..... asked Jul 12, 2018
535 views
Propositional logic
I am unable to prove following equations without using truth table 1) p -> (q v r) = (p->q) V (p->r) 2) ~(p <-> q) = p <-> ~q
I am unable to prove following equations without using truth table1) p - (q v r) = (p->q) V (p->r) 2) ~(p <- q) = p <- ~q
User Avatar
1 votes
1 votes
3 answers
3
Vicky rix asked Mar 7, 2017
903 views
propositional logic
Which of the following statements are ALWAYS TRUE ? A) ∀x [P(x)] ---> ∃x [P(x)] B) ∃x [P(x)] ---> ∀x [P(x)] C) Both A) and B) and so both are equivalent D) Neither A) nor B)
Which of the following statements are ALWAYS TRUE ?A) ∀x [P(x)] - ∃x [P(x)]B) ∃x [P(x)] - ∀x [P(x)]C) Both A) and B) and so both are equivalent D) Neither A) no...
User Avatar
0 votes
0 votes
2 answers
4
Vicky rix asked Mar 7, 2017
900 views
propositional logic
Which of the following propositional statements is TRUE ? A) ∀x ∀z ∃y [ P(x,y) ]---> ∃y ∀x ∀z [ P(x,y) ] B) ∃y ∀x ∀z [ P(x,y) ]---> ∀x ∀z ∃y [ P(x,y) ] C) Both A) and B) and so both are equivalent D) None of the above.
Which of the following propositional statements is TRUE ?A) ∀x ∀z ∃y [ P(x,y) ] - ∃y ∀x ∀z [ P(x,y) ]B) ∃y ∀x ∀z [ P(x,y) ] - ∀x ∀z ∃y [ P(x,y) ]C...
User Avatar
Link Copied!