Asked by Kid
Suppose the domain of the propositional function P(x, y) consists of pairs x and y, where x = 1, 2, or 3, and y = 1, 2, or 3. Write out the propositions below using disjunctions and conjunctions only.
∃x∀y ¬P(x, y)
The above is equivalent to ¬(∀x∃yP(x,y))
So can I just write (∀x∃yP(x,y)) using disjunctions and conjunctions only and then adjust it with the negation.
¬(∀x∃yP(x,y))
≡ ¬((P(1,1) ∨ P(1,2) ∨ P(1,3)) ∧ (P(2,1) ∨ P(2,2) ∨ P(2,3)) ∧ (P(3,1) ∨ P(3,2) ∨ P(3,3)))
≡ ¬(P(1,1) ∨ P(1,2) ∨ P(1,3)) ∨ ¬(P(2,1) ∨ P(2,2) ∨ P(2,3)) ∨ ¬(P(3,1) ∨ P(3,2) ∨ P(3,3))
≡ (¬P(1,1) ∧ ¬P(1,2) ∧ ¬P(1,3)) ∨ (¬P(2,1) ∧ ¬P(2,2) ∧ ¬P(2,3)) ∨ (¬P(3,1) ∧ ¬P(3,2) ∧ ¬P(3,3))
Is that a valid method and correct answer?
∃x∀y ¬P(x, y)
The above is equivalent to ¬(∀x∃yP(x,y))
So can I just write (∀x∃yP(x,y)) using disjunctions and conjunctions only and then adjust it with the negation.
¬(∀x∃yP(x,y))
≡ ¬((P(1,1) ∨ P(1,2) ∨ P(1,3)) ∧ (P(2,1) ∨ P(2,2) ∨ P(2,3)) ∧ (P(3,1) ∨ P(3,2) ∨ P(3,3)))
≡ ¬(P(1,1) ∨ P(1,2) ∨ P(1,3)) ∨ ¬(P(2,1) ∨ P(2,2) ∨ P(2,3)) ∨ ¬(P(3,1) ∨ P(3,2) ∨ P(3,3))
≡ (¬P(1,1) ∧ ¬P(1,2) ∧ ¬P(1,3)) ∨ (¬P(2,1) ∧ ¬P(2,2) ∧ ¬P(2,3)) ∨ (¬P(3,1) ∧ ¬P(3,2) ∧ ¬P(3,3))
Is that a valid method and correct answer?
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