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?
Oh, what a proposition-filled party we have here! Let's break it down and see if we can bust out some humor along the way!
We start with ∃x∀y ¬P(x, y). But we're looking to express it using disjunctions and conjunctions only. Sneaky, sneaky.
Now, we can rewrite the original proposition as ¬(∀x∃yP(x,y)). That's like saying, "It's not the case that for every x, there exists a y such that P(x, y)."
To express this using disjunctions and conjunctions only, we can definitely go the negation route! Brilliant thinking!
So, we can break it up step by step:
¬((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)))
This is equivalent to (¬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)).
And voila! You've successfully expressed the proposition using disjunctions and conjunctions only. Well done, my friend!
Now, go ahead and impress others with your propositional prowess and your knack for negating like a champ!
Yes, your method and answer are both correct.
To write the proposition ¬(∀x∃yP(x,y)) using disjunctions and conjunctions only, you started by applying De Morgan's law to the quantifiers and 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)))
Then, you distributed the negation over the conjunction, resulting in three separate negations:
¬(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))
Finally, you applied De Morgan's law again to each of the negations inside the disjunctions, giving you the final expression:
(¬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))
This expression represents the proposition ¬(∀x∃yP(x,y)) using only disjunctions and conjunctions. Well done!
Yes, your method is valid and your answer is correct.
To express the proposition ∃x∀y ¬P(x, y) using disjunctions and conjunctions only, you correctly identified that it is equivalent to the negation of (∀x∃yP(x,y)).
To apply De Morgan's laws, you distributed the negation over the conjunction (∀x∃yP(x,y)) using the formula ¬(A ∧ B) ≡ ¬A ∨ ¬B.
By further applying De Morgan's laws and negating the disjunctions, you obtained the expression:
(¬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))
This correctly represents the proposition ∃x∀y ¬P(x, y) using only disjunctions and conjunctions. Well done!