How to Deny Universal and Existential Quantifiers in Logic?

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Homework Statement


Let U be universe under consideration, let P(x) and Q(x) be predicate with free variable x. Find a useful denial.

1. (∀x∈U)(Q(x)∨P(x))
2. (∃x∈U)(Q(x)∧P(x)). Use implication afterwards.

Homework Equations

The Attempt at a Solution


[/B]
My answer for 1 is:

(∃x∈U)(¬Q(x)∧¬P(x))

My answer for 2 is:

(∀x∈U)(¬Q(x)∨¬P(x)), and the implication would be if Q is true then P is false.

I am unsure of if the implication is correct as I am new to this.

Are my answers correct?

Thank you.
 
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ver_mathstats said:

Homework Statement


Let U be universe under consideration, let P(x) and Q(x) be predicate with free variable x. Find a useful denial.

1. (∀x∈U)(Q(x)∨P(x))
2. (∃x∈U)(Q(x)∧P(x)). Use implication afterwards.

Homework Equations

The Attempt at a Solution


[/B]
My answer for 1 is:

(∃x∈U)(¬Q(x)∧¬P(x))

My answer for 2 is:

(∀x∈U)(¬Q(x)∨¬P(x)), and the implication would be if Q is true then P is false.

I am unsure of if the implication is correct as I am new to this.

Are my answers correct?

Thank you.
Yes, this is correct. If Q is true, then P has to be false and vice versa: If P is true, then Q has to be false. They still can both be false.
 
fresh_42 said:
Yes, this is correct. If Q is true, then P has to be false and vice versa: If P is true, then Q has to be false. They still can both be false.
Thank you for checking it over.