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Running on Zero

😃: Show that ∀xP (x) ∨ ∀xQ(x) and ∀x(P (x) ∨ Q(x)) are
not logically equivalent.
🤖: Sure! Here are some ways to show that each conditional statement in Exercise 9 is a tautology without using truth tables:

Exercise 9:

1. ∀xP(x) ∨ ∀xQ(x)

Show that it is a tautology:

• By definition, a tautology is a sentence that is always true, no matter what the values of its variables are.
• In this case, the variable x represents anything, and the predicates P and Q could represent any properties or characteristics.
• Since either P or Q (or both) must be true for any value of x, the negation of the antecedent ("it is not the case that neither P nor Q is true") is always false, making the conclusion true. Therefore, ∀xP(x) ∨ ∀xQ(x) is a tautology.

2. ∀x(P(x) ∨ Q(x))

Show that it is a tautology:

• By definition, a tautology is a sentence that is always true, no matter what the values of its variables are.
• In this case, the variable x represents anything, and the predicate P or Q could represent any property or characteristic.
• Since the antecedent ("for all x, either P(x) or Q(x) is true") is always true, the consequent ("therefore, for all x, P(x) ∨ Q(x) is true") is also always true. Therefore, ∀x(P(x) ∨ Q(x)) is a tautology.

I hope these explanations help! Let me know if you have any questions.