Which of the following has a definite truth value?
AThe predicate P(x): x > 5, with x a free variable
B∀x, x > 5, quantifying over all integers
C∃x, x² > 0, quantifying over all integers
DBoth B and C
A predicate with a free variable like 'x > 5' is neither true nor false — it's waiting for a specific value of x. Once a quantifier binds the variable, it becomes a statement with a definite truth value. ∀x, x > 5 over the integers is false (x = 3 is a counterexample). ∃x, x² > 0 is true (x = 1 works). Both B and C have definite truth values; the unquantified predicate in A does not.
Question 2 Multiple Choice
Consider both statements over the integers: (A) ∀x ∃y, y > x and (B) ∃y ∀x, y > x. Which are true?
ABoth A and B are true
BBoth A and B are false
COnly A is true — for any integer x there is always a larger y, but no single integer exceeds all integers
DOnly B is true — the largest integer exceeds all others
∀x ∃y, y > x says: for every x I give you, you can find some y larger than it — just take y = x + 1. This is true. ∃y ∀x, y > x says: there is one fixed y that is simultaneously larger than every integer. This is false — no integer is larger than all integers. The crucial difference is that in A, y can depend on x; in B, one y must work for all x at once. Swapping the quantifiers changes the meaning entirely.
Question 3 True / False
The expression 'x > 5' has no truth value until a specific value is substituted for x or a quantifier binds the variable.
TTrue
FFalse
Answer: True
This is the fundamental distinction between predicates and statements. 'x > 5' is a predicate — a sentence with a free variable. Without knowing what x is, the expression cannot be evaluated as true or false. It becomes a statement only through substitution (plug in x = 7 → true, x = 3 → false) or through quantification (∀x, x > 5 → false; ∃x, x > 5 → true). Treating a predicate as if it had a truth value is a common and consequential error in proof-writing.
Question 4 True / False
The negation of ∀x, P(x) is ∀x, ¬P(x) — to deny that most x have property P, we say most x lack property P.
TTrue
FFalse
Answer: False
The negation of ∀x, P(x) is ∃x, ¬P(x) — there exists at least one x that lacks property P. To disprove a universal claim, you only need one counterexample. Saying 'all x lack property P' (∀x, ¬P(x)) is a much stronger claim that asserts a universal negative rather than merely denying the universal positive. This negation rule is one of the most important in proof-writing.
Question 5 Short Answer
Explain why ∀x ∃y, y > x and ∃y ∀x, y > x make different claims, using the integers as your domain.
Think about your answer, then reveal below.
Model answer: The first says: for each x you choose, I can find some y larger than it — y is allowed to depend on x (e.g., y = x + 1 always works). The second says: there is one fixed y that is larger than every x simultaneously. No such integer exists — for any candidate y, the integer y itself is not less than y. So A is true and B is false.
This captures the key insight about quantifier order: ∀∃ means 'for each input, a suitable witness exists' (the witness can be tailored to the input); ∃∀ means 'one universal witness exists for all inputs simultaneously.' These are profoundly different claims. The same swap appears in analysis: the definition of pointwise continuity (∀ε ∀x ∃δ) versus uniform continuity (∀ε ∃δ ∀x) — the order of ∃δ and ∀x changes the meaning entirely.