Mixed Quantifier Practice
A free Logic lesson from the “Quantifiers and Predicates” unit, with a worked example and practice problems including step-by-step solutions.
Translate all A are B as ∀x(A(x)→B(x)) and some A are B as ∃x(A(x)∧B(x)). Negation switches every to at least one and vice versa while negating the inner claim. Quantifier order matters: ∀x∃y permits y to depend on x; ∃y∀x requires one fixed y to work for every x.
What you'll learn
- Compare all, some, none, and not all claims
- Explain the idea in plain English before using symbols
- Use examples, non-examples, or counterexamples to check the reasoning
Worked example
Problem. Let S(x) mean student and R(x) mean registered, with all people as the domain. Translate “Every student is registered.”
- Only students must satisfy registration.
- An implication expresses that restriction without saying everyone is a student.
Answer: ∀x(S(x) → R(x))
Lesson preview: this page shows 5 practice items. The interactive activity contains 6 practice items and 6 quiz items (multiple choice). Counts include repeated prompts and variants.
Practice problems
1. Let S(x) mean student and R(x) mean registered, with all people as the domain. Translate “Every student is registered.”
Choices: ∀x(R(x) → S(x)) · ∀x(S(x) → R(x)) · ∀x(S(x) ∧ R(x)) · ∃x(S(x) → R(x))
Show solution
- Only students must satisfy registration.
- An implication expresses that restriction without saying everyone is a student.
Answer: ∀x(S(x) → R(x))
2. In the same domain, translate “Some student is registered.”
Choices: ∃x(S(x) ∧ R(x)) · ∃x(S(x) → R(x)) · ∀x(S(x) ∧ R(x)) · ∀x(S(x) → R(x))
Show solution
- One witness must satisfy both properties.
- An implication could be true merely because the person is not a student.
Answer: ∃x(S(x) ∧ R(x))
3. Negate “Every student solved at least one problem.”
Choices: No student solved any problem · Every student missed a problem · At least one problem was solved by everyone · At least one student solved no problems
Show solution
- Negate the outer every to at least one.
- Negate solved at least one to solved none.
Answer: At least one student solved no problems
4. Over the integers, is ∀x∃y(y=x+1) true?
Choices: No; integers have a largest element · Yes; choose y=0 for every x · Yes; for each x choose y=x+1 · No; one fixed y must work for every x
Show solution
- The existential witness may depend on x.
- x+1 is an integer for each integer x.
Answer: Yes; for each x choose y=x+1
5. Over the integers, is ∃y∀x(y=x+1) true?
Choices: Yes; choose y=0 · No; x=0 requires y=1 while x=1 requires y=2 · Yes; choose y=x+1 separately each time · Yes; choose y=1
Show solution
- Here one fixed y must work for all x.
- The two choices of x require incompatible values.
Answer: No; x=0 requires y=1 while x=1 requires y=2
Practice this interactively with instant feedback and an AI tutor.
Practice Mixed Quantifier Practice Take the free placement check