Compound Statements

  • 12th Grade
Reviewed by Cierra Henderson
Cierra Henderson, MBA |
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Cierra is an educational consultant and curriculum developer who has worked with students in K-12 for a variety of subjects including English and Math as well as test prep. She specializes in one-on-one support for students especially those with learning differences. She holds an MBA from the University of Massachusetts Amherst and a certificate in educational consulting from UC Irvine.
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Quizzes Created: 8156 | Total Attempts: 9,588,805
| Attempts: 17 | Questions: 10 | Updated: Jan 21, 2026
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1) P → q for p = T, q = F.

Explanation

Implication is false only when p = T and q = F.

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About This Quiz
Compound Statements - Quiz

How do “and,” “or,” and “not” shape truth? In this quiz, you’ll practice working with compound statements and truth values. Take this quiz to strengthen your grasp of logical building blocks.

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2) P ↔ q for p = T, q = F.

Explanation

Values differ, so biconditional is false.

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3) ¬(p → q) for p = F, q = T.

Explanation

p → q = T; negation of T = F.

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4) (p → q) ∧ (q → p) for p = T, q = F.

Explanation

p → q = F; F ∧ anything = F.

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5) (p ↔ q) ∨ (p ∧ q) for p = T, q = T.

Explanation

p ↔ q = T; T ∨ anything = T.

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6) P → q for p = F, q = F.

Explanation

False implies false is considered true in logic.

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7) P ↔ q for p = T, q = T.

Explanation

Both values equal (T, T), so biconditional is true.

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8) ¬(p → q) for p = T, q = F.

Explanation

p → q = F; negation of F = T.

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9) (p → q) ∧ (q → p) for p = T, q = T.

Explanation

Both implications are true, so conjunction = T.

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10) (p ↔ q) ∨ (p ∧ q) for p = T, q = F.

Explanation

p ↔ q = F; p ∧ q = F; F ∨ F = F.

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Cierra Henderson |MBA |
K-12 Expert
Cierra is an educational consultant and curriculum developer who has worked with students in K-12 for a variety of subjects including English and Math as well as test prep. She specializes in one-on-one support for students especially those with learning differences. She holds an MBA from the University of Massachusetts Amherst and a certificate in educational consulting from UC Irvine.
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P → q for p = T, q = F.
P ↔ q for p = T, q = F.
¬(p → q) for p = F, q = T.
(p → q) ∧ (q → p) for p = T, q = F.
(p ↔ q) ∨ (p ∧ q) for p = T, q = T.
P → q for p = F, q = F.
P ↔ q for p = T, q = T.
¬(p → q) for p = T, q = F.
(p → q) ∧ (q → p) for p = T, q = T.
(p ↔ q) ∨ (p ∧ q) for p = T, q = F.
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