- 1.1–1.3 propositional logic and equivalences · 1.4–1.5 predicates and quantifiers · 1.6 rules of inference.
- A proposition is a declarative sentence that is either true or false, not both. Questions, commands and "x + 1 = 2" (x unknown) are NOT propositions.
🎯 How this can come up · 1 question
Which of these is a proposition?
It is false, but it is still a declarative sentence with a truth value.
- p ∧ q (and) is true only when both are true.
- p ∨ q (inclusive or) is false only when both are false.
- English "or" can be inclusive (∨) or exclusive (⊕: exactly one true).
🎯 How this can come up · 1 question
p ⊕ q is true when:
Correct: Exactly one of p, q is true.
- p → q is FALSE only when p is true and q is false. Same as ¬p ∨ q, and as its contrapositive ¬q → ¬p.
- Readings: "if p then q", "p only if q", "q whenever p", "q is necessary for p", "p is sufficient for q".
🎯 How this can come up · 2 questions
When is p → q false?
Correct: p true, q false.
If p is false, p → q is true no matter what q is.
True. A false hypothesis makes the implication vacuously true.
- From p → q: converse q → p, inverse ¬p → ¬q, contrapositive ¬q → ¬p.
- Only the contrapositive is equivalent to the original; the converse and inverse are equivalent to each other.
🎯 How this can come up · 2 questions
The contrapositive of "If it rains, then I stay home" is:
Correct: If I don’t stay home, then it doesn’t rain.
An implication is logically equivalent to its converse.
False. It is equivalent to its contrapositive.
- p ↔ q is true when p and q have the SAME truth value ("if and only if", "necessary and sufficient").
🎯 How this can come up · 1 question
p ↔ q is equivalent to:
Correct: (p → q) ∧ (q → p).
- p ↔ q is the negation of p ⊕ q.
- A compound proposition with n variables has 2ⁿ rows in its truth table.
🎯 How this can come up · 1 question
How many rows does the truth table of a proposition with 4 variables have?
2⁴ = 16.
- Precedence: ¬ first, then ∧, then ∨, then →, then ↔. So p ∨ q → ¬r means (p ∨ q) → (¬r).
🎯 How this can come up · 1 question
p ∧ q ∨ r is read as:
∧ binds tighter than ∨.
- Translate: pick atomic propositions, then join them with connectives. "You can access the Internet only if you are a CS major or not a freshman" → a → (c ∨ ¬f).
- A set of specifications is consistent if ONE assignment of truth values makes all of them true at the same time.
🎯 How this can come up · 1 question
How do you check that a list of system specifications is consistent?
Translate each into propositional logic and look for one truth assignment (a row of the truth table) that makes every specification true at once. If such a row exists the specs are consistent; if none exists they are inconsistent.
- Logic gates: inverter (¬), AND (∧), OR (∨); circuits compute compound propositions.
- Tautology: always true (p ∨ ¬p). Contradiction: always false (p ∧ ¬p). Contingency: neither.
🎯 How this can come up · 2 questions
p ∨ ¬p is a:
Correct: Tautology.
(p → q) ∧ p ∧ ¬q is a:
p and ¬q make p → q false.
- p ≡ q (logically equivalent) when p ↔ q is a tautology: same truth value in every row.
- De Morgan: ¬(p ∧ q) ≡ ¬p ∨ ¬q and ¬(p ∨ q) ≡ ¬p ∧ ¬q.
🎯 How this can come up · 1 question
¬(p ∨ q) is equivalent to:
Correct: ¬p ∧ ¬q.
- Identity, domination, idempotent, double negation, negation laws.
- Commutative, associative, distributive, absorption laws.
- Implication laws: p → q ≡ ¬p ∨ q ≡ ¬q → ¬p; p ↔ q ≡ (p → q) ∧ (q → p).
- Equivalence proof: rewrite step by step, naming the law used at each step (no truth table).
🎯 How this can come up · 1 question
Show that ¬(p → q) ≡ p ∧ ¬q without a truth table.
¬(p → q) ≡ ¬(¬p ∨ q) (implication law) ≡ ¬¬p ∧ ¬q (De Morgan) ≡ p ∧ ¬q (double negation).
- Satisfiable: SOME assignment makes it true. Unsatisfiable: none does (its negation is a tautology).
🎯 How this can come up · 1 question
A compound proposition is unsatisfiable exactly when its negation is a tautology.
True.
- A propositional function P(x) becomes a proposition when x gets a value or is quantified.
- ∀x P(x): true when P(x) holds for EVERY x in the domain. ∃x P(x): true when it holds for AT LEAST ONE.
- One counterexample makes ∀x P(x) false.
🎯 How this can come up · 1 question
Domain: all integers. ∀x (x² ≥ x) is:
x² ≥ x holds for every integer (it fails only for 0 < x < 1).
- ∃x P(x) is false only when P(x) is false for every x. Over an empty domain ∃ is false and ∀ is true.
- ∀ and ∃ bind tighter than all connectives: ∀x P(x) ∨ Q(x) means (∀x P(x)) ∨ Q(x).
- "Every student has a computer": ∀x (S(x) → C(x)). "Some student has a computer": ∃x (S(x) ∧ C(x)).
🎯 How this can come up · 1 question
Domain: all people. "Some students are lazy" is:
Correct: ∃x (S(x) ∧ L(x)).
- ¬∀x P(x) ≡ ∃x ¬P(x) and ¬∃x P(x) ≡ ∀x ¬P(x) (De Morgan for quantifiers).
🎯 How this can come up · 1 question
¬∀x (x > 0) is equivalent to:
Correct: ∃x (x ≤ 0).
- Nested quantifiers: order matters when they differ. ∀x ∃y (x + y = 0) is true over the integers; ∃y ∀x (x + y = 0) is false.
🎯 How this can come up · 1 question
Domain: integers. Which is TRUE?
Each x has its own y = −x; no single y works for all x.
- ∀x∀y and ∃x∃y can be swapped freely; ∀x∃y and ∃y∀x cannot.
- An argument is valid if the conclusion must be true whenever all premises are true (the conditional (p₁ ∧ … ∧ pₙ) → q is a tautology).
- The 8 rules: modus ponens, modus tollens, hypothetical syllogism, disjunctive syllogism, addition, simplification, conjunction, resolution.
- Modus ponens: p, p → q ∴ q.
🎯 How this can come up · 1 question
"If it snows, school closes. It snows. Therefore school closes." uses:
Correct: Modus ponens.
- Modus tollens: ¬q, p → q ∴ ¬p.
🎯 How this can come up · 1 question
From p → q and ¬q we may conclude:
Modus tollens.
- Hypothetical syllogism: p → q, q → r ∴ p → r.
- Disjunctive syllogism: p ∨ q, ¬p ∴ q.
🎯 How this can come up · 1 question
"Either the file is corrupt or the disk is full. The disk is not full. So the file is corrupt." uses:
Correct: Disjunctive syllogism.
- Addition: p ∴ p ∨ q.
- Simplification: p ∧ q ∴ p.
- Conjunction: p, q ∴ p ∧ q.
- Resolution: p ∨ q, ¬p ∨ r ∴ q ∨ r (the clashing literal p cancels).
🎯 How this can come up · 1 question
From p ∨ q and ¬p ∨ r, resolution gives:
Correct: q ∨ r.
- Build an argument as a numbered list of steps, each a premise or a rule applied to earlier steps.
🎯 How this can come up · 1 question
Premises: p → q, q → r, p. Show r with rules of inference.
1. p → q (premise) 2. q → r (premise) 3. p → r (hypothetical syllogism 1, 2) 4. p (premise) 5. r (modus ponens 3, 4).
No slides match this filter.







































































![Slide 131: Rule of inference Tautology Name p q p [ p ( p q )] q Modusponens](figures/slide-131.jpg)
