1.1 Propositional Logic

  • Proposition - A statement that has a truth value.
  • Truth table and how to construct one.
  • Logical operations - and (∧\land), or (∨\lor), negation (¬\neg).
  • Implication (If assumption is false, the conclusion is always true).

    p→qp\to q, if pp, then qq.
    If pp (assumption) is false, p→qp\to q (conclusion) is always true.

  • Bi-implication (Double implication, if and only if) p⟺q≡(p→q∧q→p)p \iff q \equiv (p \to q \land q \to p).
  • Contrapositive - The contrapositive of p→q is ¬q→¬pp\to q \text{ is } \lnot q \to \lnot p.

Homework

p34: 31d, 32de, 38, 39

31. Construct a truth table for each of these compound propositions.
d). (p∨q)→(p∧q)(p \lor q) \to (p \land q)

Solution

pp qq p∨qp \lor q p∧qp \land q (p∨q)→(p∧q)(p \lor q) \to (p \land q)
T T T T T
T F T F F
F T T F F
F F F F T

32. Construct a truth table for each of these compound propositions.
d. (p∧q)→(p∨q)(p \land q) \to (p \lor q)

Solution

pp qq p∧qp \land q p∨qp \lor q (p∧q)→(p∨q)(p \land q) \to (p \lor q)
T T T T T
T F F T T
F T F T T
F F F F T

e. (q→¬p)↔(p↔q)(q \to \neg p) \bi (p \bi q)

Solution

pp qq q→¬pq \to \neg p p↔qp \bi q (q→¬p)↔(p↔q)(q \to \neg p) \bi (p \bi q)
T T F T F
T F T F F
F T T F F
F F T T T

38. Construct a truth table for ((p→q)→r)→s((p \to q) \to r) \to s.

Solution

pp qq rr ss p→qp \to q (p→q)→r(p \to q) \to r ((p→q)→r)→s((p \to q) \to r) \to s
T T T T T T T
T T T F T T F
T T F T T F T
T T F F T F T
T F T T F T T
T F T F F T F
T F F T F T T
T F F F F T F
F T T T T T T
F T T F T T F
F T F T T F T
F T F F T F T
F F T T T T T
F F T F T T F
F F F T F T T
F F F F F T F

39. Construct a truth table for (p↔q)↔(r↔s)(p \bi q) \bi (r \bi s).

Solution

pp qq rr ss p↔qp \bi q r↔sr \bi s (p↔q)↔(r↔s)(p \bi q) \bi (r \bi s)
T T T T T T T
T T T F T F F
T T F T T F F
T T F F T T T
T F T T F T F
T F T F F F T
T F F T F F T
T F F F F T F
F T T T F T F
F T T F F F T
F T F T F F T
F T F F F T F
F F T T T T T
F F T F T F F
F F F T T F F
F F F F T T T