Solutions for Exercise Set 1007

Note. You can use /\ for ∧, \/ for ∨ and -> for → to simplify typing answers.

  1. Show that each of the following is a tautology using truth tables.

    There are different styles of truth tables. Here, I show a style where full compound propositions are shown.

    1. ¬(pq) → p

      p q pq ¬(pq) ¬(pq) → p
      F F T F T
      F T T F T
      T F F T T
      T T T F T
    2. ¬(pq) ↔ ¬p ∧ ¬q

      p q pq ¬(pq) ¬p ¬q ¬p ∧ ¬q ¬(pq) ↔ ¬p ∧ ¬q
      F F F T T T T T
      F T T F T F F T
      T F T F F T F T
      T T T F F F F T
    3. p ∧ (pq)) → q

      p q ¬p pq ¬p ∧ (pq) p ∧ (pq)) → q
      F F T F F T
      F T T T T T
      T F F T F T
      T T F T F T
    4. ((pq) → r) → ((p ∧ ¬r) → ¬q)

      p q r pq (pq) → r ¬r p ∧ ¬r ¬q (p ∧ ¬r) → ¬q ((pq) → r) → ((p ∧ ¬r) → ¬q)
      F F F F T T F T T T
      F F T F T F F T T T
      F T F F T T F F T T
      F T T F T F F F T T
      T F F F T T T T T T
      T F T F T F F T T T
      T T F T F T T F F T
      T T T T T F F F T T


  2. Convert each of the following into an equivalent form that does not involve → and does not contain double-negations (such as ¬¬p).
    1. ¬p → ¬q

      p ∨ ¬q

    2. (pa) → r

      ¬(¬pa) ∨ r

      or

      (p ∧ ¬a) ∨ r

      (by DeMorgan's Law)

    3. (pq) → ¬ p

      ¬(pq) ∨ ¬p

      or

      p ∧ ¬q) ∨ ¬p

      (by DeMorgan's Law) or

      ¬p

      (by the Absorption Law).