SE 504 Spring 2019
HW #2: Problem 7 Sample Solution

{P: x = X} if x ≥ 0 ⟶ skip [] x ≤ 0 ⟶ x = -x fi {Q: x = |X|}

Solution: By the Hoare Triple Selection Command Law, it suffices to prove these three things:

• x = X ⟹ x≥0 ∨ x≤0
• {P ∧ x≥0} skip {Q}
• {P ∧ x≤0} x := -x {Q}

The first one is trivial because, by "number theory", the consequent reduces to true, which (by (3.72)) makes the whole implication reduce to true. In gory detail:

x = X ⟹ x≥0 ∨ x≤0 = < number theory > x = X ⟹ true <-------- (3.72) with p := x=X

As for the second and third, by the Hoare Triple skip Law and the Hoare Triple Assignment Law, they are equivalent to, respectively,

Assume P (i.e., x=X) and x≥0 Q = < defn of Q > x = |X| = < assumption x=X > x = |x| = < given absolute value theorem > x ≥ 0 = < assumption > true

Assume P (i.e., x=X) and x≤0. Q(x:=-x) = < defn of Q > (x = |X|)(x:=-x) = < textual substitution > -x = |X| = < assumption x=X > -x = |x| = < given absolute value theorem > x ≤ 0 = < assumption > true