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wp(do x > 0 → L: x := x-1 od, (x<0 ∧ wp(x := -x; goto L, post)) ∨ (x ≥ 0 ∧ post)
= { sequential composition, goto, assignment rules }
= { repetition rule }
the strongest solution of Z: [ Z ≡ x > 0 ∧ Z(x ← x-1) ∨ x < 0 ∧ Z(x ← x-1) (x ← -x) ∨ x=0 ∧ post]
= { substitution }
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