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Proof step 25 from proof abs-pos (previous step)

abs(i) / gcd(i, j) = 0, 0 < j ⊦ j = 0, i = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 26:
⊦ ¬ j < 0 → abs(j) = j

Rule argument

seq: ⊦ ¬ j < 0 → abs(j) = j subst: [j] → [i] rotate: F spec: int-abs inst: name: Abs-pos

Additional information

Used simplifier rules: ⊦ i < 0 → abs(i) = ~ i;% ⊦ gcd(i, j) = 0 ↔ i = 0 ∧ j = 0;%j ≠ 0 ⊦ ~ i / j = ~(i / j);% ⊦ ~ i = 0 ↔ i = 0;% ⊦ i ≠ 0 → i / gcd(i, i0) ≠ 0; Used terms: i