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Proof step 4 from proof div-gcd-neg-not-zero (previous step)

i ≠ 0 → (¬ abs(i0) < gcd(i0, i) ↔ i0 ≠ 0), i0 / gcd(i, i0) = 0, i0 < 0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
i0 / gcd(i, i0) = 0, i0 < 0, i ≠ 0 → i0 ≠ 0 → ¬ 0 < gcd(i0, i) + i0 ⊦

Rule argument

Additional information

Used simplifier rules: gcd-zero-03: ⊦ i ≠ 0 → gcd(i, 0) = abs(i);%gcd-com: ⊦ gcd(i, j) = gcd(j, i);%neg-13: ⊦ ~ i < j ↔ 0 < j + i; from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%abs-06: ⊦ 0 < abs(i) ↔ i ≠ 0; from specification int-abs