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Proof step 18 from proof gcd-divides-both (previous step)

~ abs(i0) < 0, j | ~ abs(i0) ⊦ j | i0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
j | abs(i0) ⊦ i0 = 0, j | i0

Rule argument

Additional information

Used simplifier rules: div-neg: ⊦ i | ~ j ↔ i | j;% ⊦ 0 < abs(i) ↔ i ≠ 0; (s)% ⊦ ~ i < 0 ↔ 0 < i; (s)