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Proof step 14 from proof gcd-divides-both (previous step)

abs(i0) = abs(i), i < 0, j | abs(i) ⊦ j | i ∧ j | i0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
i + abs(i0) = 0, i < 0, j | i ⊦ j | i0

Rule argument

Additional information

Used simplifier rules: div-neg: ⊦ i | ~ j ↔ i | j;%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs% ⊦ ~ i = j ↔ i + j = 0; (s)% ⊦ i < 0 → abs(i) = ~ i; (s)