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Proof step 15 from proof gcd-rec-sub-01 (previous step)

i < 0, i0 < 0 ⊦ gcd(abs(i), abs(i0 + i)) = gcd(i, i0), i = 0, i0 = 0, 0 < i ∧ i < i0 + i, 0 < abs(i) ∧ abs(i) < abs(i0 + i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
i < 0, i0 < 0, abs(i0 + i) + i ≤ 0 ⊦ gcd(i, abs(i0 + i)) = gcd(i, i0), i = 0, i0 = 0

Rule argument

Additional information

Used simplifier rules: gcd-neg: ⊦ gcd(~ i, j) = gcd(i, j);%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs% ⊦ i < j → ¬ j < i; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ ~ i < j ↔ 0 < j + i; (s)% ⊦ 0 < ~ i ↔ i < 0; (s)% ⊦ i < 0 → abs(i) = ~ i; (s)