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Proof step 50 from proof gcd-rec-sub-01 (previous step)

0 < i, i0 + i ≤ i ⊦ gcd(i, i0 + i) = gcd(i, i0), i0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 51:
0 < i, i0 < 0 ⊦ gcd(i, i0 + i) = gcd(i, i0), i0 = 0

Rule argument

Additional information

Used simplifier rules: forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)% ⊦ i + j ≤ i ↔ j ≤ 0; (s)