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Proof step 57 from proof gcd-rec-sub-01 (previous step)

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

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 3

Additional information

Used simplifier rules: gcd-neg: ⊦ gcd(~ i, j) = gcd(i, j);%gcd-com: ⊦ gcd(i, j) = gcd(j, i);% ⊦ 0 < j → abs(j) = j; (s)% ⊦ i < 0 → abs(i) = ~ i; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i < j → (j ≤ i ↔ false); (as)