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Proof step 60 from proof gcd-rec-sub-01 (previous step)

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

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information

Used simplifier rules: gcd-neg: ⊦ gcd(~ i, j) = gcd(i, j);%gcd-com: ⊦ gcd(i, j) = gcd(j, i);%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs