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Proof step 62 from proof gcd-rec-sub-03 (previous step)

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

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: i0 = 0

Additional information

Used simplifier rules: gcd-same: ⊦ i ≠ 0 → gcd(i, i) = abs(i);%gcd-zero-03: ⊦ i ≠ 0 → gcd(i, 0) = abs(i);%add-zero: ⊦ i + 0 = i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs Used formulas: i0 = 0