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Proof step 6 from proof gcd-rec-sub-01 (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 7:
⊦ i = 0 ∨ i0 = 0, gcd(i, i0 + i) = gcd(i, i0), 0 < i ∧ i < i0 + i

Rule argument

fma: i = 0 ∨ i0 = 0

Additional information

Used simplifier rules: gcd-zero-04: ⊦ gcd(i, 0) = abs(i);%gcd-com: ⊦ gcd(i, j) = gcd(j, i);%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 Used formulas: i = 0 ∨ i0 = 0