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Proof step 1 from proof gcd-zero-05 (conclusion)

⊦ gcd(i0, j0) + gcd(i, j) = 0 ↔ gcd(i0, j0) = 0 ∧ gcd(i, j) = 0

User applied rule case distinction

And obtained the following new goals

Proof step 2:
gcd(i0, j0) + gcd(i, j) = 0 ⊦ gcd(i0, j0) = 0 ∧ gcd(i, j) = 0

Rule argument

pos: right 1

Additional information

Used simplifier rules: gcd-zero-zero: ⊦ gcd(i, j) = 0 ↔ i = 0 ∧ j = 0;%gcd-zero: ⊦ gcd(0, 0) = 0;%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2