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Proof step 1 from proof gcd-mult (conclusion)

⊦ gcd(i * i1, j * i1) = gcd(i, j) * abs(i1)

User applied rule cut formula

And obtained the following new goals

Proof step 2:
⊦ i1 = 0, gcd(i * i1, j * i1) = gcd(i, j) * abs(i1)

Rule argument

fma: i1 = 0

Additional information

Used simplifier rules: gcd-zero: ⊦ gcd(0, 0) = 0;%zero-left: ⊦ 0 * i = 0; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult Used formulas: i1 = 0