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Proof step 4 from proof gcd-mult (previous step)

⊦ i = 0, i1 = 0, gcd(i * i1, j * i1) = gcd(i, j) * abs(i1)

User applied rule cut formula

And obtained the following new goals

Proof step 5:
⊦ j = 0, i = 0, i1 = 0, gcd(i * i1, j * i1) = gcd(i, j) * abs(i1)

Rule argument

fma: j = 0

Additional information

Used simplifier rules: gcd-zero-04: ⊦ gcd(i, 0) = abs(i);%gcd-zero-03: ⊦ i ≠ 0 → gcd(i, 0) = abs(i);%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%abs-eqall: ⊦ abs(i * j) = abs(i) * abs(j); from specification int-mult Used formulas: j = 0