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Proof step 2 from proof gcd-mult-pos-01 (previous step)

0 < i1, 0 < i, 0 ≤ j ⊦ gcd(i1 * i, i1 * j) = gcd(i, j) * i1

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: gcd-mult: ⊦ gcd(i * i1, j * i1) = gcd(i, j) * abs(i1);% ⊦ 0 < i → (i * j = i * j0 ↔ j = j0); (s)%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult% ⊦ 0 < j → abs(j) = j; (s)