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Proof step 1 from proof gcd-mult-pos-01 (conclusion)

⊦ 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);%eq-pos: ⊦ 0 < i → (i * j = i * j0 ↔ j = j0); from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs