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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs