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Proof step 3 from proof gcd-mult (
previous step
)
⊦ i = 0, i1 = 0, gcd(i * i1, j * i1) = gcd(i, j) * abs(i1)
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 4
:
⊦ i = 0, i1 = 0, gcd(i * i1, j * i1) = gcd(i, j) * abs(i1)
Rule argument
Additional information