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Proof step 1 from proof i2n-mult (conclusion)

⊦ 0 ≤ i ∧ 0 ≤ j → i→n(i * j) = i→n(i) * i→n(j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
0 ≤ i, 0 ≤ j ⊦ i→n(i * j) = i→n(i) * i→n(j)

Rule argument

Additional information

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