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Proof step 10 from proof i2n (previous step)

j < 0, 0 ≤ i ⊦ i→n(i) * i→n(j) = i→n(i * j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
j < 0, j < i, 0 ≤ i ⊦ i→n(i) * i→n(j) = i→n(i * j)

Rule argument

Additional information

Used simplifier rules: forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs