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Proof step 5 from proof i2n (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
i < j, i < 0, 0 ≤ j ⊦ i→n(i) * i→n(j) = i→n(i * j)

Rule argument

Additional information

Used simplifier rules: ⊦ ~1 < i ↔ 0 ≤ i; (s)%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%numlef-01: ⊦ (* i) ≤ j → i - 1 < j; (forward) from specification int-abs% ⊦ i < j ↔ ¬ j ≤ i; (ss)