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Proof step 5 from proof i2n (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs