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Proof step 13 from proof i2n (previous step)

0 ≤ i ⊦ 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 14:
0 ≤ i, 0 ≤ j ⊦ i→n(i) * i→n(j) = i→n(i * j)

Rule argument

Additional information

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