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Proof step 6 from proof three (previous step)

j ≤ 1, i * j = 3, 0 < i, 1 < i ⊦ i = 3 ∧ j = 1, 3 ≤ i ∧ 1 < j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
i * j = 3, j ≤ 1, 0 < i, 1 < i, j < i ⊦ i = 3 ∧ j = 1

Rule argument

Additional information

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