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Proof step 7 from proof three (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
i * j = 3, 0 < i, 1 < i, j < i, j ≤ 1 ⊦ i = 3 ∧ j = 1, 3 ≤ i ∧ 1 < j

Rule argument

Additional information

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