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Proof step 13 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 14:
i * j = 3, 0 < j, i < 3, 1 < j, 1 < i, 0 < i ⊦ i = 3

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1% ⊦ i ≤ j ↔ ¬ j < i; (ss)