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Proof step 10 from proof three (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
i * j = 3, j < i, 1 < i, 0 < i, j ≤ 0 ⊦ j = 1, 3 ≤ i ∧ 1 < j

Rule argument

Additional information

Used simplifier rules: one: ⊦ 1 * i = i;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%j0 ≠ 0, j ≤ j0 ⊦ i ≤ (* j) → i ≤ (* j0); (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)