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Proof step 5 from proof three (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
i * j = 3, 0 < i, 1 < i ⊦ i = 3 ∧ j = 1, 3 ≤ i ∧ 1 < j

Rule argument

Additional information

Used simplifier rules: commutativity: ⊦ i * j = j * i;% ⊦ i + 0 = i; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2