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Proof step 17 from proof three (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
i * j = 3, j < 3, 0 < i, i < 3, 0 < j, 1 < j, 1 < i ⊦ i = 3, j = 3

Rule argument

Additional information

Used simplifier rules: ls-same-01: ⊦ 0 < i → (i < i * j ↔ 1 < j);%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%numlelsdel-01: i0 ≠ 0, i0 ≤ i ⊦ (* i) < j → (* i0) < j; from specification int-abs%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs