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Proof step 12 from proof three (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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;%eq-same-02: ⊦ 0 < i → (i * j = i ↔ j = 1);%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%numlelsdel-01: i0 ≠ 0, i0 ≤ i ⊦ (* i) < j → (* i0) < j; from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs