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Proof step 10 from proof less-mult-div-01 (previous step)

i * i1 = i0 * j + j0, ∃ i0. j = i1 + i0 ∧ 0 < i0, i ≤ i0, 0 < i, 0 < j1, j0 < j, 1 < j, 0 < j, 0 < i0, 0 ≤ j0, 0 < i1 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
i * i1 = i0 * (i1 + i2) + j0, 1 < i1 + i2, 0 < i2, 0 < i1, 0 ≤ j0, 0 < i0, j0 < i1 + i2, 0 < j1, 0 < i, i ≤ i0 ⊦

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%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%zero-02: ⊦ 0 < i ∧ 0 < j → 0 < i + j; from specification int-basic2