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Proof step 24 from proof div-ls-mone (previous step)

i0 < i0 * j + 2 * j, i0 < (i0 * j + j0) + 2 * j, 0 < (i0 * j + j0) + 2 * j, ∃ i1. ~1 = i0 + i1 ∧ 0 < i1, j0 < 0, j0 < j, i0 < j, i0 * j + j0 < j, (i0 * j + j0) + 1 < j, 0 < j + j0, i0 < j + j0, j0 + 1 < j, 1 < j, 0 < j, i0 < 0, 0 < i0 * j + 2 * j, 1 < i0 * j + 2 * j, j0 + 1 < i0 * j + 2 * j, j0 < i0 * j + 2 * j, i0 + 1 < i0 * j + 2 * j, i0 + 1 < j, i0 + 1 < (i0 * j + j0) + 2 * j, i0 + 1 < j + j0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
~1 = i0 + i, j0 < i, 0 < i, i0 + 1 < j + j0, i0 + 1 < (i0 * j + j0) + 2 * j, i0 + 1 < j, i0 + 1 < i0 * j + 2 * j, j0 < i0 * j + 2 * j, j0 + 1 < i0 * j + 2 * j, 1 < i0 * j + 2 * j, 0 < i0 * j + 2 * j, i0 < 0, 0 < j, 1 < j, j0 + 1 < j, i0 < j + j0, 0 < j + j0, (i0 * j + j0) + 1 < j, i0 * j + j0 < j, i0 < j, j0 < j, j0 < 0, 0 < (i0 * j + j0) + 2 * j, i0 < (i0 * j + j0) + 2 * j, i0 < i0 * j + 2 * j, i0 + 1 < i, i0 < i, j0 + 1 < i ⊦

Rule argument

Additional information

Used simplifier rules: succ-01: ⊦ i +1 = i + 1; from specification int-basic2%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1