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Proof step 21 from proof less-mult-div-01 (previous step)

¬ i3 * i1 < 0 ∧ ¬ i2 * (i + i3) + j0 < 0, i3 * i1 = 0 ∧ i2 * (i + i3) + j0 = 0, 1 < i1 + i2, 0 < i2, 0 < i1, 0 ≤ j0, j0 < i1 + i2, 0 < j1, 0 < i, 0 < i + i3, 0 ≤ i3 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%lenls: ⊦ i ≤ j → ¬ j < i; from specification int-abs%zero-ls: ⊦ 0 < i → (i * j < 0 ↔ j < 0); from specification int-mult%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%zero-neq-01: ⊦ 0 < i → (i * j = 0 ↔ j = 0); from specification int-mult