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Proof step 3 from proof div-mult-nls-zero (previous step)

(i * i0) / j < 0, 0 < j, 0 ≤ i, 0 ≤ i0 ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: div-ls-zero seq: 0 < j ⊦ i / (* j) < 0 ↔ i + j ≤ 0 subst: [j, i] → [j, i * i0] flags: <all paths>

Additional information

Used simplifier rules: forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%forward-lele: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward) from specification int-abs% ⊦ i + j < j ↔ i < 0; (s)% ⊦ 0 ≤ i → (i * j < 0 ↔ j < 0 ∧ i ≠ 0); (s)% ⊦ i ≤ j → ¬ j < i; (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: j, i * i0