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Proof step 9 from proof numle-07 (previous step)

¬ 0 < j ⊦ j * i1 ≤ j ↔ ¬ (1 - i1) * j < 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: zero-ls-03: ⊦ i ≤ 0 → (i * j < 0 ↔ ¬ ¬(0 < j ∧ i ≠ 0));%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%lsone: ⊦ i < 1 ↔ i ≤ 0; 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%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%sub-ls-01: ⊦ i < j - i0 ↔ i + i0 < j; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs