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Proof step 5 from proof numls-07 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
j ≤ 0 ⊦ i1 * j < j ↔ 1 < i1 ∧ j ≠ 0

Rule argument

Additional information

Used simplifier rules: zero-gr-03: ⊦ i ≤ 0 → (0 < i * j ↔ ¬ ¬(j < 0 ∧ 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%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: ⊦ i - j < i0 ↔ i < j + i0; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs