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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
j ≤ 0 ⊦ j * i1 ≤ j ↔ ¬ i1 ≤ 0 ∨ j = 0

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;% ⊦ i < 1 ↔ i ≤ 0; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + 0 = i; (s)% ⊦ i < j - i0 ↔ i + i0 < j; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)