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Proof step 7 from proof numls-07 (previous step)

i1 * j < j, j ≤ 0 ⊦ 0 < (1 - i1) * j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
j < 0, 0 < i1, 1 < i1, (1 - i1) * j ≤ 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;%ls-same-02: ⊦ i < 0 → (i * j < i ↔ 1 < j);% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs% ⊦ i < j ↔ ¬ j ≤ i; (ss)