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Proof step 6 from proof zero-ls-03 (previous step)

i ≤ 0, i < 0 → i * j < 0 → 0 < j ⊦ i * j < 0 ↔ 0 < j ∧ i ≠ 0

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: zero-left: ⊦ 0 * i = 0;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%zero-ls: ⊦ 0 < i → (i * j < 0 ↔ j < 0);%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)% ⊦ i < 0 ∧ 0 ≤ j → i < j; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i ≤ j → ¬ j < i; (s)