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Proof step 3 from proof zero-ls-02 (previous step)

0 ≤ i ⊦ i * j < 0 ↔ j < 0 ∧ i ≠ 0

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: Mult-zero: ⊦ i * 0 = 0;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%zero-ls: ⊦ 0 < i → (i * j < 0 ↔ j < 0);%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 < j ↔ ¬ j ≤ i; (ss)% ⊦ i < j → (j ≤ i ↔ false); (as)