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Proof step 2 from proof zero-ge-03 (previous step)

i ≤ 0 ⊦ 0 ≤ i * j ↔ i = 0 ∨ j ≤ 0

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: zero-ge-01: ⊦ i < 0 → (0 ≤ i * j ↔ j ≤ 0);%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%zero-ls-03: ⊦ i ≤ 0 → (i * j < 0 ↔ ¬ ¬(0 < j ∧ i ≠ 0));%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs