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Proof step 2 from proof zero-ge-01 (previous step)

i < 0 ⊦ 0 ≤ i * j ↔ j ≤ 0

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

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