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Proof step 2 from proof zero-ge-02 (previous step)

0 ≤ i ⊦ 0 ≤ i * j ↔ i = 0 ∨ 0 ≤ j

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-02: ⊦ 0 ≤ i → (i * j < 0 ↔ ¬ ¬(j < 0 ∧ 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