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Proof step 6 from proof numls-07 (previous step)

j ≤ 0 ⊦ i1 * j < j ↔ 1 < i1 ∧ j ≠ 0

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: zero-ls-03: ⊦ i ≤ 0 → (i * j < 0 ↔ ¬ ¬(0 < j ∧ i ≠ 0));%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%ls-same-02: ⊦ i < 0 → (i * j < i ↔ 1 < j);%ls-same-02: ⊦ i < 0 → (i * j < i ↔ 1 < j);%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%numlelsdel-01: i0 ≠ 0, i0 ≤ i ⊦ (* i) < j → (* i0) < j; from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs