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Proof step 13 from proof abs-ls-same-01 (previous step)

0 ≤ j ⊦ abs(i) * j < i ↔ ¬ i ≤ 0 ∧ ¬ 0 < j

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-abs inst: name: anti seq: ⊦ i ≤ j → (j ≤ i ↔ i = j) subst: [i, j] → [j, 0] flags: rewrite

Additional information

Used simplifier rules: mods-nls: ⊦ 0 < i → ¬ i * abs(j) < j;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%zero-left: ⊦ 0 * i = 0;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs Used terms: j, 0