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Proof step 2 from proof mods-nls (previous step)

i * abs(j) < j, 0 < i ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-abs inst: name: Abs-neg seq: ⊦ j < 0 → abs(j) = 0 - j subst: [j] → [j] flags: rewrite

Additional information

Used simplifier rules: minus-left: ⊦ ~ i * j = ~(i * j);%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%zero-gr: ⊦ 0 < i → (0 < i * j ↔ 0 < j);%neg-06: ⊦ 0 - i = ~ i; from specification int-abs%neg-13: ⊦ ~ i < j ↔ 0 < j + i; from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%ls-01: ⊦ i + j < i + i0 ↔ j < i0; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%ls-02: ⊦ i < i + j ↔ 0 < j; from specification int-basic2%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1 Used terms: j