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Proof step 2 from proof div-nls-01 (previous step)

i0 < i / j, abs(i) ≤ i0, i ≤ 0, j < 0, abs(i) < i / j ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ 0 ≤ i ∧ 0 < j ∨ i < 0 ∧ j < 0 → i / j = abs(i) / abs(j)
Proof step 4:
0 ≤ i ∧ 0 < j ∨ i < 0 ∧ j < 0, i0 < abs(i) / abs(j), abs(i) ≤ i0, i ≤ 0, j < 0, abs(i) < abs(i) / abs(j) ⊦

Rule argument

spec: inst: name: abs-intro seq: ⊦ 0 ≤ i ∧ 0 < j ∨ i < 0 ∧ j < 0 → i / j = abs(i) / abs(j) subst: [j, i] → [j, i] flags: rewrite, current

Additional information

Used simplifier rules: div-zero-ngr-01: ⊦ 0 ≤ i ∧ j < 0 → ¬ 0 < i / j;%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%Abs-pos: ⊦ ¬ j < 0 → abs(j) = j; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs Used terms: j, i