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Proof step 4 from proof one (previous step)

i < 0 ⊦ i / 1 = i

User applied rule insert elim lemma

And obtained the following new goals

Rule argument

spec: int-div inst: name: div-elim2 seq: ⊦ 0 < j ∧ i < 0 → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j) subst: [i, j, i0, j0] → [i, 1, i0, j] flags: rewrite, current

Additional information

Used simplifier rules: one: ⊦ 1 * i = i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%eq: ⊦ i = i + j ↔ j = 0; 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%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%neg-13: ⊦ ~ i < j ↔ 0 < j + i; from specification int-abs%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs Used terms: i, 1, i0, j