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

i ≤ j, j < 0, i < 0 ⊦ i / j = (i - j) / j + 1

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
0 < j, j ≤ i ⊦ i / j = (i - j) / j + 1
Proof step 4:
~ i / ~ j = (~ i - ~ j) / ~ j + 1, i ≤ j, j < 0, i < 0 ⊦ i / j = (i - j) / j + 1

Rule argument

spec: int-div inst: name: div-rec seq: 0 < j, j ≤ i ⊦ i / j = (i - j) / j + 1 subst: [j, i] → [~ j, ~ i] flags: current, []

Additional information

Used simplifier rules: neg: ⊦ 0 < ~ i ↔ i < 0; from specification int-abs%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%negle-13: ⊦ ~ i ≤ j ↔ 0 ≤ i + j; from specification int-abs%neg-15: ⊦ i + ~ j = i - j; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; from specification int-abs%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2 Used terms: ~ j, ~ i