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Proof step 31 from proof equal-minus-one (previous step)

1 / i ^ ~ j = ~1, j < 0 ⊦ i = 0, i = ~1, j = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 32:
0 < j, j ≤ i ⊦ i / j = (i - j) / j + 1
Proof step 33:
1 / i ^ ~ j = ~1, j < 0 ⊦ i = ~1, i = 0, j = 0, 0 < i ^ ~ j ∧ i ^ ~ j ≤ 1

Rule argument

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

Additional information

Used simplifier rules: ⊦ i + 1 = ~1 ↔ i + 1 + 1 = 0; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; (s)% ⊦ 0 < i → (i ≤ 1 ↔ i = 1); (s)% ⊦ i = j → i - j = 0; (s)%j ≠ 0 ⊦ 0 / j = 0; (s) Used terms: i ^ ~ j, 1