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Proof step 10 from proof div-div-rec (previous step)

j + 1 = 2 ^ i * (i0 + 1) + j0, 2 ^ i * i0 + j0 < 0, j0 ≤ 0, j0 < i, i0 ≤ 0, i0 < i, 0 ≤ i, j < 0, 0 < i, abs(j0) < 2 ^ i ⊦ (i0 - 1) / 2 = (j + 1 - 2 ^ i * 2) / (2 ^ i * 2)

User applied rule insert elim lemma

And obtained the following new goals

Proof step 11:
i0 - 1 = i1 * 2 + j1 ∧ j1 ≤ 0 ∧ abs(j1) < 2 ∧ i1 ≤ 0 ∧ 0 < 2 ∧ i0 - 1 < 0, j + 1 = 2 ^ i * (i0 + 1) + j0, 2 ^ i * i0 + j0 < 0, j0 ≤ 0, j0 < i, i0 ≤ 0, i0 < i, 0 ≤ i, j < 0, 0 < i, abs(j0) < 2 ^ i ⊦ i1 = (j + 1 - 2 ^ i * 2) / (2 ^ i * 2)

Rule argument

spec: int-div inst: name: div-elim-negpos seq: ⊦ 0 < j ∧ i < 0 → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j ∧ i0 ≤ 0) subst: [i, j, i0, j0] → [i0 - 1, 2, i1, j1] flags:

Additional information

Used simplifier rules: sub-ls: ⊦ i - j < i0 ↔ i < j + i0; from specification int-abs%lsone: ⊦ i < 1 ↔ i ≤ 0; from specification int-abs%⊦ 0 < j ∧ i < 0 → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j ∧ i0 ≤ 0); (elim) Used terms: i0 - 1, 2, i1, j1