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Proof step 66 from proof div-div-rec (previous step)

2 ^ i * 2 * i1 + j0 < 0, 2 ^ i * (2 * (i1 + 1) - 1) + j0 < i, 2 ^ i * (2 * (i1 + 1) - 1) + j0 ≤ 0, 2 * i1 < i, 0 ≤ i, 2 ^ i * 2 ≤ abs(j0 - 2 ^ i), j0 ≤ 2 ^ i, j0 < 2 ^ i + i, 0 < i, abs(j0) < 2 ^ i, i1 ≤ 0, i1 < i, j0 ≤ 0, j0 < i ⊦ i1 = (2 ^ i * (2 * i1 - 1) + j0) / (2 ^ i * 2)

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-abs inst: name: Abs-neg seq: ⊦ j < 0 → abs(j) = 0 - j subst: [j] → [j0 - 2 ^ i] flags: rewrite

Additional information

Used simplifier rules: two-zero: ⊦ 0 ≤ j → 2 ^ j ≠ 0;%Abs-pos: ⊦ ¬ j < 0 → abs(j) = j; from specification int-abs%sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; from specification int-abs%succ-rmerge: ⊦ i * j + i = i * (j + 1); from specification int-mult%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%sub-ls: ⊦ i - j < i0 ↔ i < j + i0; from specification int-abs%add-zero: ⊦ i + 0 = i; from specification int-basic2%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%forward-lele: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward) from specification int-abs%Sub-def: ⊦ i + (j - i) = j; from specification int-abs%div-mult-01: j ≠ 0, i0 ≠ 0 ⊦ (i * i0) / (j * i0) = i / j; from specification int-div%div-mult-03: j ≠ 0 ⊦ (i * j) / j = i; from specification int-div Used terms: j0 - 2 ^ i