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Proof step 63 from proof div-div-rec (previous step)

abs(j1) = 1, 0 < i, abs(j0) < 2 ^ i, j1 ≤ 0, j1 < i, i1 ≤ 0, i1 < i, j0 ≤ 0, j0 < i, 2 * i1 + j1 < 0, 2 ^ i * ((2 * i1 + j1) + 1) + j0 < 0, (2 * i1 + j1) + 1 < i, 0 ≤ i, 2 ^ i * (j1 + 2 * (i1 + 1)) + j0 ≤ 0, 2 ^ i * (j1 + 2 * (i1 + 1)) + j0 < i, 2 ^ i * 2 ≤ abs(2 ^ i * j1 + j0), 2 ^ i * j1 + j0 ≤ 0, 2 ^ i * j1 + j0 < i ⊦ i1 = (2 ^ i * (j1 + 2 * i1) + j0) / (2 ^ i * 2)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 64:
⊦ j < 0 → abs(j) = 0 - j
Proof step 65:
j1 < 0, 0 - j1 = 1, 0 < i, abs(j0) < 2 ^ i, j1 ≤ 0, j1 < i, i1 ≤ 0, i1 < i, j0 ≤ 0, j0 < i, 2 * i1 + j1 < 0, 2 ^ i * ((2 * i1 + j1) + 1) + j0 < 0, (2 * i1 + j1) + 1 < i, 0 ≤ i, 2 ^ i * (j1 + 2 * (i1 + 1)) + j0 ≤ 0, 2 ^ i * (j1 + 2 * (i1 + 1)) + j0 < i, 2 ^ i * 2 ≤ abs(2 ^ i * j1 + j0), 2 ^ i * j1 + j0 ≤ 0, 2 ^ i * j1 + j0 < i ⊦ i1 = (2 ^ i * (j1 + 2 * i1) + j0) / (2 ^ i * 2)

Rule argument

spec: int-abs inst: name: Abs-neg seq: ⊦ j < 0 → abs(j) = 0 - j subst: [j] → [j1] flags: rewrite

Additional information

Used simplifier rules: Abs-pos: ⊦ ¬ j < 0 → abs(j) = j; from specification int-abs%one-01: ⊦ i = 1 → i * j = j; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%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-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%forward-lele: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward) from specification int-abs Used terms: j1