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

j < 0, 0 < i ⊦ (j - (2 ^ i - 1)) / 2 ^ i = ((j - (2 ^ (i - 1) - 1)) / 2 ^ (i - 1) - 1) / 2

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ j < 0 ∧ 0 < i → ((j - (2 ^ i - 1)) / 2 ^ i - 1) / 2 = (j - (2 ^ (i + 1) - 1)) / 2 ^ (i + 1)

Rule argument

spec: <toplevel> name: div-div-rec seq: ⊦ j < 0 ∧ 0 < i → ((j - (2 ^ i - 1)) / 2 ^ i - 1) / 2 = (j - (2 ^ (i + 1) - 1)) / 2 ^ (i + 1) subst: [i, j] → [i - 1, j] flags: <all paths>

Additional information

Used simplifier rules: ⊦ i - (j - j0) = i + j0 - j; (s)% ⊦ i + (j - i) = j; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i < j - i0 ↔ i + i0 < j; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ 0 < i → (i ≤ 1 ↔ i = 1); (s)% ⊦ i = j → i - j = 0; (s)% ⊦ (* i) + i0 - (* j) = (i - j) + i0; (s)% ⊦ i + 0 = i; (s)% ⊦ i / 1 = i; (s)%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%i < 0 ⊦ (* i) + j = j - ~ i; (s) Used terms: i - 1, j