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Proof step 5 from proof div-div-rec (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ i < 0 ∧ 0 < j → ¬ j ≤ i

Rule argument

spec: int-abs name: zero-gels-02 seq: ⊦ i < 0 ∧ 0 < j → ¬ j ≤ i subst: [i, j] → [j + 1, 2 ^ i] flags: <all paths>

Additional information

Used simplifier rules: pot2-pos: ⊦ 0 < 2 ^ i ↔ 0 ≤ i;%zero: ⊦ 0 < j → (i ^ j = 0 ↔ i = 0);%two-lez-01: ⊦ 2 ^ i ≤ 0 ↔ i < 0;% ⊦ i + (* j) < (* j0) ↔ i < j0 - 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 ↔ ¬ j ≤ i; (ss)% ⊦ i < 0 → (~1 ≤ i ↔ i = ~1); (s)% ⊦ 0 - i = ~ i; (s)%j ≠ 0 ⊦ ~ i / j = ~(i / j); (s)%i ≠ 0 ⊦ i / i = 1; (s)% ⊦ i < j → ¬ j < i; (s) Used terms: j + 1, 2 ^ i