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Proof step 21 from proof shiftzero-max-neg-01 (previous step)

1 ≤ n, 2 ^ (n→i(n) - j) ≤ (2 ^ n→i(n) - 1) / 2 ^ j, j ≤ n→i(n), 0 < j, ∀ n, i, j0. i→n(j0) < i→n(j) → 0 < j0 ∧ i < 0 ∧ 2 ^ (n→i(n) - j0) ≤ >>>(n, i, j0) ∧ j0 ≤ n→i(n) ∧ # int2bits(i) ≤ n → n = 0 ⊦ j = 0, n = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 22:
⊦ j ≤ i ∧ 0 ≤ j → (2 ^ i - 1) / 2 ^ j = 2 ^ i / 2 ^ j - 1
Proof step 23:
j ≤ n→i(n) ∧ 0 ≤ j, 1 ≤ n, 2 ^ (n→i(n) - j) ≤ 2 ^ n→i(n) / 2 ^ j - 1, j ≤ n→i(n), 0 < j, ∀ n, i, j0. i→n(j0) < i→n(j) → 0 < j0 ∧ i < 0 ∧ 2 ^ (n→i(n) - j0) ≤ >>>(n, i, j0) ∧ j0 ≤ n→i(n) ∧ # int2bits(i) ≤ n → n = 0 ⊦ j = 0, n = 0

Rule argument

seq: ⊦ j ≤ i ∧ 0 ≤ j → (2 ^ i - 1) / 2 ^ j = 2 ^ i / 2 ^ j - 1 subst: [j, i] → [j, n→i(n)] rotate: F spec: int-pot inst: name: twopot-mone-div

Additional information

Used simplifier rules: ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ i < j → ¬ j < i; Used terms: j, n→i(n)