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Proof step 18 from proof shiftzero-max-neg-01 (previous step)

1 ≤ n, 0 < j, 2 ^ (n→i(n) - j) ≤ (2 ^ n→i(n) - 1) >> j, j ≤ n→i(n), ∀ 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 19:
⊦ 0 ≤ i ∧ 0 ≤ j → i >> j = i / 2 ^ j
Proof step 20:
0 ≤ 2 ^ n→i(n) - 1 ∧ 0 ≤ j, 1 ≤ n, 0 < j, 2 ^ (n→i(n) - j) ≤ (2 ^ n→i(n) - 1) / 2 ^ j, j ≤ n→i(n), ∀ 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

name: shiftright-pos-div rotate: F subst: [j, i] → [j, 2 ^ n→i(n) - 1]

Additional information

Used simplifier rules: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;% ⊦ 1 ≤ i ↔ 0 < i;% ⊦ 0 < 2 ^ i ↔ 0 ≤ i;% ⊦ 0 ≤ n→i(n);% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i < j → ¬ j < i; Used terms: j, 2 ^ n→i(n) - 1