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Proof step 20 from proof int2bits-len-range (previous step)

j = ~(2 ^ i0), i→n(i) + 1 < # int2bits(~(j + 1)), 0 < 2 ^ i0, j < 0, 2 ^ i0 + j ≤ i0, i0 ≤ i, j < i0, 0 < 2 ^ i, j + 1 < 2 ^ i0, 2 ^ i0 + j < 2 ^ i, 0 ≤ i, 2 ^ i0 + j ≤ i, j < 2 ^ i, j + 1 < 2 ^ i, j < i, j < 2 ^ i0, 0 < 2 ^ i0 * 2 + j, 2 ^ i0 + j ≤ 0, 0 ≤ i0, 0 ≤ 2 ^ i + j ⊦ j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
0 ≤ 2 ^ i0 + 2 ^ i, 0 ≤ 2 ^ i0 + 2 ^ i0, 0 < i + 2 ^ i0, 1 < 2 ^ i0 + 2 ^ i, 0 ≤ i, 1 < 2 ^ i0 + 2 ^ i0, 0 < 2 ^ i, 0 < i0 + 2 ^ i0, i0 ≤ i, 0 ≤ i0, 0 < 2 ^ i0, i→n(i) + 1 < # int2bits(~(1 - 2 ^ i0)) ⊦

Rule argument

Additional information

Used simplifier rules: int2bits-not-empty: ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ (* i) < j ∧ 0 < i → 0 ≤ j ∧ 0 < j; (forward)% ⊦ 0 ≤ i → (2 ^ i ≤ 2 ^ j ↔ i ≤ j);% ⊦ i + 0 = i;% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 1 * i = i;% ⊦ i * j - i = i * (j - 1);% ⊦ 0 < i ∧ 0 < j → 0 < i + j;% ⊦ i - j < i0 ↔ i < j + i0;% ⊦ ~ i < j ↔ 0 < j + i;% ⊦ ~ i + i = 0;% ⊦ ~ i < 0 ↔ 0 < i;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + ~ j = i - j;% ⊦ 0 ≤ j → 2 ^ j ≠ 0;% ⊦ ~ i = 0 ↔ i = 0;