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Proof step 18 from proof int2bits-len-range (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
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

Rule argument

Additional information

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