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Proof step 28 from proof int2bits-rec-2pot-neg-02 (previous step)

i→n(i) + 1 ≤ # uint2bits(~(j % 2 ^ i)), 0 ≤ i, j < 0, j + 2 ^ i < i, j + 2 ^ i ≤ 0, 0 < i ⊦ j % 2 ^ i = 0, sgn(bitsinvert(uint2bits(~(j / 2 ^ i))) +1 + uint2bits(~(j % 2 ^ i)).rest) = int2bits(j / 2 ^ i - 1) + fillsgn(i→n(i) + 1, int2bits(j % 2 ^ i)).rest, true ∈ uint2bits(~(j % 2 ^ i)).rest

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 29:
i→n(i) < # uint2bits(~(j % 2 ^ i)), 0 < i, j + 2 ^ i ≤ 0, j + 2 ^ i < i, j < 0, 0 ≤ i ⊦ sgn(bitsinvert(uint2bits(~(j / 2 ^ i))) +1 + uint2bits(~(j % 2 ^ i)).rest) = int2bits(j / 2 ^ i - 1) + fillsgn(i→n(i) + 1, int2bits(j % 2 ^ i)).rest, j % 2 ^ i = 0, uint2bits(~(j % 2 ^ i)) = [], true ∈ uint2bits(~(j % 2 ^ i)).rest

Rule argument

Additional information

Used simplifier rules: ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + 1 ≤ n ↔ m < n;% ⊦ m + n = n + m;