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Proof step 31 from proof int2bits-rec-2pot-neg-02 (previous step)

¬ 0 < ~(j % 2 ^ i), 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 32:
i→n(i) < # uint2bits(~(j % 2 ^ i)), 0 < i, j + 2 ^ i ≤ 0, j + 2 ^ i < i, j < 0, 0 ≤ i, j + 2 ^ i < j % 2 ^ i, 0 < j % 2 ^ 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: ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ 0 < ~ i ↔ i < 0;