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Proof step 13 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 + fillsgn(i→n(i) + 1, uint2bits(~(j % 2 ^ i))).rest) = int2bits(j / 2 ^ i - 1) + fillsgn(i→n(i) + 1, int2bits(j % 2 ^ i)).rest, true ∈ fillsgn(i→n(i) + 1, uint2bits(~(j % 2 ^ i))).rest

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
0 ≤ i, j < 0, j + 2 ^ i < i, j + 2 ^ i ≤ 0, 0 < i, # uint2bits(~(j % 2 ^ i)) ≤ i→n(i) ⊦ j % 2 ^ i = 0, sgn(bitsinvert(uint2bits(~(j / 2 ^ i))) +1 + fillsgn(i→n(i) + 1, uint2bits(~(j % 2 ^ i))).rest) = int2bits(j / 2 ^ i - 1) + fillsgn(i→n(i) + 1, int2bits(j % 2 ^ i)).rest, true ∈ fillsgn(i→n(i) + 1, uint2bits(~(j % 2 ^ i))).rest

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ ¬ m < n ↔ n ≤ m;% ⊦ m + 1 ≤ n ↔ m < n;