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Proof step 7 from proof int2bits-rec-2pot-neg-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
0 ≤ i, j < 0, j + 2 ^ i < i, j + 2 ^ i ≤ 0, 0 < i ⊦ sgn((bitsinvert(uint2bits(~(j / 2 ^ i))) + bitsinvert(fillsgn(i→n(i) + 1, uint2bits(~(j % 2 ^ i))).rest))+1) = int2bits(j / 2 ^ i - 1) + fillsgn(i→n(i) + 1, int2bits(j % 2 ^ i)).rest, j % 2 ^ i = 0

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j;% ⊦ i ≤ ~ j ↔ i + j ≤ 0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bitsinvert(bits + bits0) = bitsinvert(bits) + bitsinvert(bits0);%j ≠ 0 ⊦ ~ i % j = ~(i % j);% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ 0 < i → i→n(i + 1) = i→n(i) + 1;%j ≠ 0 ⊦ ~ i / j = ~(i / j);% ⊦ 0 ≤ j → 2 ^ j ≠ 0;