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Proof step 6 from proof uint2bits-add-high-01 (previous step)

0 ≤ i ∧ 0 < j - 1 ∧ 0 ≤ 2 ^ i + j0 ∧ 2 ^ i + j0 < 2 ^ i, 1 < j, 0 < j, j0 < 0, 0 < i, 2 ^ (i - 1) + j0 < 0, 0 ≤ j0 + 2 ^ i ⊦ uint2bits(j - 1) + fillsgn(i→n(i) + 1, uint2bits(2 ^ i + j0)).rest = uint2bits(j - 1) + fillsgn(i→n(i), int2bits(j0).rest)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
1 < j, 0 < j, j0 < 0, 0 < i, 2 ^ (i - 1) + j0 < 0, 0 ≤ j0 + 2 ^ i, 0 ≤ i ⊦ fillsgn(i→n(i) + 1, uint2bits(2 ^ i + j0)).rest = fillsgn(i→n(i), int2bits(j0).rest)

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ i + j < j ↔ i < 0;% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j;% ⊦ i + j = j + i;%i0 ≠ 0, i0 ≤ i ⊦ (* i) < j → (* i0) < j;% ⊦ i < j - i0 ↔ i + i0 < j;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + bits0 = bits + bits1 ↔ bits0 = bits1;