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Proof step 1 from proof uint2bits-add-high-02 (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ ~ i ≤ j ↔ 0 ≤ i + j;% ⊦ i < ~ j ↔ i + j < 0;% ⊦ i < 0 → abs(i) = ~ i;