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Proof step 1 from proof int2bits-add-high (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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