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

⊦ j < 0 ∧ j0 < 0 ∧ abs(j0) < 2 ^ i ∧ 0 ≤ i → # int2bits(~(j * 2 ^ i + j0)) = # int2bits(~ j) + i→n(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
0 ≤ i, j < 0, j0 < 0, 0 < 2 ^ i + j0 ⊦ # int2bits(~(j * 2 ^ i + j0)) = # int2bits(~ j) + i→n(i)

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 < j + i;% ⊦ i < 0 → abs(i) = ~ i;