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Proof step 9 from proof int2bits-len-abs (previous step)

j < 0, # int2bits(j) + 1 < # int2bits(~ j) ⊦ 2 ^ i + j ≠ 0 ∧ ¬ 0 ≤ j, j ≠ 2 ^ i0 ∧ ¬ j ≤ 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
2 ^ i + j = 0, # int2bits(j) + 1 < # int2bits(~ j), j < 0 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i < 0 ∧ 0 ≤ j → i ≤ j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);