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Proof step 9 from proof int2bits-len-neg-02 (previous step)

¬(¬(∃ i. 2 ^ i = ~ j) ∧ 0 ≤ ~ j), j < 0 ⊦ # int2bits(~ j - 1) = # int2bits(~ j), ∃ i. 2 ^ i + j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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