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Proof step 4 from proof int2bits-len-neg-02 (previous step)

j < 0, 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 5:
j < 0 ⊦ # int2bits(~(j + 1)) = # int2bits(~ j), ∃ i. 2 ^ i + j = 0

Rule argument

Additional information

Used simplifier rules: ⊦ ~ i - j = ~(i + j);