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Proof step 1 from proof short-law-compl (conclusion)

⊦ i2s(bcompl(i)) = i2s(bcompl(s2i(i2s(i))))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ i2s(~(i + 1)) = i2s(~(s2i(i2s(i)) + 1))

Rule argument

Additional information

Used simplifier rules: ⊦ ~ i - j = ~(i + j);% ⊦ bcompl(i) = ~ i - 1;