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Proof step 2 from proof byte-law-compl (previous step)

⊦ i2b(~(i + 1)) = i2b(~(b2i(i2b(i)) + 1)), i = b2i(i2b(i))

User applied rule cut formula

And obtained the following new goals

Proof step 3:
~(i + 1) = ~1 - i ⊦ i2b(~(i + 1)) = i2b(~(b2i(i2b(i)) + 1)), i = b2i(i2b(i))

Rule argument

fma: ~(i + 1) = ~1 - i

Additional information

Used simplifier rules: ⊦ ~ i = j ↔ i + j = 0;% ⊦ i + (j - i) = j;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i; Used formulas: ~(i + 1) = ~1 - i