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Proof step 3 from proof byte-law-compl (
previous step
)
~(i + 1) = ~1 - i ⊦ i2b(~(i + 1)) = i2b(~(b2i(i2b(i)) + 1)), i = b2i(i2b(i))
User applied rule insert equation
And obtained the following new goals
Proof step 4
:
⊦ i2b(~1 - i) = i2b(~(b2i(i2b(i)) + 1)), i = b2i(i2b(i))
Rule argument
pos: left 1 flags: discard
Additional information