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Proof step 1 from proof int2bits-restsgn (conclusion)

¬(∃ i. 2 ^ i = j), 0 ≤ j ⊦ sgn(int2bits(2 ^ n→i(# int2bits(j)) - j).rest) = int2bits(2 ^ n→i(# int2bits(j)) - j).rest

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information