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Proof step 6 from proof sis-on-bits (previous step)

16 ≠ # fillsgn(16, int2bits(i)), ~32768 ≤ i, i ≤ 32767 ⊦ i = bits2int(lastn(16, fillsgn(16, int2bits(i))))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information