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Proof step 17 from proof sis-on-bits (previous step)

i = ~1, 16 < # int2bits(i), i ≤ 0, ~1 ≤ i ⊦ i = bits2int(lastn(16, int2bits(i))), 16 = # fillsgn(16, int2bits(i))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
16 < # int2bits(~1) ⊦ 16 = # fillsgn(16, int2bits(~1)), ~1 = bits2int(lastn(16, int2bits(~1)))

Rule argument

Additional information

Used simplifier rules: ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)