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Proof step 20 from proof byte-law-righthift-not (previous step)

¬ # int2bits(128) ≤ i→n(bits2int(bitsand(int2bits(1), int2bits(31)))), i2b(bits2int(sublist(0, # int2bits(128) - i→n(bits2int(bitsand(int2bits(1), int2bits(31)))), int2bits(128)))) = i2b ((# int2bits(~128) ≤ i→n(bits2int(bitsand(int2bits(1), int2bits(31)))) ⊃ ~1 ;bits2int(sublist(0, # int2bits(~128) - i→n(bits2int(bitsand(int2bits(1), int2bits(31)))), int2bits(~128))) )) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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