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Proof step 24 from proof byte-law-righthift-not (previous step)

# int2bits(~128) ≤ i→n(bits2int(bitsand(int2bits(1), int2bits(31)))), i2b(bits2int(sublist(0, n, int2bits(128)))) = i2b(~1), # int2bits(128) = n + i→n(bits2int(bitsand(int2bits(1), int2bits(31)))), i→n(bits2int(bitsand(int2bits(1), int2bits(31)))) < # int2bits(128) ⊦ n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
i2b(bits2int(sublist(0, n, int2bits(128)))) = i2b(~1), # int2bits(128) = n + i→n(bits2int(bitsand(int2bits(1), int2bits(31)))), # int2bits(~128) ≤ i→n(bits2int(bitsand(int2bits(1), int2bits(31)))), i→n(bits2int(bitsand(int2bits(1), int2bits(31)))) < # int2bits(128), # int2bits(~128) < # int2bits(128) ⊦ n = 0

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ∧ n = n0 → m < n0; (forward)% ⊦ m ≤ n ∧ n < n0 → m < n0; (forward)