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Proof step 35 from proof byte-law-righthift-not (previous step)

# int2bits(~128) = n0 + i→n(bits2int(bitsand(int2bits(1), int2bits(31)))) ∧ i→n(bits2int(bitsand(int2bits(1), int2bits(31)))) ≤ # int2bits(~128), # int2bits(128) = n + i→n(bits2int(bitsand(int2bits(1), int2bits(31)))), i2b(bits2int(sublist(0, n, int2bits(128)))) = i2b(bits2int(sublist(0, n0, int2bits(~128)))), i→n(bits2int(bitsand(int2bits(1), int2bits(31)))) < # int2bits(~128), i→n(bits2int(bitsand(int2bits(1), int2bits(31)))) < # int2bits(128) ⊦ n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m < m + n ↔ n ≠ 0;% ⊦ m < n ∧ n = n0 → m < n0; (forward)% ⊦ m < n → m ≤ n;