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Proof step 34 from proof byte-law-righthift-not (previous step)

# int2bits(128) = n + i→n(bits2int(bitsand(int2bits(1), int2bits(31)))), i2b(bits2int(sublist(0, n, int2bits(128)))) = i2b(bits2int(sublist(0, # int2bits(~128) - i→n(bits2int(bitsand(int2bits(1), int2bits(31)))), 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 elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [# int2bits(~128), i→n(bits2int(bitsand(int2bits(1), int2bits(31)))), n0] rotate: T spec: nat inst: name: elim-sub

Additional information

Used simplifier rules: ⊦ ¬ m ≤ n ↔ n < m;% ⊦ m < n → ¬ n < m;%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: # int2bits(~128), i→n(bits2int(bitsand(int2bits(1), int2bits(31)))), n0