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Proof step 60 from proof bytes2int-append (previous step)

¬ abs(bits2int(bytes2bits(bytes0))) < 2 ^ (8 * n→i(# bytes0)), bytes2bits(bytes).first, 0 ≤ bits2int(bytes2bits(bytes0)) ⊦ bytes = [], bytes0 = [], bytes2bits(bytes) + bytes2bits(bytes0) = [], bitsinvert(bytes2bits(bytes)) + bitsinvert(bytes2bits(bytes0)) = [], bits2int(bytes2bits(bytes0)) % 2 ^ (8 * n→i(# bytes0)) = bits2int(bytes2bits(bytes0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 61:
bytes2bits(bytes).first, 0 ≤ bits2int(bytes2bits(bytes0)), 2 ^ (8 * n→i(# bytes0)) ≤ bits2int(bytes2bits(bytes0)) ⊦ bytes = [], bytes0 = [], bytes2bits(bytes) + bytes2bits(bytes0) = [], bitsinvert(bytes2bits(bytes)) + bitsinvert(bytes2bits(bytes0)) = [], bits2int(bytes2bits(bytes0)) % 2 ^ (8 * n→i(# bytes0)) = bits2int(bytes2bits(bytes0))

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ i < j ↔ j ≤ i;% ⊦ 0 ≤ i → abs(i) = i;