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Proof step 12 from proof bytes2int-int2bytes-len-eq (previous step)

¬ 255 < bytes2int(bytes.butlast) * 256 + band(b2i(bytes.last), 255), # bytes = # int2bytes(bytes2int(bytes.butlast) * 256 + band(b2i(bytes.last), 255)), ∀ bytes0. # bytes0 < # bytes → true → # bytes0 = # int2bytes(bytes2int(bytes0)) → bytes0 = int2bytes(bytes2int(bytes0)) ⊦ bytes = int2bytes(bytes2int(bytes.butlast) * 256 + band(b2i(bytes.last), 255)), # bytes = 1, bytes = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
# bytes = # int2bytes(bytes2int(bytes.butlast) * 256 + band(b2i(bytes.last), 255)), bytes2int(bytes.butlast) * 256 + band(b2i(bytes.last), 255) ≤ 255, ∀ bytes0. # bytes0 < # bytes → true → # bytes0 = # int2bytes(bytes2int(bytes0)) → bytes0 = int2bytes(bytes2int(bytes0)) ⊦ bytes = int2bytes(bytes2int(bytes.butlast) * 256 + band(b2i(bytes.last), 255)), # bytes = 1, bytes = []

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ i < j ↔ j ≤ i;