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Proof step 4 from proof bytes2bytes-identity (previous step)

¬(bytes ≠ [] ∧ b2i(bytes.first) ≠ 0 ∧ b2i(bytes.first) ≠ ~1), # bytes = # int2bytes(bytes2int(bytes)) ⊦ int2bytes(bytes2int(bytes)) = bytes

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
# bytes = # int2bytes(bytes2int(bytes)) ⊦ int2bytes(bytes2int(bytes)) = bytes, bytes = [], b2i(bytes.first) ≠ ~1 ∧ b2i(bytes.first) ≠ 0

Rule argument

Additional information

Used simplifier rules: ⊦ # bytes = 0 ↔ bytes = [];% ⊦ # [] = 0;