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Proof step 5 from proof bytes2bytes-length (previous step)

¬(bytes ≠ [] ∧ b2i(bytes.first) ≠ 0 ∧ b2i(bytes.first) ≠ ~1), # bytes < # int2bytes(bytes2int(bytes)), ∀ bytes0. # bytes0 < # bytes → # bytes0 < # int2bytes(bytes2int(bytes0)) → bytes0 = [] ⊦ bytes = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
# bytes < # int2bytes(bytes2int(bytes)), ∀ bytes0. # bytes0 < # bytes → # bytes0 < # int2bytes(bytes2int(bytes0)) → bytes0 = [] ⊦ bytes = [], b2i(bytes.first) ≠ 0 ∧ b2i(bytes.first) ≠ ~1

Proof step 5 from proof bytes2bytes-length (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information