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Proof step 9 from proof bytes2bytes-other (previous step)

bytes2bits(bytes) ≠ [], ∀ bytes0. # bytes0 < # bytes → bytes2bits(bytes0) ≠ [] → bytes0 = [] ∨ b2i(bytes0.first) = 0 ∨ b2i(bytes0.first) = ~1 ∨ bits2bytes(sgn(bytes2bits(bytes0))) = bytes0 ⊦ bytes = [] ∨ b2i(bytes.first) = 0 ∨ b2i(bytes.first) = ~1 ∨ bits2bytes(sgn(bytes2bits(bytes))) = bytes

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
∀ bytes0. # bytes0 < # bytes → bytes2bits(bytes0) ≠ [] → bytes0 = [] ∨ b2i(bytes0.first) = 0 ∨ b2i(bytes0.first) = ~1 ∨ bits2bytes(sgn(bytes2bits(bytes0))) = bytes0 ⊦ bytes = [], b2i(bytes.first) = 0, b2i(bytes.first) = ~1, bits2bytes(sgn(bytes2bits(bytes))) = bytes

Rule argument

Additional information

Used simplifier rules: bytes2bits-empty-01: ⊦ bytes2bits(bytes) = [] ↔ bytes = [];