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Proof step 12 from proof bytes2bytes-other (previous step)

∀ bytes0. # bytes0 < # bytes → bytes2bits(bytes0) ≠ [] → bytes0 = [] ∨ b2i(bytes0.first) = 0 ∨ b2i(bytes0.first) = ~1 ∨ bits2bytes(sgn(bytes2bits(bytes0))) = bytes0 ⊦ ∀ bytes0, by. bytes ≠ bytes0 + by ', 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 13:
∀ bytes1. # bytes1 < #(bytes0 + by ') → bytes2bits(bytes1) ≠ [] → bytes1 = [] ∨ b2i(bytes1.first) = 0 ∨ b2i(bytes1.first) = ~1 ∨ bits2bytes(sgn(bytes2bits(bytes1))) = bytes1 ⊦ b2i((bytes0 + by ').first) = 0, b2i((bytes0 + by ').first) = ~1, bits2bytes(sgn(bytes2bits(bytes0) + byte2bits(by))) = bytes0 + by '

Rule argument

Additional information

Used simplifier rules: bytes2bits-append: ⊦ bytes2bits(bytes + bytes0) = bytes2bits(bytes) + bytes2bits(bytes0);%bytes2bits-one: ⊦ bytes2bits(by ') = byte2bits(by);% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;