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Proof step 18 from proof bytes2bytes-other (previous step)

¬ bytes0 ≠ [], ∀ bytes1. # bytes1 < #(bytes0 + by ') → bytes2bits(bytes1) ≠ [] → bytes1 = [] ∨ b2i(bytes1.first) = 0 ∨ b2i(bytes1.first) = ~1 ∨ bits2bytes(sgn(bytes2bits(bytes1))) = bytes1 ⊦ bits2bytes(sgn(bytes2bits(bytes0) + byte2bits(by))) = bytes0 + by ', b2i((bytes0 + by ').first) = ~1, b2i((bytes0 + by ').first) = 0, true ∈ bytes2bits(bytes0) ∧ false ∈ bytes2bits(bytes0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
∀ bytes1. # bytes1 < #([] + by ') → bytes2bits(bytes1) ≠ [] → bytes1 = [] ∨ b2i(bytes1.first) = 0 ∨ b2i(bytes1.first) = ~1 ∨ bits2bytes(sgn(bytes2bits(bytes1))) = bytes1 ⊦ bits2bytes(sgn(byte2bits(by))) = by ', b2i(by) = ~1, b2i(by) = 0

Rule argument

Additional information

Used simplifier rules: bytes2bits-empty: ⊦ bytes2bits([]) = [];% ⊦ ¬ boolvar ∈ [];% ⊦ by '.first = by;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ [] + bytes = bytes;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ [] + bits = bits;