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Proof step 70 from proof bytes2bytes-zero (previous step)

8 = # byte2bits(by1), b2i(by) = 0, b2i(by0) < 0, ∀ by0, by, bytes1. # bytes1 < #(bytes0 + by1 ') → b2i(by) = 0 ∧ b2i(by0) < 0 → bits2bytes(sgn(byte2bits(by) + byte2bits(by0) + bytes2bits(bytes1))) = by ' + by0 ' + bytes1 ⊦ bits2bytes(sgn(byte2bits(by) + byte2bits(by0) + bytes2bits(bytes0))) + i2b(bits2int(byte2bits(by1))) ' = by ' + by0 ' + bytes0 + by1 ', byte2bits(by) + byte2bits(by0) + bytes2bits(bytes0) = [], ¬ true ∈ byte2bits(by) ∧ ¬ true ∈ byte2bits(by0) ∧ ¬ true ∈ bytes2bits(bytes0), ¬ false ∈ byte2bits(by) ∧ ¬ false ∈ byte2bits(by0) ∧ ¬ false ∈ bytes2bits(bytes0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 71:
b2i(by) = 0, b2i(by0) < 0, ∀ by0, by, bytes1. # bytes1 < #(bytes0 + by1 ') → b2i(by) = 0 ∧ b2i(by0) < 0 → bits2bytes(sgn(byte2bits(by) + byte2bits(by0) + bytes2bits(bytes1))) = by ' + by0 ' + bytes1 ⊦ byte2bits(by) + byte2bits(by0) + bytes2bits(bytes0) = [], bits2bytes(sgn(byte2bits(by) + byte2bits(by0) + bytes2bits(bytes0))) = by ' + by0 ' + bytes0, ¬ true ∈ byte2bits(by) ∧ ¬ true ∈ byte2bits(by0) ∧ ¬ true ∈ bytes2bits(bytes0), ¬ false ∈ byte2bits(by) ∧ ¬ false ∈ byte2bits(by0) ∧ ¬ false ∈ bytes2bits(bytes0)

Rule argument

Additional information

Used simplifier rules: byte2bits-length: ⊦ # byte2bits(by) = 8;%byte2bits-value: ⊦ bits2int(byte2bits(by)) = b2i(by);% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ bytes + bytes0 = bytes1 + bytes0 ↔ bytes = bytes1;% ⊦ i2b(b2i(by)) = by;