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Proof step 49 from proof bytes2bytes-zero (previous step)

b2i(by) = 0, b2i(by0) < 0, ∀ by0, by, bytes0. # bytes0 < # bytes → b2i(by) = 0 ∧ b2i(by0) < 0 → bits2bytes(sgn(byte2bits(by) + byte2bits(by0) + bytes2bits(bytes0))) = by ' + by0 ' + bytes0 ⊦ ∀ bytes0, by. bytes ≠ bytes0 + by ', bits2bytes(sgn(byte2bits(by) + byte2bits(by0) + bytes2bits(bytes))) = by ' + by0 ' + bytes

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 50:
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) + byte2bits(by1))) = by ' + by0 ' + bytes0 + by1 '

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;