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Proof step 18 from proof bits2bits-sgn (previous step)

8 < # bits, ∀ bits0. # bits0 < # bits → true → bits0 = [] ∨ sgn(bytes2bits(bits2bytes(bits0))) = sgn(bits0) ⊦ sgn(bytes2bits(bits2bytes(firstn(# bits - 8, bits)) + i2b(bits2int(lastn(8, bits))))) = sgn(bits), bits = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: bytes2bits-append: ⊦ bytes2bits(bytes + bytes0) = bytes2bits(bytes) + bytes2bits(bytes0);%bytes2bits-one: ⊦ bytes2bits(by ') = byte2bits(by);% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ bytes + by = bytes + by ';