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Proof step 5 from proof bytes2bytes-mone (previous step)

1 < # bytes, b2i(bytes.first) = ~1, 0 ≤ b2i(bytes.rest.first) ⊦ bits2bytes(int2bits(bits2int(bytes2bits(bytes)))) = bytes

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
b2i(bytes.first) = ~1, 0 ≤ b2i(bytes.rest.first) ⊦ bits2bytes(sgn(bytes2bits(bytes))) = bytes, # bytes = 1, bytes = []

Rule argument

Additional information

Used simplifier rules: bytes2bits-empty-01: ⊦ bytes2bits(bytes) = [] ↔ bytes = [];% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);%bits ≠ [] ⊦ int2bits(bits2int(bits)) = sgn(bits);% ⊦ # [] = 0;