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Proof step 16 from proof bytes2bytes-mone (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
b2i(bytes0.first) = ~1, bytes0.rest = [], 0 ≤ b2i(by), ∀ bytes1. # bytes1 < #(bytes0 + by ') → b2i(bytes1.first) = ~1 ∧ 0 ≤ b2i(bytes1.rest.first) → bits2bytes(sgn(bytes2bits(bytes1))) = bytes1 ∨ # bytes1 = 1 ∨ bytes1 = [] ⊦ bytes0 = [], bits2bytes(sgn(bytes2bits(bytes0) + byte2bits(by))) = bytes0 + by ', true ∈ bytes2bits(bytes0) ∧ false ∈ bytes2bits(bytes0)

Rule argument

Additional information

Used simplifier rules: ⊦ by '.first = by;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ [] + bytes = bytes;% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)