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Proof step 10 from proof bytes2bytes-mone (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
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 = [] ⊦ bytes0 = [], bits2bytes(sgn(bytes2bits(bytes0) + byte2bits(by))) = bytes0 + by '

Rule argument

Additional information

Used simplifier rules: bytes2bits-append: ⊦ bytes2bits(bytes + bytes0) = bytes2bits(bytes) + bytes2bits(bytes0);%bytes2bits-one: ⊦ bytes2bits(by ') = byte2bits(by);% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ bytes ≠ [] → (bytes + bytes0).rest = bytes.rest + bytes0;% ⊦ bytes ≠ [] → (bytes + bytes0).first = bytes.first;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ # by ' = 0 +1;% ⊦ #(bytes + bytes0) = # bytes + # bytes0;