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Proof step 45 from proof bytes2bytes-mone (previous step)

bytes0 = by0 + bytes, b2i(by0) = ~1, 0 ≤ b2i(bytes.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 = [], bytes = [], true ∈ bytes2bits(bytes0) ∧ false ∈ bytes2bits(bytes0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 46:
b2i(by0) = ~1, 0 ≤ b2i(bytes.first), ∀ bytes1. # bytes1 < #((by0 + bytes) + by ') → b2i(bytes1.first) = ~1 ∧ 0 ≤ b2i(bytes1.rest.first) → bits2bytes(sgn(bytes2bits(bytes1))) = bytes1 ∨ # bytes1 = 1 ∨ bytes1 = [] ⊦ bytes = [], bits2bytes(sgn((byte2bits(by0) + bytes2bits(bytes)) + byte2bits(by))) = (by0 ' + bytes) + by ', (true ∈ byte2bits(by0) ∨ true ∈ bytes2bits(bytes)) ∧ (false ∈ byte2bits(by0) ∨ false ∈ bytes2bits(bytes))

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)% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ by ' + bytes ≠ [];% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ by + bytes = by ' + bytes;