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Proof step 66 from proof bytes2bytes-mone (previous step)

8 ≠ # fillsgn(8, int2bits(b2i(by))), b2i(bytes0.first) = ~1, 0 ≤ b2i((bytes0.rest + by ').first), false ∈ bytes2bits(bytes0), true ∈ bytes2bits(bytes0), ∀ 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))) + i2b(bits2int(lastn(8, sgn(bytes2bits(bytes0)) + fillsgn(8, int2bits(b2i(by)))))) ' = bytes0 + by '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 67:
b2i(bytes0.first) = ~1, true ∈ bytes2bits(bytes0), false ∈ bytes2bits(bytes0), 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 = [] ⊦ 8 = # fillsgn(8, int2bits(b2i(by))), bits2bytes(sgn(bytes2bits(bytes0))) + i2b(bits2int(lastn(8, sgn(bytes2bits(bytes0)) + fillsgn(8, int2bits(b2i(by)))))) ' = bytes0 + by ', bytes0 = []

Proof step 66 from proof bytes2bytes-mone (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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