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Proof step 71 from proof bytes2bytes-mone (previous step)

int2bits(b2i(by)) ≠ [], 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(int2bits(b2i(by)))) ' = bytes0 + by '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 72:
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))) = bytes0

Rule argument

Additional information

Used simplifier rules: ⊦ int2bits(j) ≠ [];% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ bytes + bytes0 = bytes1 + bytes0 ↔ bytes = bytes1;% ⊦ i2b(b2i(by)) = by;% ⊦ bits2int(int2bits(i)) = i;