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Proof step 23 from proof bytes2bytes-other (previous step)

int2bits(b2i(by)) ≠ [], ∀ bytes1. # bytes1 < #([] + by ') → bytes2bits(bytes1) ≠ [] → bytes1 = [] ∨ b2i(bytes1.first) = 0 ∨ b2i(bytes1.first) = ~1 ∨ bits2bytes(sgn(bytes2bits(bytes1))) = bytes1 ⊦ bits2bytes(sgn(int2bits(b2i(by)))) = by ', b2i(by) = ~1, b2i(by) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
∀ bytes1. # bytes1 < #([] + by ') → bytes2bits(bytes1) ≠ [] → bytes1 = [] ∨ b2i(bytes1.first) = 0 ∨ b2i(bytes1.first) = ~1 ∨ bits2bytes(sgn(bytes2bits(bytes1))) = bytes1 ⊦ b2i(by) = ~1, b2i(by) = 0, bits2bytes(int2bits(b2i(by))) = by '

Rule argument

Additional information

Used simplifier rules: ⊦ int2bits(j) ≠ [];% ⊦ sgn(int2bits(i)) = int2bits(i);