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Proof step 8 from proof int2bits-neg-03 (previous step)

bits2int(int2bits(j)) = bits2int(int2bits(2 ^ n→i(# int2bits(~ j)) + j).rest), j < 0 ⊦ int2bits(j) ≠ [] ∧ int2bits(2 ^ n→i(# int2bits(~ j)) + j).rest ≠ [], int2bits(j) = sgn(int2bits(2 ^ n→i(# int2bits(~ j)) + j).rest)

User applied rule case distinction

And obtained the following new goals

Proof step 9:
bits2int(int2bits(j)) = bits2int(int2bits(2 ^ n→i(# int2bits(~ j)) + j).rest), j < 0 ⊦ int2bits(2 ^ n→i(# int2bits(~ j)) + j).rest ≠ [], int2bits(j) = sgn(int2bits(2 ^ n→i(# int2bits(~ j)) + j).rest)

Rule argument

pos: right 1

Additional information

Used simplifier rules: int2bits-not-empty: ⊦ int2bits(j) ≠ [];