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Proof step 6 from proof int2bytes-minus-one (previous step)

¬ # int2bits(i) ≤ 8, bits2bytes(int2bits(i)) = i2b(~1) ' ⊦ i = ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
bits2bytes(int2bits(i)) = i2b(~1) ', 8 < # int2bits(i) ⊦ i = ~1

Rule argument

Additional information

Used simplifier rules: ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ ¬ m ≤ n ↔ n < m;