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Proof step 9 from proof int2bytes-minus-one (previous step)

# int2bits(i) ≤ 8, i2b(bits2int(int2bits(i))) ' = i2b(~1) ' ⊦ i = ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
i2b(i) = i2b(~1), # int2bits(i) ≤ 8 ⊦ i = ~1

Rule argument

Additional information

Used simplifier rules: ⊦ by ' = by0 ' ↔ by = by0;% ⊦ bits2int(int2bits(i)) = i;