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Proof step 24 from proof int2bytes-byte-neg (previous step)

i < ~1, ~256 < i, i < ~128, i < 0, # int2bits(i) ≤ 9 ⊦ i2b(i + 2 ^ (n→i(# int2bits(i)) - 1)) = i2b(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
~256 < i, i < ~128, i < 0, # int2bits(i) ≤ 9 ⊦ i2b(i + 2 ^ (n→i(# int2bits(i)) - 1)) = i2b(i)

Rule argument

Additional information

Used simplifier rules: j0 ≠ 0, j < j0 ⊦ i < (* j) → i < (* j0);