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Proof step 12 from proof int2bytes-minus-one (previous step)

¬(~128 ≤ i ∧ i ≤ 127 ∧ ~128 ≤ ~1 ∧ ~1 ≤ 127 ∧ i ≠ ~1), i2b(i) = i2b(~1), # int2bits(i) ≤ 8 ⊦ i = ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
i2b(i) = i2b(~1), # int2bits(i) ≤ 8 ⊦ i = ~1, ¬ i < ~128 ∧ ¬ 127 < i

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ i ≤ j ↔ j < i;