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Proof step 1 from proof byte-law-ls-not (conclusion)

⊦ (i = 127 ∧ j = 128 → ¬(i < j → b2i(i2b(i)) < b2i(i2b(j)))) ∧ (i = 128 ∧ j = 1 → ¬(b2i(i2b(i)) < b2i(i2b(j)) → i < j)) ∧ (i = 255 ∧ j = ~1 → ¬(b2i(i2b(i)) = b2i(i2b(j)) → i = j))

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: num-ibbi: ~128 ≤ i, i ≤ 127 ⊦ b2i(i2b((* i))) = i;%bib-upper: ⊦ 127 < i ∧ i < 256 → b2i(i2b(i)) = i - 256;