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Proof step 8 from proof byte-mult-neg (previous step)

¬(i * 256 + j < 0 ∧ (i * 256 + j) % 256 < ~128), ¬(i * 256 + j < 0 ∧ ~128 ≤ (i * 256 + j) % 256), i < 0, j < 0 ⊦ b2i(i2b(i * 256 + j)) = b2i(i2b(j))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
i < 0, j < 0 ⊦ i2b(i * 256 + j) = i2b(j), ¬ 0 ≤ i * 256 + j ∧ ¬ (i * 256 + j) % 256 < ~128, ¬ 0 ≤ i * 256 + j ∧ ¬ ~128 ≤ (i * 256 + j) % 256

Rule argument

Additional information

Used simplifier rules: byte-equal: ⊦ b2i(byi) = b2i(byj) ↔ byi = byj;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ ¬ i < j ↔ j ≤ i;