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Proof step 16 from proof byte-mult-neg (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
~128 ≤ (i * 256 + j) % 256, i < 0, j < 0, i * 256 + j < 0 ⊦ (i * 256 + j) % 256 = b2i(i2b(j))

Rule argument

Additional information

Used simplifier rules: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)