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Proof step 8 from proof byte-mult-256 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
j < 0 ⊦ i = 0, i2b(i * 256 + j) = i2b(j)

Rule argument

Additional information

Used simplifier rules: byte-equal: ⊦ b2i(byi) = b2i(byj) ↔ byi = byj;% ⊦ i + j = j + i;% ⊦ i * j = j * i;% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i < 0 ∧ 0 ≤ j → i < j;