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Proof step 11 from proof byte-mult-256 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
j < 0, 0 < i ⊦ i = 0, i2b(i * 256 + j) = i2b(j)

Rule argument

Additional information

Used simplifier rules: ⊦ i + j = j + i;% ⊦ i * j = j * i;% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ ¬ i < j ↔ j ≤ i;