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Proof step 10 from proof byte-mult-neg (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
0 ≤ i * 256 + j, j < 0, i < 0 ⊦ i2b(i * 256 + j) = i2b(j)

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j;