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Proof step 17 from proof byte-law-add-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
j < 0, 0 ≤ i ⊦ i2b(i + j) = i2b(b2i(i2b(i)) + j)

Rule argument

Additional information

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