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Proof step 1 from proof byte-law-add-pos (conclusion)

⊦ 0 ≤ i ∧ 0 ≤ j → i2b(i + j) = i2b(b2i(i2b(i)) + j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
0 ≤ i, 0 ≤ j ⊦ i2b(i + j) = i2b(b2i(i2b(i)) + j)

Rule argument

Additional information

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