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Proof step 5 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 6:
i < 0 ⊦ i2b(i + j) = i2b(b2i(i2b(i)) + j)

Rule argument

Additional information

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