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Proof step 7 from proof byte-law-add-pos (previous step)

0 < i % 256, 0 ≤ j, 0 ≤ i, 127 < i % 256 ⊦ i2b(i + j) = i2b(j + i % 256 - 256)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [j + i % 256, 256, i0] rotate: T spec: int-abs inst: name: elim-sub

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: j + i % 256, 256, i0