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Proof step 24 from proof byte-law-add-one (previous step)

0 < j0, j0 < 256, j < 0, 0 ≤ j0, 0 ≤ i0, 127 < j0 ⊦ i2b(j0 + j) = i2b(j + j0 - 256)

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [j + j0, 256, i] rotate: T spec: int-abs inst: name: elim-sub

Additional information

Used simplifier rules: byte-add-256: ⊦ i2b(i + 256) = i2b(i);% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: j + j0, 256, i