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Proof step 19 from proof getsetShort-theorem (previous step)

j < 0, 0 ≤ ub2i(i2b(j)), ub2i(i2b(j)) < 256, ~32768 ≤ j ⊦ 256 * b2i(i2b((j - 255) / 256)) + ub2i(i2b(j)) = j, ((j - 255) / 256)∈byte

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 20:
j = i + 255, j < 0, 0 ≤ ub2i(i2b(j)), ub2i(i2b(j)) < 256, ~32768 ≤ j ⊦ 256 * b2i(i2b(i / 256)) + ub2i(i2b(j)) = j, (i / 256)∈byte

Rule argument

spec: int-abs inst: name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [j, 255, i] flags:

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: j, 255, i