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Proof step 61 from proof getsetShort-theorem (previous step)

ub2i(i2b(j)) < 256, ~32768 ≤ j, j < 0, 0 ≤ ub2i(i2b(j)), j / 256 ≤ 128, ~127 ≤ j / 256 ⊦ ub2i(i2b(j)) + 256 * (j / 256 - 1) = j, j % 256 = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 62:
j / 256 = i + 1, ub2i(i2b(j)) < 256, ~32768 ≤ j, j < 0, 0 ≤ ub2i(i2b(j)), j / 256 ≤ 128, ~127 ≤ j / 256 ⊦ ub2i(i2b(j)) + 256 * i = j, j % 256 = 0

Rule argument

spec: int-abs inst: name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [j / 256, 1, i] flags:

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: j / 256, 1, i