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Proof step 76 from proof byte-mult-256 (previous step)

i0 + i = ~1, j0 < 256, 0 ≤ j0, 0 ≤ i0, i < 0 ⊦ i = 0, i2b((i0 + i + 1) * 256 + j0 - 256) = i2b(j0)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 77:
(i0 + i + 1) * 256 + j0 = i1 + 256, i0 + i = ~1, j0 < 256, 0 ≤ j0, 0 ≤ i0, i < 0 ⊦ i = 0, i2b(i1) = i2b(j0)

Rule argument

seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [(i0 + i + 1) * 256 + j0, 256, i1] rotate: T spec: int-abs inst: name: elim-sub

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: (i0 + i + 1) * 256 + j0, 256, i1