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Proof step 8 from proof add (previous step)

0 ≤ j, 0 < i0, ∀ j. 0 ≤ i0 ∧ 0 ≤ j → i→n(i0) + i→n(j) = i→n(i0 + j) ⊦ i→n(i0 - 1) + i→n(j) = i→n(j + i0 - 1)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
i0 = i1 + 1, 0 ≤ j, 0 < i0, ∀ j. 0 ≤ i0 ∧ 0 ≤ j → i→n(i0) + i→n(j) = i→n(i0 + j) ⊦ i→n(i1) + i→n(j) = i→n(j + i0 - 1)

Rule argument

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

Additional information

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