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Proof step 4 from proof in-1dim-15 (previous step)

0 ≤ j, j < i, 0 < i, ∀ st, val, r, j, i0. i→n(i0) < i→n(i) → (j < i0 ∧ 0 ≤ j ∧ 0 < i0 → r -- j ' ∈ addarray(r, val, i0, st)) ⊦ i = j + 1, r -- j ' ∈ addarray(r, val, i - 1, st)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
i = i1 + 1, 0 ≤ j, j < i, 0 < i, ∀ st, val, r, j, i0. i→n(i0) < i→n(i) → (j < i0 ∧ 0 ≤ j ∧ 0 < i0 → r -- j ' ∈ addarray(r, val, i0, st)) ⊦ i = j + 1, r -- j ' ∈ addarray(r, val, i1, st)

Rule argument

[speclemmaarg: ,i0 = i - j ↔ i = i0 + j ,[i, j, i0] → [i, 1, i1] ,T ,int-abs , ,elim-sub ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ i0 = i - j ↔ i = i0 + j])], [EXTRATERMS : (i, 1, i1)]]