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Proof step 10 from proof in-1dim-other-index-01 (previous step)

0 < j, r -- i ' ∈ (addarray(r, val, j - 1, st)[r -- (j - 1) ', val]), i < 0 ∨ j ≤ i, ∀ st, val, r, i, j0. i→n(j0) < i→n(j) → (true → (i < 0 ∨ j0 ≤ i → (r -- i ' ∈ addarray(r, val, j0, st) ↔ r -- i ' ∈ st))) ⊦ r -- i ' ∈ st

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
j = i0 + 1, 0 < j, r -- i ' ∈ (addarray(r, val, i0, st)[r -- i0 ', val]), i < 0 ∨ j ≤ i, ∀ st, val, r, i, j0. i→n(j0) < i→n(j) → (true → (i < 0 ∨ j0 ≤ i → (r -- i ' ∈ addarray(r, val, j0, st) ↔ r -- i ' ∈ st))) ⊦ r -- i ' ∈ st

Rule argument

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

Additional information

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