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Proof step 11 from proof at-1dim-other-index-01 (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
j = i0 + 1, 0 < j, i < 0 ∨ j ≤ i, ∀ st, val, r, i, j0. i→n(j0) < i→n(j) → (i < 0 ∨ j0 ≤ i → (addarray(r, val, j0, st)[r -- i ']) = st[r -- i ']) ⊦ (addarray(r, val, i0, st)[r -- i0 ', val][r -- i ']) = st[r -- i ']

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)]]