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Proof step 18 from proof get-get (previous step)

1 < j, n→i(n) < j, ∀ j0, n, r, i, st. i→n(j0) < i→n(j) → (n→i(n) < j0 → (getarray(r, i, j0, st)[n]) = st[r -- (i + n→i(n)) ']) ⊦ (getarray(r, i, j - 1, st) + st[r -- (i + j - 1) '] ')[n] = st[r -- (i + n→i(n)) ']

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 17:
j = i0 + 1, 1 < j, n→i(n) < j, ∀ j0, n, r, i, st. i→n(j0) < i→n(j) → (n→i(n) < j0 → (getarray(r, i, j0, st)[n]) = st[r -- (i + n→i(n)) ']) ⊦ (getarray(r, i, i0, st) + st[r -- (i + j - 1) '] ')[n] = st[r -- (i + n→i(n)) ']

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)]]