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Proof step 14 from proof at=get (previous step)

i < j, 0 ≤ i, 1 < j, 0 < j, ∀ r, st, i, j0. i→n(j0) < i→n(j) → (true → (0 ≤ i ∧ i < j0 → st[r -- i '] = (getarray(r, 0, j0, st)[i→n(i)]))) ⊦ st[r -- i '] = (getarray(r, 0, j - 1, st) + st[r -- (j - 1) '] ')[i→n(i)]

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 13:
j = i0 + 1, i < j, 0 ≤ i, 1 < j, 0 < j, ∀ r, st, i, j0. i→n(j0) < i→n(j) → (true → (0 ≤ i ∧ i < j0 → st[r -- i '] = (getarray(r, 0, j0, st)[i→n(i)]))) ⊦ st[r -- i '] = (getarray(r, 0, i0, st) + st[r -- i0 '] ')[i→n(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)]]