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Proof step 11 from proof getarray-addarray-05 (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

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