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Proof step 20 from proof putarray-addarray (previous step)

0 ≤ j + n→i(# vals), j ≤ i, i ≤ j + n→i(# vals), 0 ≤ j, 0 ≤ i, ∀ st, val, r, j, i, vals0. # vals0 < # vals → ( 0 ≤ i ∧ 0 ≤ j ∧ i ≤ j + n→i(# vals0) ∧ j ≤ i ∧ 0 ≤ j + n→i(# vals0) → putArray(vals0, r, j, addarray(r, val, i, st)) = addarray(r, mklist(val, i→n(j)) + vals0, 0, st)) ⊦ mklist(val, i→n(j)) + vals = [], vals = [], putArray(vals, r, j, addarray(r, val, i, st)) = (addarray(r, mklist(val, i→n(j)) + vals .butlast, 0, st)[r -- (n→i(# vals) + j - 1) ', vals .last])

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 19:
n→i(# vals) + j = i0 + 1, 0 ≤ j + n→i(# vals), j ≤ i, i ≤ j + n→i(# vals), 0 ≤ j, 0 ≤ i, ∀ st, val, r, j, i, vals0. # vals0 < # vals → ( 0 ≤ i ∧ 0 ≤ j ∧ i ≤ j + n→i(# vals0) ∧ j ≤ i ∧ 0 ≤ j + n→i(# vals0) → putArray(vals0, r, j, addarray(r, val, i, st)) = addarray(r, mklist(val, i→n(j)) + vals0, 0, st)) ⊦ mklist(val, i→n(j)) + vals = [], vals = [], putArray(vals, r, j, addarray(r, val, i, st)) = (addarray(r, mklist(val, i→n(j)) + vals .butlast, 0, st)[r -- i0 ', vals .last])

Rule argument

[speclemmaarg: ,i0 = i - j ↔ i = i0 + j ,[i, j, i0] → [n→i(# vals) + j, 1, i0] ,T ,int-abs , ,elim-sub ]

Additional information

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