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Proof step 58 from proof multg-at-ref (previous step)

0 ≤ i0, 0 < i0, 2 ≤ i0, i→n(i0 * (countrefs2(ints2) + 1)) ≤ 1 + # refs0, ∀ sk, i, ints0, st, ty, r0, r, refs, ints1. ints#(ints1) < ints#(i0 + ints2) → (i→n(countrefs2(ints1)) ≤ # refs → r = r0 ∨ (addarraymultlist(r, ty, st, refs, ints1, ints0, i)[r0 -- sk]) = st[r0 -- sk] ∨ r0 ∈ refs) ⊦ r = r0, ints2 = [], (addarraymultlist (r1, ty, addarraymultlist (r, ty, st[r -- (i0 - 1) ', refval(r1)][r -- _type, typeval(mktype_from_dims(ty, (1 + n→i(# ints2)) + i))][r -- _length, intval(ints0 .first)], refs0, (i0 - 1) ' + ints2, ints0, i), restn(i→n(countrefs2((i0 - 1) ' + ints2)), refs0), ints2, ints0 .rest, i) [r0 -- sk ] ) = st[r0 -- sk], r0 = r1, r0 ∈ refs0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 57:
i0 = i1 + 1, 0 ≤ i0, 0 < i0, 2 ≤ i0, i→n(i0 * (countrefs2(ints2) + 1)) ≤ 1 + # refs0, ∀ sk, i, ints0, st, ty, r0, r, refs, ints1. ints#(ints1) < ints#(i0 + ints2) → (i→n(countrefs2(ints1)) ≤ # refs → r = r0 ∨ (addarraymultlist(r, ty, st, refs, ints1, ints0, i)[r0 -- sk]) = st[r0 -- sk] ∨ r0 ∈ refs) ⊦ r = r0, ints2 = [], (addarraymultlist (r1, ty, addarraymultlist (r, ty, st[r -- i1 ', refval(r1)][r -- _type, typeval(mktype_from_dims(ty, (1 + n→i(# ints2)) + i))][r -- _length, intval(ints0 .first)], refs0, i1 ' + ints2, ints0, i), restn(i→n(countrefs2(i1 ' + ints2)), refs0), ints2, ints0 .rest, i) [r0 -- sk ] ) = st[r0 -- sk], r0 = r1, r0 ∈ refs0

Rule argument

[speclemmaarg: ,i0 = i - j ↔ i = i0 + j ,[i, j, i0] → [i0, 1, i1] ,T ,int-abs , ,elim-sub ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ i0 = i - j ↔ i = i0 + j])], [EXTRATERMS : (i0, 1, i1)]]