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Proof step 62 from proof multg-at-ref (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 61:
ints = i0 + ints2, i→n(countrefs2(ints)) ≤ # refs, 2 ≤ i0, 0 ≤ i0, 0 < i0, ∀ sk, i, ints0, st, ty, r0, r, refs, ints1. ints#(ints1) < ints#(ints) → (i→n(countrefs2(ints1)) ≤ # refs → r = r0 ∨ (addarraymultlist(r, ty, st, refs, ints1, ints0, i)[r0 -- sk]) = st[r0 -- sk] ∨ r0 ∈ refs) ⊦ r = r0, # ints = 1, (addarraymultlist (refs .first, ty, addarraymultlist (r, ty, st[r -- (i0 - 1) ', refval(refs .first)][r -- _type, typeval(mktype_from_dims(ty, i + n→i(# ints)))][r -- _length, intval(ints0 .first)], refs .rest, (i0 - 1) ' + ints2, ints0, i), restn(i→n(countrefs2((i0 - 1) ' + ints2)), refs .rest), ints2, ints0 .rest, i) [r0 -- sk ] ) = st[r0 -- sk], ints = [], r0 ∈ refs

Rule argument

[speclemmaarg: ints ≠ [] ,i = ints .first ∧ ints0 = ints .rest ↔ ints = i + ints0 ,[ints, i, ints0] → [ints, i0, ints2] ,T ,list-data ,intlista ,elim-list-c ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ints ≠ [] ⊦ i = ints .first ∧ ints0 = ints .rest ↔ ints = i + ints0])], [EXTRATERMS : (ints, i0, ints2)]]