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Proof step 15 from proof multg-at-ref (previous step)

ints .first = 1, i→n(countrefs2(ints)) ≤ # refs, ∀ 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) ⊦ ints = [], r = r0, (addarraymultlist(r, ty, st, refs, ints, ints0, i)[r0 -- sk]) = st[r0 -- sk], # ints = 1, r0 ∈ refs

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 14:
ints = i0 + ints2, i0 = 1, i→n(countrefs2(ints)) ≤ # refs, ∀ 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) ⊦ ints = [], r = r0, (addarraymultlist(r, ty, st, refs, ints, ints0, i)[r0 -- sk]) = st[r0 -- sk], # ints = 1, 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)]]