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Proof step 68 from proof multi-lengthin (previous step)

# refs = i→n(countrefs(i ' + ints .rest)), r ∈ refs, posints(ints), i ≤ ints .first, 0 ≤ i, 0 ≤ ints .first, ∀ st, ty, acc, ints, i, r, refs0. # refs0 < # refs → ( true → ( r ∈ refs0 ∧ # refs0 = i→n(countrefs(i ' + ints .rest)) ∧ ints ≠ [] ∧ 0 ≤ i ∧ posints(ints) ∧ ¬ 0 ∈ ints .butlast ∧ i ≤ ints .first → r -- _length ∈ addmultiarrays(refs0, i, acc, ty, ints, st) ∧ is_integervalue(addmultiarrays(refs0, i, acc, ty, ints, st)[r -- _length]) ∧ 0 ≤ (addmultiarrays(refs0, i, acc, ty, ints, st)[r -- _length]) .intval)) ⊦ ints = [], 0 ∈ ints .butlast, r -- _length ∈ addmultiarrays(refs, i, acc, ty, ints, st) ∧ is_integervalue(addmultiarrays(refs, i, acc, ty, ints, st)[r -- _length]) ∧ ¬ (addmultiarrays(refs, i, acc, ty, ints, st)[r -- _length]) .intval < 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

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

Additional information

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