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Proof step 36 from proof multi-refin (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 35:
ints = i0 + ints0, # refs = i→n(countrefs(i ' + ints0)), rk ∈ st, posints(ints), i ≤ i0, 0 ≤ i, 0 ≤ i0, ∀ ty, acc, ints, i, st, rk, refs0. # refs0 < # refs → ( true → ( rk ∈ st ∧ # refs0 = i→n(countrefs(i ' + ints .rest)) ∧ ints ≠ [] ∧ 0 ≤ i ∧ posints(ints) ∧ ¬ 0 ∈ ints .butlast ∧ i ≤ ints .first → rk ∈ addmultiarrays(refs0, i, acc, ty, ints, st))) ⊦ ints = [], rk ∈ addmultiarrays(refs, i, acc, ty, ints, st), 0 ∈ ints .butlast

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)]]