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Proof step 7 from proof addarrays-length-01 (previous step)

1 + n→i(# refs) = countrefs(ints), newref(r, st), is_newref_list(refs, st), posints(ints) ⊦ (addarrays(r ' + refs, ty, ints, 0, st)[r -- _length]) = intval(ints .first), ints = [], r = jvmref, r ∈ refs

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
ints = i + ints0, 1 + n→i(# refs) = countrefs(ints), newref(r, st), is_newref_list(refs, st), posints(ints) ⊦ (addarrays(r ' + refs, ty, ints, 0, st)[r -- _length]) = intval(i), ints = [], r = jvmref, r ∈ refs

Rule argument

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

Additional information

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