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Proof step 16 from proof cyclic-array (previous step)

st[r -- i '] = refval(refs .first), dups(refs), validrefs(str, refs, st[refs .first -- _type] .type, st), is_array(str, r, st), i < st[r -- _length] .intval, is_reftype(st[r -- _type] .type .type), st[refs .first -- _type] .type ≤ st[r -- _type] .type .type, 0 ≤ i, 0 < st[r -- _length] .intval ⊦ cyclic(str, r, st)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 15:
refs = r0 + refs0 ∧ refs ≠ [], st[r -- i '] = refval(r0), dups(refs), validrefs(str, refs, st[r0 -- _type] .type, st), is_array(str, r, st), i < st[r -- _length] .intval, is_reftype(st[r -- _type] .type .type), st[r0 -- _type] .type ≤ st[r -- _type] .type .type, 0 ≤ i, 0 < st[r -- _length] .intval ⊦ cyclic(str, r, st)

Rule argument

[speclemmaarg: refs ≠ [] ,r = refs .first ∧ refs0 = refs .rest ↔ refs = r + refs0 ,[refs, r, refs0] → [refs, r0, refs0] ,T ,list-data ,references-olista ,elim-list-c ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ dups([])], [CSIMPELIM : refs ≠ [] ⊦ r = refs .first ∧ refs0 = refs .rest ↔ refs = r + refs0])], [EXTRATERMS : (refs, r0, refs0)]]