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Proof step 15 from proof cyclic-subset-01 (previous step)

st ⊆ st0, dups(refs), validrefs(str, refs, st[refs .first -- _type] .type, st) ⊦ junkref(str, refs .first, st), cyclic(str, refs .first, st0)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 14:
refs = r + refs0 ∧ refs ≠ [], st ⊆ st0, dups(refs), validrefs(str, refs, st[r -- _type] .type, st) ⊦ junkref(str, r, st), cyclic(str, r, st0)

Rule argument

[speclemmaarg: refs ≠ [] ,r = refs .first ∧ refs0 = refs .rest ↔ refs = r + refs0 ,[refs, r, refs0] → [refs, r, 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, r, refs0)]]