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Proof step 34 from proof cyclic-put-nopathex (previous step)

r1 ∈ refs, validrefs(str, refs + r1 ', st[refs .first -- _type] .type, st) ⊦ sk = _type, path∃(str, refs .first, r0, st), junkref(str, refs .first, st), dups(refs), ∃ refs0, r1. ¬ dups(refs0) ∧ r1 ∈ refs0 ∧ refs0 .first = refs .first ∧ validrefs(str, refs0 + r1 ', st[refs0 .first -- _type] .type, st[r0 -- sk, val])

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 33:
refs = r + refs1 ∧ refs ≠ [], r1 ∈ refs, validrefs(str, refs + r1 ', st[r -- _type] .type, st) ⊦ sk = _type, path∃(str, r, r0, st), junkref(str, r, st), dups(refs), ∃ refs0, r1. ¬ dups(refs0) ∧ r1 ∈ refs0 ∧ refs0 .first = r ∧ validrefs(str, refs0 + r1 ', st[refs0 .first -- _type] .type, st[r0 -- sk, val])

Rule argument

[speclemmaarg: refs ≠ [] ,r = refs .first ∧ refs0 = refs .rest ↔ refs = r + refs0 ,[refs, r, refs0] → [refs, r, refs1] ,T ,list-data ,references-olista ,elim-list-c ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ r ∈ []], [CSIMPELIM : refs ≠ [] ⊦ r = refs .first ∧ refs0 = refs .rest ↔ refs = r + refs0])], [EXTRATERMS : (refs, r, refs1)]]