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Proof step 6 from proof validrefs-append-01 (previous step)

refs .last = refs0 .first, validrefs(str, refs0, ty0, st), validrefs(str, refs, ty, st) ⊦ refs = [], refs0 = [], validrefs(str, refs + refs0 .rest, ty, st)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
refs0 = r + refs1, refs .last = r, validrefs(str, refs0, ty0, st), validrefs(str, refs, ty, st) ⊦ refs = [], refs0 = [], validrefs(str, refs + refs1, ty, st)

Rule argument

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

Additional information

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