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Proof step 67 from proof ex-validrefs-fits-stronger (previous step)

validrefs(str, refs, ty, st) ⊦ refs = [], ∃ sks. fits(refs .first, sks, st) ∧ 1 + # sks = # refs ∧ (∀ n. n ≠ 0 ∧ n ≤ # sks → st[refs .first -- firstn(n, sks)] .refval = refs[n])

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 66:
refs = r + refs0, validrefs(str, refs, ty, st) ⊦ refs = [], ∃ sks. fits(r, sks, st) ∧ 1 + # sks = # refs ∧ (∀ n. n ≠ 0 ∧ n ≤ # sks → st[r -- firstn(n, sks)] .refval = refs[n])

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