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Proof step 12 from proof junkref-put-norefval (previous step)

junkrefs(str, refs, (st[r1 -- sk, val][refs .first -- _type]) .type, st[r1 -- sk, val]), is_fskey(sk), okval(val, sk .fs .type) ⊦ sk = _length, sk .fs .type = void_type, refs = [], is_referencevalue(val), ∃ refs0. refs0 ≠ [] ∧ refs0 .first = refs .first ∧ junkrefs(str, refs0, st[refs .first -- _type] .type, st)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 11:
refs = r + refs1, junkrefs(str, refs, (st[r1 -- sk, val][r -- _type]) .type, st[r1 -- sk, val]), is_fskey(sk), okval(val, sk .fs .type) ⊦ sk = _length, sk .fs .type = void_type, refs = [], is_referencevalue(val), ∃ refs0. refs0 ≠ [] ∧ refs0 .first = r ∧ junkrefs(str, refs0, st[r -- _type] .type, st)

Rule argument

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

Additional information

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