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Proof step 26 from proof junkref-subset (previous step)

junkrefs(str, refs, st1[refs .first -- _type] .type, st1), st ⊆ st1 ⊦ refs = [], ∃ 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 25:
refs = r + refs1, junkrefs(str, refs, st1[r -- _type] .type, st1), st ⊆ st1 ⊦ refs = [], ∃ 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)]]