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Proof step 91 from proof hassharing-validrefs-dups (previous step)

dups(refs0), path∃(str, r, r, st), validrefs(str, r ' + refs0, st[r -- _type] .type, st), validrefs(str, refs0, st[refs0 .first -- _type] .type, st), ∀ st, str, refs, r0. # (r0 ' + refs) < # (r ' + refs0) → (validrefs(str, r0 ' + refs, st[r0 -- _type] .type, st) → hasSharing(r0, st) ∨ ¬ r0 ∈ refs ∧ ¬ dups(refs)) ⊦ refs0 = [], refs0 .first ∈ refs0 .rest, dups(refs0 .rest), hasSharing(r, st), r ∈ refs0

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

[speclemmaarg: refs ≠ [] ,r = refs .first ∧ refs0 = refs .rest ↔ refs = r + refs0 ,[refs, r, refs0] → [refs0, r1, refs1] ,T ,list-data ,references-olista ,elim-list-c ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r + refs = r ' + refs], [CSIMPSEQ : ⊦ r ' + refs ≠ []], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ ¬ r ∈ refs → (dups(r ' + refs) ↔ dups(refs))], [CSIMPSEQ : ⊦ ¬ r ∈ r0 ' + refs ↔ r ≠ r0 ∧ ¬ r ∈ refs], [CSIMPELIM : refs ≠ [] ⊦ r = refs .first ∧ refs0 = refs .rest ↔ refs = r + refs0])], [EXTRATERMS : (refs0, r1, refs1)]]