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Proof step 9 from proof validrefs-nodups-01 (previous step)

validrefs(str, refs, ty, st), dups(refs), ∀ st, ty, str, refs0. # refs0 < # refs → ( validrefs(str, refs0, ty, st) ∧ refs0 ≠ [] → (∃ refs1. validrefs(str, refs1, ty, st) ∧ refs1 ≠ [] ∧ refs1 .last = refs0 .last ∧ refs0 .first = refs1 .first ∧ ¬ dups(refs1))) ⊦ refs = [], ∃ refs0. validrefs(str, refs0, ty, st) ∧ refs0 ≠ [] ∧ refs0 .last = refs .last ∧ refs .first = refs0 .first ∧ ¬ dups(refs0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ dups(refs) ↔ (∃ r, refs0, refs1, refs2. refs = refs0 + (r + (refs1 + (r + refs2))))
Proof step 8:
validrefs(str, refs, ty, st), ∃ r, refs0, refs1, refs2. refs = refs0 + (r + (refs1 + (r + refs2))), ∀ st, ty, str, refs0. # refs0 < # refs → ( validrefs(str, refs0, ty, st) ∧ refs0 ≠ [] → (∃ refs1. validrefs(str, refs1, ty, st) ∧ refs1 ≠ [] ∧ refs1 .last = refs0 .last ∧ refs0 .first = refs1 .first ∧ ¬ dups(refs1))) ⊦ refs = [], ∃ refs0. validrefs(str, refs0, ty, st) ∧ refs0 ≠ [] ∧ refs0 .last = refs .last ∧ refs .first = refs0 .first ∧ ¬ dups(refs0)

Rule argument

[speclemmaarg: , dups(refs) ↔ (∃ r, refs0, refs1, refs2. refs = refs0 + (r + (refs1 + (r + refs2)))) ,[refs] → [refs] ,F ,list-dup ,references ,dups ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (refs)]]