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Proof step 13 from proof validrefs-nodups (previous step)

refs1 .last = jvmref, refs .last = jvmref, validrefs(str, refs1 .first ' + refs, ty, st), validrefs(str, refs1 .first ' + refs0 + refs1 .first ' + refs, ty, st), refs1 ⊆ refs1 .first ' + refs, validrefs(str, refs1, ty, st), ∀ st, ty, str, refs2. # refs2 < # (refs1 .first + (refs0 + (refs1 .first + refs))) → ( validrefs(str, refs2, ty, st) ∧ refs2 ≠ [] ∧ refs2 .last = jvmref → (∃ refs3. validrefs(str, refs3, ty, st) ∧ refs3 ≠ [] ∧ refs3 .last = jvmref ∧ refs2 .first = refs3 .first ∧ ¬ dups(refs3) ∧ refs3 ⊆ refs2)) ⊦ refs1 = [], refs = [], dups(refs1), ∃ refs2. validrefs(str, refs2, ty, st) ∧ refs2 ≠ [] ∧ refs2 .last = jvmref ∧ refs1 .first = refs2 .first ∧ ¬ dups(refs2) ∧ refs2 ⊆ refs2 .first ' + refs0 + refs2 .first ' + refs

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 12:
refs1 = r + refs4, refs1 .last = jvmref, refs .last = jvmref, validrefs(str, r ' + refs, ty, st), validrefs(str, r ' + refs0 + r ' + refs, ty, st), refs1 ⊆ r ' + refs, validrefs(str, refs1, ty, st), ∀ st, ty, str, refs2. # refs2 < # (r + (refs0 + (r + refs))) → ( validrefs(str, refs2, ty, st) ∧ refs2 ≠ [] ∧ refs2 .last = jvmref → (∃ refs3. validrefs(str, refs3, ty, st) ∧ refs3 ≠ [] ∧ refs3 .last = jvmref ∧ refs2 .first = refs3 .first ∧ ¬ dups(refs3) ∧ refs3 ⊆ refs2)) ⊦ refs1 = [], refs = [], dups(refs1), ∃ refs2. validrefs(str, refs2, ty, st) ∧ refs2 ≠ [] ∧ refs2 .last = jvmref ∧ r = refs2 .first ∧ ¬ dups(refs2) ∧ refs2 ⊆ refs2 .first ' + refs0 + refs2 .first ' + refs

Rule argument

[speclemmaarg: refs ≠ [] ,r = refs .first ∧ refs0 = refs .rest ↔ refs = r + refs0 ,[refs, r, refs0] → [refs1, r, refs4] ,T ,list-data ,references ,elim-list-c ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : refs ≠ [] ⊦ r = refs .first ∧ refs0 = refs .rest ↔ refs = r + refs0])], [EXTRATERMS : (refs1, r, refs4)]]