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Proof step 10 from proof disjoint-nopathex (previous step)

validrefs(str, refs, st[refs .first -- _type] .type, st), path∃(str, r0, refs .last, st), ∀ refs1, refs0. refs1 = [] ∨ refs0 = [] ∨ refs1 .first ≠ refs .first ∧ refs1 .first ≠ r0 ∨ refs0 .first ≠ refs .first ∧ refs0 .first ≠ r0 ∨ refs1 .first = refs0 .first ∨ refs1 .last ≠ refs0 .last ∨ refs1 .last = jvmref ∨ ¬ validrefs(str, refs1, st[refs1 .first -- _type] .type, st) ∨ ¬ validrefs(str, refs0, st[refs0 .first -- _type] .type, st) ⊦ refs .first = r0, refs .last = jvmref, refs = []

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
refs = r + refs2, validrefs(str, refs, st[r -- _type] .type, st), path∃(str, r0, refs .last, st), ∀ refs1, refs0. refs1 = [] ∨ refs0 = [] ∨ refs1 .first ≠ r ∧ refs1 .first ≠ r0 ∨ refs0 .first ≠ r ∧ refs0 .first ≠ r0 ∨ refs1 .first = refs0 .first ∨ refs1 .last ≠ refs0 .last ∨ refs1 .last = jvmref ∨ ¬ validrefs(str, refs1, st[refs1 .first -- _type] .type, st) ∨ ¬ validrefs(str, refs0, st[refs0 .first -- _type] .type, st) ⊦ r = r0, refs .last = jvmref, refs = []

Rule argument

[speclemmaarg: refs ≠ [] ,r = refs .first ∧ refs0 = refs .rest ↔ refs = r + refs0 ,[refs, r, refs0] → [refs, r, refs2] ,T ,list-data ,references ,elim-list-c ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : refs ≠ [] ⊦ r = refs .first ∧ refs0 = refs .rest ↔ refs = r + refs0])], [EXTRATERMS : (refs, r, refs2)]]