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Proof step 12 from proof hasSharing-different-disjoint (previous step)

refs0 .last = refs .last, validrefs(str, refs0, st[refs0 .first -- _type] .type, st), validrefs(str, refs, st[refs .first -- _type] .type, st), path∃(str, r, r2, st), path∃(str, r, r1, st) ⊦ refs .last = jvmref, refs0 .first = refs .first, refs = [], refs0 = [], hasSharing(r, st), path∃(str, r1, r2, st), path∃(str, r2, r1, st), refs0 .first ≠ r1 ∧ refs0 .first ≠ r2, refs .first ≠ r1 ∧ refs .first ≠ r2

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 11:
refs0 = r0 + refs1, refs0 .last = refs .last, validrefs(str, refs0, st[r0 -- _type] .type, st), validrefs(str, refs, st[refs .first -- _type] .type, st), path∃(str, r, r2, st), path∃(str, r, r1, st) ⊦ refs .last = jvmref, r0 = refs .first, refs = [], refs0 = [], hasSharing(r, st), path∃(str, r1, r2, st), path∃(str, r2, r1, st), r0 ≠ r1 ∧ r0 ≠ r2, refs .first ≠ r1 ∧ refs .first ≠ r2

Rule argument

[speclemmaarg: refs ≠ [] ,r = refs .first ∧ refs0 = refs .rest ↔ refs = r + refs0 ,[refs, r, refs0] → [refs0, r0, refs1] ,T ,list-data ,references ,elim-list-c ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : refs ≠ [] ⊦ r = refs .first ∧ refs0 = refs .rest ↔ refs = r + refs0])], [EXTRATERMS : (refs0, r0, refs1)]]