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Proof step 10 from proof hasSharing-different-disjoint (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
refs = r3 + refs0, (r0 ' + refs1) .last = refs .last, validrefs(str, refs, st[r3 -- _type] .type, st), path∃(str, r, r2, st), path∃(str, r, r1, st), validrefs(str, r0 ' + refs1, st[r0 -- _type] .type, st) ⊦ refs .last = jvmref, r0 = r3, refs = [], hasSharing(r, st), path∃(str, r1, r2, st), path∃(str, r2, r1, st), r0 ≠ r1 ∧ r0 ≠ r2, r3 ≠ r1 ∧ r3 ≠ r2

Rule argument

[speclemmaarg: refs ≠ [] ,r = refs .first ∧ refs0 = refs .rest ↔ refs = r + refs0 ,[refs, r, refs0] → [refs, r3, refs0] ,T ,list-data ,references ,elim-list-c ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : refs ≠ [] ⊦ r = refs .first ∧ refs0 = refs .rest ↔ refs = r + refs0])], [EXTRATERMS : (refs, r3, refs0)]]