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Proof step 134 from proof hasSharing-dd-help (previous step)

validrefs(str, r ' + refs0, st[r -- _type] .type, st), validrefs(str, r2 ' + refs2 + r0 ', st[r2 -- _type] .type, st), validrefs(str, refs0 .last ' + refs1 + r0 ', st[refs0 .last -- _type] .type, st), path∃(str, r, r2, st) ⊦ refs0 = [], r2 = refs0 .last, r0 = jvmref, refs0 .last = r0, path∃(str, refs0 .last, r2, st), path∃(str, r2, refs0 .last, st), hasSharing(r, st)

User applied rule insert elim lemma

And obtained the following new goals

Proof step 133:
refs0 = refs + r1 ', validrefs(str, r ' + refs0, st[r -- _type] .type, st), validrefs(str, r2 ' + refs2 + r0 ', st[r2 -- _type] .type, st), validrefs(str, r1 ' + refs1 + r0 ', st[r1 -- _type] .type, st), path∃(str, r, r2, st) ⊦ refs0 = [], r2 = r1, r0 = jvmref, r1 = r0, path∃(str, r1, r2, st), path∃(str, r2, r1, st), hasSharing(r, st)

Rule argument

[speclemmaarg: refs ≠ [] ,r = refs .last ∧ refs0 = refs .butlast ↔ refs = refs0 + r ' ,[refs, r, refs0] → [refs0, r1, refs] ,T ,list-last ,references ,last-elim ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : refs ≠ [] ⊦ r = refs .last ∧ refs0 = refs .butlast ↔ refs = refs0 + r '])], [EXTRATERMS : (refs0, r1, refs)]]