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Proof step 10 from proof pathex-subset (previous step)

st ⊆ st0, validrefs(str, refs, mkclasstype(`java.lang.Object' .class), st), validref(str, refs .first, ty, st) ⊦ refs = [], path∃(str, refs .first, refs .last, st0)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
refs = r + refs0, st ⊆ st0, validrefs(str, refs, mkclasstype(`java.lang.Object' .class), st), validref(str, r, ty, st) ⊦ refs = [], path∃(str, r, refs .last, st0)

Rule argument

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

Additional information

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