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Proof step 14 from proof unchanged-pathex-append (previous step)

validrefs(str, refs, mkclasstype(`java.lang.Object' .class), st), unchanged(str, refs .first, st, st1), validrefs(str, refs .last ' + refs0, mkclasstype(`java.lang.Object' .class), st), r1 = refs .last ∨ r1 ∈ refs0 ⊦ refs = [], path∃(str, refs .last, (refs .last ' + refs0) .last, st1), unchangedNorec(r1, st, st1)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 13:
refs = r + refs1, validrefs(str, refs, mkclasstype(`java.lang.Object' .class), st), unchanged(str, r, st, st1), validrefs(str, refs .last ' + refs0, mkclasstype(`java.lang.Object' .class), st), r1 = refs .last ∨ r1 ∈ refs0 ⊦ refs = [], path∃(str, refs .last, (refs .last ' + refs0) .last, st1), unchangedNorec(r1, st, st1)

Rule argument

[x0lemmaarg: spec-name: list-data instance-name: references lemma-name: elim-list-c antecedent: refs ≠ [] succedent: r = refs .first ∧ refs0 = refs .rest ↔ refs = r + refs0 substlist: [refs, r, refs0] → [refs, r, refs1] add preconditions?: F is rewrite lemma?: F current spec?: F rotate lemma?: T replace all?: T paths: () dummy: () ]

Additional information

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