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Proof step 10 from proof disjoint-newref (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

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

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, r0, refs0] 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, r0, refs0)]]