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Proof step 10 from proof disjoint-newref-01 (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

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

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, r2, 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, r2, refs0)]]