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Proof step 67 from proof disjoint-nojunk-newref-subset (previous step)

(r2 ' + refs1) .last = refs .last, validrefs(str, refs, st1[refs .first -- _type] .type, st1), st ⊆ st1, newref(r, st), validrefs(str, r2 ' + refs1, st1[r2 -- _type] .type, st1), ∀ r0. path∃(str, r, r0, st1) → newref(r0, st) ∨ r0 = jvmref ⊦ r = jvmref, refs .last = jvmref, r2 = refs .first, refs = [], junkref(str, r1, st), r2 ≠ r ∧ r2 ≠ r1, refs .first ≠ r ∧ refs .first ≠ r1

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 66:
refs = r3 + refs0, (r2 ' + refs1) .last = refs .last, validrefs(str, refs, st1[r3 -- _type] .type, st1), st ⊆ st1, newref(r, st), validrefs(str, r2 ' + refs1, st1[r2 -- _type] .type, st1), ∀ r0. path∃(str, r, r0, st1) → newref(r0, st) ∨ r0 = jvmref ⊦ r = jvmref, refs .last = jvmref, r2 = r3, refs = [], junkref(str, r1, st), r2 ≠ r ∧ r2 ≠ r1, r3 ≠ r ∧ r3 ≠ 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, r3, 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, r3, refs0)]]