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Proof step 42 from proof validrefs-addobj-junkref (previous step)

newref(r, st), validrefs(str, r1 ' + acc, ty0, addobj(r, class, fis, st)), r ∈ acc, ∀ str0, ty, r0, class0, fis0, st0. validrefs(str0, acc, ty, addobj(r0, class0, fis0, st0)) ∧ newref(r0, st0) ∧ r0 ∈ acc → junkref(str0, acc .first, st0) ⊦ r = r1, r = jvmref, junkref(str, r1, st)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 41:
acc = r2 + refs ∧ acc ≠ [], newref(r, st), validrefs(str, r1 ' + acc, ty0, addobj(r, class, fis, st)), r ∈ acc, ∀ str0, ty, r0, class0, fis0, st0. validrefs(str0, acc, ty, addobj(r0, class0, fis0, st0)) ∧ newref(r0, st0) ∧ r0 ∈ acc → junkref(str0, r2, st0) ⊦ r = r1, r = jvmref, junkref(str, r1, st)

Rule argument

[speclemmaarg: refs ≠ [] ,r = refs .first ∧ refs0 = refs .rest ↔ refs = r + refs0 ,[refs, r, refs0] → [acc, r2, refs] ,T ,list-data ,references-olista ,elim-list-c ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ refs + [] = refs], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ ¬ r ∈ []], [CSIMPELIM : refs ≠ [] ⊦ r = refs .first ∧ refs0 = refs .rest ↔ refs = r + refs0])], [EXTRATERMS : (acc, r2, refs)]]