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Proof step 13 from proof validrefs-has-null (previous step)

acc .first = st[r -- fs '] .refval, is_obj(str, r, st), jvmref ∈ acc, st[r -- _type] .type ≤ ty, fs ' ∈ instfields(st[r -- _type] .type .class), is_referencevalue(st[r -- fs ']), validrefs(str, acc + refs0, fs .type, st), ∀ refs, str0, ty0, st0. refs ≠ [] → ¬ validrefs(str0, acc + refs, ty0, st0) ⊦ r = jvmref, refs0 = []

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 12:
acc = r0 + refs1 ∧ acc ≠ [], r0 = st[r -- fs '] .refval, is_obj(str, r, st), jvmref ∈ acc, st[r -- _type] .type ≤ ty, fs ' ∈ instfields(st[r -- _type] .type .class), is_referencevalue(st[r -- fs ']), validrefs(str, acc + refs0, fs .type, st), ∀ refs, str0, ty0, st0. refs ≠ [] → ¬ validrefs(str0, acc + refs, ty0, st0) ⊦ r = jvmref, refs0 = []

Rule argument

[speclemmaarg: refs ≠ [] ,r = refs .first ∧ refs0 = refs .rest ↔ refs = r + refs0 ,[refs, r, refs0] → [acc, r0, refs1] ,T ,list-data ,references-olista ,elim-list-c ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ r ∈ []], [CSIMPELIM : refs ≠ [] ⊦ r = refs .first ∧ refs0 = refs .rest ↔ refs = r + refs0])], [EXTRATERMS : (acc, r0, refs1)]]