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Proof step 49 from proof junkrefs-notin (previous step)

is_array(str, r0, st), st[r0 -- _type] .type ≤ ty, ∀ str, r, refs0, ty, st. (junkrefs(str, acc + r ' + refs0, ty, st) ↔ junkrefs(str, acc + r ', ty, st)) ∨ r ∈ st ⊦ r0 = jvmref, (∃ i. 0 ≤ i ∧ i < st[r0 -- _length] .intval ∧ is_referencevalue(st[r0 -- i ']) ∧ (acc + r ') .first = st[r0 -- i '] .refval ∧ junkrefs(str, acc + r ' + refs0, st[r0 -- _type] .type .type, st)) ↔ junkrefs(str, (r0 ' + acc) + r ', ty, st), r ∈ st

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 48:
acc + r ' = r1 + refs ∧ acc + r ' ≠ [], is_array(str, r0, st), st[r0 -- _type] .type ≤ ty, ∀ str, r, refs0, ty, st. (junkrefs(str, acc + r ' + refs0, ty, st) ↔ junkrefs(str, acc + r ', ty, st)) ∨ r ∈ st ⊦ r0 = jvmref, (∃ i. 0 ≤ i ∧ i < st[r0 -- _length] .intval ∧ is_referencevalue(st[r0 -- i ']) ∧ r1 = st[r0 -- i '] .refval ∧ junkrefs(str, acc + r ' + refs0, st[r0 -- _type] .type .type, st)) ↔ junkrefs(str, (r0 ' + acc) + r ', ty, st), r ∈ st

Rule argument

[speclemmaarg: refs ≠ [] ,r = refs .first ∧ refs0 = refs .rest ↔ refs = r + refs0 ,[refs, r, refs0] → [acc + r ', r1, refs] ,T ,list-data ,references-olista ,elim-list-c ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ refs + refs0 ≠ [] ↔ refs = [] ∧ refs0 = []], [CSIMPSEQ : ⊦ r ' ≠ []], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPELIM : refs ≠ [] ⊦ r = refs .first ∧ refs0 = refs .rest ↔ refs = r + refs0])], [EXTRATERMS : (acc + r ', r1, refs)]]