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Proof step 21 from proof validrefs-not-in (previous step)

acc .first = st[r0 -- i '] .refval, is_array(str, r0, st), st[r0 -- _type] .type ≤ ty, i < st[r0 -- _length] .intval, is_referencevalue(st[r0 -- i ']), validrefs(str, acc, st[r0 -- _type] .type .type, st), 0 ≤ i, 0 < st[r0 -- _length] .intval, ∀ r, str, ty, st. r ∈ acc ∧ validrefs(str, acc, ty, st) → r = jvmref ∨ is_valid_ref(r, st) ∧ type∃(str, st[r -- _type] .type) ⊦ acc = [], r0 = jvmref, r = jvmref, r ≠ r0 ∧ ¬ r ∈ acc, is_valid_ref(r, st) ∧ type∃(str, st[r -- _type] .type)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 20:
acc = r1 + refs, r1 = st[r0 -- i '] .refval, is_array(str, r0, st), st[r0 -- _type] .type ≤ ty, i < st[r0 -- _length] .intval, is_referencevalue(st[r0 -- i ']), validrefs(str, acc, st[r0 -- _type] .type .type, st), 0 ≤ i, 0 < st[r0 -- _length] .intval, ∀ r, str, ty, st. r ∈ acc ∧ validrefs(str, acc, ty, st) → r = jvmref ∨ is_valid_ref(r, st) ∧ type∃(str, st[r -- _type] .type) ⊦ acc = [], r0 = jvmref, r = jvmref, r ≠ r0 ∧ ¬ r ∈ acc, is_valid_ref(r, st) ∧ type∃(str, st[r -- _type] .type)

Rule argument

[speclemmaarg: refs ≠ [] ,r = refs .first ∧ refs0 = refs .rest ↔ refs = r + refs0 ,[refs, r, refs0] → [acc, r1, refs] ,T ,list-data ,references-olista ,elim-list-c ]

Additional information

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