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Proof step 28 from proof validrefs-addarray (previous step)

acc .first = st[r0 -- i '] .refval, newref(r, st), st[r0 -- _type] .type ≤ ty, is_array(str, r0, st), i < st[r0 -- _length] .intval, is_referencevalue(st[r0 -- i ']), validrefs(str, acc, st[r0 -- _type] .type .type, addarray(r, ty0, vals, st)), 0 ≤ i, 0 < st[r0 -- _length] .intval, ∀ str, ty, r, ty0, vals, st. validrefs(str, acc, ty, addarray(r, ty0, vals, st)) ∧ newref(r, st) → r = jvmref ∨ r ∈ acc ∨ validrefs(str, acc, ty, st) ⊦ r = jvmref, r = r0, r0 = jvmref, acc = [], r ∈ acc, ∃ i0. ¬ i0 < 0 ∧ ¬ st[r0 -- _length] .intval ≤ i0 ∧ is_referencevalue(st[r0 -- i0 ']) ∧ acc .first = st[r0 -- i0 '] .refval ∧ validrefs(str, acc, st[r0 -- _type] .type .type, st)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

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)]]