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Proof step 80 from proof validref-refs-last-valid-subset (previous step)

acc + r1 ' = st0[r0 -- i '] .refval ' + refs, 0 < st0[r0 -- _length] .intval, 0 ≤ i, validrefs(str, acc + r1 ', st0[r0 -- _type] .type .type, st0), is_referencevalue(st0[r0 -- i ']), i < st0[r0 -- _length] .intval, is_array(str, r0, st0), st ⊆ st0, validref(str, r, st[r -- _type] .type, st), st0[r0 -- _type] .type ≤ ty, ∀ str, r, st, st0, r1, ty. validref(str, r, st[r -- _type] .type, st) ∧ st ⊆ st0 ∧ r ∈ acc ∧ validrefs(str, acc + r1 ', ty, st0) → validref(str, r1, st[r1 -- _type] .type, st) ⊦ r0 = jvmref, validref(str, r1, st[r1 -- _type] .type, st), r ≠ r0 ∧ ¬ r ∈ acc, r1 ≠ st0[r0 -- i '] .refval ∧ ¬ r1 ∈ refs, ¬ st0[r0 -- i '] .refval ∈ acc ∧ st0[r0 -- i '] .refval ≠ r1

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

[speclemmaarg: ,is_referencevalue(val) → (r = val .refval ↔ val = refval(r)) ,[val, r] → [st0[r0 -- i '], r2] ,T ,javavalue , ,elim-13 ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ is_referencevalue(val) → ( r = val .refval ↔ val = refval(r))])], [EXTRATERMS : (st0[r0 -- i '], r2)]]