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Proof step 60 from proof validrefs-subset-01 (previous step)

0 ≤ i, 0 < st[r0 -- _length] .intval, st0[r0 -- _type] .type ≤ ty, is_array(str, r0, st), st ⊆ st0, st[r0 -- _type] .type ≤ ty, i < st[r0 -- _length] .intval, is_referencevalue(st[r0 -- i ']), validrefs(str, st[r0 -- i '] .refval ' + refs, st[r0 -- _type] .type .type, st), ∀ str0, ty0, st1, st2. validrefs(str0, st[r0 -- i '] .refval ' + refs, ty0, st1) ∧ st1 ⊆ st2 ∧ ¬ junkref(str0, st[r0 -- i '] .refval, st1) → validrefs(str0, st[r0 -- i '] .refval ' + refs, ty0, st2) ⊦ r0 = jvmref, junkref(str, r0, st), ∃ i0. ¬ i0 < 0 ∧ ¬ st0[r0 -- _length] .intval ≤ i0 ∧ is_referencevalue(st0[r0 -- i0 ']) ∧ st[r0 -- i '] .refval = st0[r0 -- i0 '] .refval ∧ validrefs(str, st[r0 -- i '] .refval ' + refs, st0[r0 -- _type] .type .type, st0)

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] → [st[r0 -- i '], r] ,T ,javavalue , ,elim-13 ]

Additional information

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