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Proof step 50 from proof junkrefs-subset (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 49:
st1[r -- i '] = refval(r0), st ⊆ st1, st[r -- _type] .type ≤ ty, 0 ≤ i, 0 < st1[r -- _length] .intval, st1[r -- _type] .type ≤ ty, is_array(str, r, st), i < st1[r -- _length] .intval, is_referencevalue(st1[r -- i ']), junkrefs(str, r0 ' + refs, st1[r -- _type] .type .type, st1), ∀ str0, ty0, st0, st2. junkrefs(str0, r0 ' + refs, ty0, st0) ∧ st2 ⊆ st0 → junkrefs(str0, r0 ' + refs, ty0, st2) ⊦ r = jvmref, ∃ i0. ¬ i0 < 0 ∧ ¬ st[r -- _length] .intval ≤ i0 ∧ is_referencevalue(st[r -- i0 ']) ∧ r0 = st[r -- i0 '] .refval ∧ junkrefs(str, r0 ' + refs, st[r -- _type] .type .type, st)

Rule argument

[speclemmaarg: ,is_referencevalue(val) → (r = val .refval ↔ val = refval(r)) ,[val, r] → [st1[r -- i '], r0] ,T ,javavalue , ,elim-13 ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ is_referencevalue(val) → ( r = val .refval ↔ val = refval(r))])], [EXTRATERMS : (st1[r -- i '], r0)]]