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Proof step 22 from proof validrefs-postfix (previous step)

acc + r0 ' = st[r -- i '] .refval ' + refs1, 0 < st[r -- _length] .intval, 0 ≤ i, validrefs(str, acc + r0 ' + refs, st[r -- _type] .type .type, st), is_referencevalue(st[r -- i ']), i < st[r -- _length] .intval, st[r -- _type] .type ≤ ty, is_array(str, r, st), ∀ str, refs0, st, ty. ¬ validrefs(str, refs0, st[refs0 .first -- _type] .type, st) ∧ refs0 ≠ [] → ¬ validrefs(str, acc + refs0, ty, st) ⊦ r = jvmref, validrefs(str, r0 ' + refs, st[r0 -- _type] .type, st), r0 ≠ st[r -- i '] .refval ∧ ¬ r0 ∈ refs1, ¬ st[r -- i '] .refval ∈ acc ∧ st[r -- i '] .refval ≠ r0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 21:
st[r -- i '] = refval(r1), acc + r0 ' = r1 ' + refs1, 0 < st[r -- _length] .intval, 0 ≤ i, validrefs(str, acc + r0 ' + refs, st[r -- _type] .type .type, st), is_referencevalue(st[r -- i ']), i < st[r -- _length] .intval, st[r -- _type] .type ≤ ty, is_array(str, r, st), ∀ str, refs0, st, ty. ¬ validrefs(str, refs0, st[refs0 .first -- _type] .type, st) ∧ refs0 ≠ [] → ¬ validrefs(str, acc + refs0, ty, st) ⊦ r = jvmref, validrefs(str, r0 ' + refs, st[r0 -- _type] .type, st), r0 ≠ r1 ∧ ¬ r0 ∈ refs1, ¬ r1 ∈ acc ∧ r1 ≠ r0

Rule argument

[speclemmaarg: ,is_referencevalue(val) → (r = val .refval ↔ val = refval(r)) ,[val, r] → [st[r -- i '], r1] ,T ,javavalue , ,elim-13 ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ is_referencevalue(val) → ( r = val .refval ↔ val = refval(r))])], [EXTRATERMS : (st[r -- i '], r1)]]