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Proof step 82 from proof pathex-put-nopathex (previous step)

0 ≤ i, 0 < (st[r0 -- sk, val][r -- _length]) .intval, is_array(str, r, st[r0 -- sk, val]), i < (st[r0 -- sk, val][r -- _length]) .intval, is_referencevalue(st[r0 -- sk, val][r -- i ']), validrefs(str, (st[r0 -- sk, val][r -- i ']) .refval ' + refs1, st[r -- _type] .type .type, st[r0 -- sk, val]), ∀ val0, sk0, r3, st0, r2, str, refs. # refs < # ((st[r0 -- sk, val][r -- i ']) .refval + refs1) → ( validrefs(str, r2 ' + refs, st0[r2 -- _type] .type, st0[r3 -- sk0, val0]) → sk0 = _type ∨ path∃(str, r2, r3, st0) ∨ junkref(str, r2, st0) ∨ path∃(str, r2, (r2 ' + refs) .last, st0)) ⊦ sk = _type, r = jvmref, junkref(str, r, st), path∃(str, r, r0, st), path∃(str, r, ((st[r0 -- sk, val][r -- i ']) .refval ' + refs1) .last, st)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 81:
(st[r0 -- sk, val][r -- i ']) = refval(r1), 0 ≤ i, 0 < (st[r0 -- sk, val][r -- _length]) .intval, is_array(str, r, st[r0 -- sk, val]), i < (st[r0 -- sk, val][r -- _length]) .intval, is_referencevalue(st[r0 -- sk, val][r -- i ']), validrefs(str, r1 ' + refs1, st[r -- _type] .type .type, st[r0 -- sk, val]), ∀ val0, sk0, r3, st0, r2, str, refs. # refs < # (r1 + refs1) → ( validrefs(str, r2 ' + refs, st0[r2 -- _type] .type, st0[r3 -- sk0, val0]) → sk0 = _type ∨ path∃(str, r2, r3, st0) ∨ junkref(str, r2, st0) ∨ path∃(str, r2, (r2 ' + refs) .last, st0)) ⊦ sk = _type, r = jvmref, junkref(str, r, st), path∃(str, r, r0, st), path∃(str, r, (r1 ' + refs1) .last, st)

Rule argument

[speclemmaarg: ,is_referencevalue(val) → (r = val .refval ↔ val = refval(r)) ,[val, r] → [st[r0 -- sk, val][r -- i '], r1] ,T ,javavalue , ,elim-13 ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ is_referencevalue(val) → ( r = val .refval ↔ val = refval(r))])], [EXTRATERMS : (st[r0 -- sk, val][r -- i '], r1)]]