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Proof step 46 from proof junkrefs-put-nopathex-equiv (previous step)

0 ≤ i, 0 < st[r1 -- _length] .intval, is_array(str, r1, st), st[r1 -- _type] .type ≤ ty, is_valid_ref(str, r0, st), i < st[r1 -- _length] .intval, is_referencevalue(st[r1 -- i ']), junkrefs(str, st[r1 -- i '] .refval ' + refs, st[r1 -- _type] .type .type, st), ∀ str0, r, st0, sk0, ty0, val0. ¬ path∃(str0, st[r1 -- i '] .refval, r, st0) ∧ is_valid_ref(str0, r, st0) ∧ sk0 ≠ _type → (junkrefs(str0, st[r1 -- i '] .refval ' + refs, ty0, st0[r -- sk0, val0]) ↔ junkrefs(str0, st[r1 -- i '] .refval ' + refs, ty0, st0)) ⊦ r0 = r1, r1 = jvmref, sk = _type, path∃(str, r1, r0, st), junkrefs(str, r1 ' + st[r1 -- i '] .refval ' + refs, ty, st[r0 -- sk, val])

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 45:
st[r1 -- i '] = refval(r2), 0 ≤ i, 0 < st[r1 -- _length] .intval, is_array(str, r1, st), st[r1 -- _type] .type ≤ ty, is_valid_ref(str, r0, st), i < st[r1 -- _length] .intval, is_referencevalue(st[r1 -- i ']), junkrefs(str, r2 ' + refs, st[r1 -- _type] .type .type, st), ∀ str0, r, st0, sk0, ty0, val0. ¬ path∃(str0, r2, r, st0) ∧ is_valid_ref(str0, r, st0) ∧ sk0 ≠ _type → (junkrefs(str0, r2 ' + refs, ty0, st0[r -- sk0, val0]) ↔ junkrefs(str0, r2 ' + refs, ty0, st0)) ⊦ r0 = r1, r1 = jvmref, sk = _type, path∃(str, r1, r0, st), junkrefs(str, r1 ' + r2 ' + refs, ty, st[r0 -- sk, val])

Rule argument

[speclemmaarg: ,is_referencevalue(val) → (r = val .refval ↔ val = refval(r)) ,[val, r] → [st[r1 -- i '], r2] ,T ,javavalue , ,elim-13 ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ is_referencevalue(val) → ( r = val .refval ↔ val = refval(r))])], [EXTRATERMS : (st[r1 -- i '], r2)]]