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Proof step 10 from proof okclass-03 (previous step)

is_referencevalue(val) ⊦ sk = _type, (val .refval = jvmref ∨ val .refval ∈ st[r0 -- sk, val0]) ∧ (val .refval ≠ jvmref → val .refval -- _type ∈ st[r0 -- sk, val0] ∧ st[val .refval -- _type] = typeval(mkclasstype(class1))) ↔ (val .refval = jvmref ∨ val .refval ∈ st) ∧ (val .refval ≠ jvmref → val .refval -- _type ∈ st ∧ st[val .refval -- _type] = typeval(mkclasstype(class1)))

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
val = refval(r), is_referencevalue(val) ⊦ sk = _type, (r = jvmref ∨ r ∈ st[r0 -- sk, val0]) ∧ (r ≠ jvmref → r -- _type ∈ st[r0 -- sk, val0] ∧ st[r -- _type] = typeval(mkclasstype(class1))) ↔ (r = jvmref ∨ r ∈ st) ∧ (r ≠ jvmref → r -- _type ∈ st ∧ st[r -- _type] = typeval(mkclasstype(class1)))

Rule argument

[speclemmaarg: is_referencevalue(val) ,r = val .refval ↔ val = refval(r) ,[val, r] → [val, r] ,T ,javavalue , ,elim-refval ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : is_referencevalue(val) ⊦ r = val .refval ↔ val = refval(r)])], [EXTRATERMS : (val, r)]]