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Proof step 27 from proof okarray-okarray (previous step)

okarray(val, ty, i, st) ⊦ okarray(val .refval, ty, st)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 1:
⊦ ¬ is_referencevalue(val) → ¬ okarray(val, ty, i, st)
Proof step 26:
val = refval(r) ∧ is_referencevalue(val), okarray(val, ty, i, st) ⊦ okarray(r, ty, st)

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)]]