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Proof step 7 from proof nonnullrefs-in (previous step)

fis0[sk] = refval(r), is_referencevalue(val), sk ∈ fis0, ∀ sk, r. fis0[sk] = refval(r) ∧ sk ∈ fis0 → r = jvmref ∨ r ∈ fis0 .vals .nonnullrefs ⊦ r = jvmref, sk = sk0, val .refval = jvmref, r = val .refval, r ∈ fis0 .vals .nonnullrefs

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
val = refval(r0), fis0[sk] = refval(r), is_referencevalue(val), sk ∈ fis0, ∀ sk, r. fis0[sk] = refval(r) ∧ sk ∈ fis0 → r = jvmref ∨ r ∈ fis0 .vals .nonnullrefs ⊦ r = jvmref, sk = sk0, r0 = jvmref, r = r0, r ∈ fis0 .vals .nonnullrefs

Rule argument

[x0lemmaarg: spec-name: javavalue instance-name: lemma-name: elim-13 antecedent: succedent: is_referencevalue(val) → (r = val .refval ↔ val = refval(r)) substlist: [val, r] → [val, r0] add preconditions?: F is rewrite lemma?: F current spec?: F rotate lemma?: T replace all?: T paths: () dummy: () ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ is_referencevalue(val) → ( r = val .refval ↔ val = refval(r))])], [EXTRATERMS : (val, r0)]]