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Proof step 23 from proof nonnullrefs-dups (previous step)

((sk1 × val) ' + fis0)[sk0] = refval(r), ((sk1 × val) ' + fis0)[sk] = refval(r), sk0 ∈ (sk1 × val) ' + fis0, sk ∈ (sk1 × val) ' + fis0, ∀ sk0, r, sk. fis0[sk0] = refval(r) ∧ fis0[sk] = refval(r) ∧ sk0 ∈ fis0 ∧ sk ∈ fis0 → sk = sk0 ∨ r = jvmref ∨ dups(fis0 .vals .nonnullrefs) ⊦ sk = sk0, r = jvmref, dups((val ' + fis0 .vals) .nonnullrefs)

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

[x0lemmaarg: spec-name: valuesfuns instance-name: lemma-name: nonnullrefs-no antecedent: succedent: ¬ is_referencevalue(val) ∨ val .refval = jvmref → (val ' + vals) .nonnullrefs = vals .nonnullrefs substlist: [vals, val] → [fis0 .vals, val] add preconditions?: F is rewrite lemma?: T current spec?: F rotate lemma?: F replace all?: T paths: ((1, 0, 2, 1, 0)) dummy: () ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (fis0 .vals, val)]]