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Proof step 6 from proof nonnullrefs-in (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
fis0[sk] = refval(r), 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

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : javavalue, , test-19, ⊦ is_referencevalue(refval(r))])]]