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Proof step 12 from proof nonnullrefs-dups (previous step)

val = refval(r1), ((sk1 × val) ' + fis0)[sk0] = refval(r), ((sk1 × val) ' + fis0)[sk] = refval(r), is_referencevalue(val), sk0 ∈ (sk1 × val) ' + fis0, sk ∈ (sk1 × val) ' + fis0, ∀ sk2, r0, sk3. fis0[sk2] = refval(r0) ∧ fis0[sk3] = refval(r0) ∧ sk2 ∈ fis0 ∧ sk3 ∈ fis0 → sk3 = sk2 ∨ r0 = jvmref ⊦ sk = sk0, r = jvmref, r1 = jvmref, dups(fis0 .vals .nonnullrefs), r1 ∈ fis0 .vals .nonnullrefs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
((sk1 × refval(r1)) ' + fis0)[sk0] = refval(r), ((sk1 × refval(r1)) ' + fis0)[sk] = refval(r), sk0 ∈ (sk1 × refval(r1)) ' + fis0, sk ∈ (sk1 × refval(r1)) ' + fis0, ∀ sk2, r0, sk3. fis0[sk2] = refval(r0) ∧ fis0[sk3] = refval(r0) ∧ sk2 ∈ fis0 ∧ sk3 ∈ fis0 → sk3 = sk2 ∨ r0 = jvmref ⊦ sk = sk0, r = jvmref, r1 = jvmref, dups(fis0 .vals .nonnullrefs), r1 ∈ fis0 .vals .nonnullrefs

Rule argument

[emptyarg]

Additional information

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