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Proof step 17 from proof nonnullrefs-in-key (previous step)

val = refval(r0), is_referencevalue(val), r ∈ fis0 .vals .nonnullrefs → fis0[sk] = refval(r) ∧ sk ∈ fis0 ⊦ r = jvmref, r0 = jvmref, dups(fis0 .keys), sk0 ∈ fis0 .keys, ∃ sk. ((sk0 × val) ' + fis0)[sk] = refval(r) ∧ (sk = sk0 ∨ sk ∈ fis0), r ≠ r0 ∧ ¬ r ∈ fis0 .vals .nonnullrefs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
r ∈ fis0 .vals .nonnullrefs → fis0[sk] = refval(r) ∧ sk ∈ fis0 ⊦ r = jvmref, r0 = jvmref, dups(fis0 .keys), sk0 ∈ fis0 .keys, ∃ sk. ((sk0 × refval(r0)) ' + fis0)[sk] = refval(r) ∧ (sk = sk0 ∨ sk ∈ fis0), r ≠ r0 ∧ ¬ r ∈ fis0 .vals .nonnullrefs

Rule argument

[emptyarg]

Additional information

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