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Proof step 30 from proof nonnullrefs-in-key (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 29:
∀ r0. r0 ∈ fis0 .vals .nonnullrefs → r0 = jvmref ∨ (∃ sk. fis0[sk] = refval(r0) ∧ sk ∈ fis0) ⊦ r = jvmref, dups(fis0 .keys), sk0 ∈ fis0 .keys, ¬ r ∈ val ' .nonnullrefs ∧ ¬ r ∈ fis0 .vals .nonnullrefs, ∃ sk. ((sk0 × val) ' + fis0)[sk] = refval(r) ∧ (sk = sk0 ∨ sk ∈ fis0)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : assoclist, fieldinit, inone, ⊦ sk ∈ (sk0 × val) ' ↔ sk = sk0])]]