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Proof step 19 from proof nonnullrefs-in-key (previous step)

¬ (¬ is_referencevalue(val) ∨ val .refval = jvmref), r ∈ fis0 .vals .nonnullrefs → fis0[sk] = refval(r) ∧ 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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
is_referencevalue(val), r ∈ fis0 .vals .nonnullrefs → fis0[sk] = refval(r) ∧ sk ∈ fis0 ⊦ r = jvmref, val .refval = jvmref, dups(fis0 .keys), sk0 ∈ fis0 .keys, ∃ sk. ((sk0 × val) ' + fis0)[sk] = refval(r) ∧ (sk = sk0 ∨ sk ∈ fis0), r ≠ val .refval ∧ ¬ r ∈ fis0 .vals .nonnullrefs

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : list, references, in-17, ⊦ r ∈ r0 ' ↔ r = r0], [CSIMPNAMEDSEQ : valuesfuns, , nonnullrefs-one-yes, ⊦ is_referencevalue(val) ∧ val .refval ≠ jvmref → val ' .nonnullrefs = val .refval '])]]