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Proof step 26 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 ∈ [] ∧ ¬ 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 20:
⊦ fis[sk] = refval(r) ∧ sk ∈ fis ∧ r ≠ jvmref → r ∈ fis .vals .nonnullrefs
Proof step 25:
fis0[sk] = refval(r), sk ∈ fis0, ¬ is_referencevalue(val) ∨ val .refval = jvmref ⊦ r = jvmref, dups(fis0 .keys), sk0 ∈ fis0 .keys, ∃ sk. ((sk0 × val) ' + fis0)[sk] = refval(r) ∧ (sk = sk0 ∨ sk ∈ fis0)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : list, references, in-empty, ⊦ ¬ r ∈ []])]]