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Proof step 3 from proof nonnullrefs-in (previous step)

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

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : list, references, a, ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPNAMEDSEQ : list, references, in-06, ⊦ ¬ r ∈ r0 ' + refs ↔ r ≠ r0 ∧ ¬ r ∈ refs], [CSIMPNAMEDSEQ : list, javavalues, a, ⊦ (vals + vals0) + vals1 = vals + vals0 + vals1], [CSIMPNAMEDSEQ : valuesfuns, , nonnullrefs-ok, ⊦ r ≠ jvmref → (refval(r) ' + vals) .nonnullrefs = r ' + vals .nonnullrefs], [CSIMPNAMEDSEQ : assoclist, fieldinit, infirst, ⊦ sk ∈ (sk × val) ' + fis], [CSIMPNAMEDSEQ : list, fieldinit-assoclista, a, ⊦ (fis + fis0) + fis1 = fis + fis0 + fis1], [CSIMPNAMEDSEQ : assoclist, fieldinit, assoc-yes, ⊦ ((sk × val) ' + fis)[sk] = val])]]