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Proof step 7 from proof nonnullrefs-in-key (previous step)

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

User applied rule exists right

And closed the goal!

Rule argument

[EXRARG : pos: right 4, [extquanttermlist: (sk0)abort:F, discard:T, computed:T]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : assoclist, fieldinit, assoc-yes, ⊦ ((sk × val) ' + fis)[sk] = val], [CSIMPNAMEDSEQ : list, fieldinit-assoclista, a, ⊦ (fis + fis0) + fis1 = fis + fis0 + fis1], [CSIMPNAMEDSEQ : javavalue, , inj-03, ⊦ refval(r) = refval(r0) ↔ r = r0])], [EXTRATERMS : (sk0)]]