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Proof step 20 from proof nonnullrefs-dups-key (previous step)

fis0[sk] = refval(r1), sk ∈ fis0 ⊦ r1 = jvmref, dups(fis0 .keys), sk3 ∈ fis0 .keys, dups(fis0 .vals .nonnullrefs), ∃ r0, sk1, sk2. sk1 ≠ sk2 ∧ r0 ≠ jvmref ∧ ((sk3 × refval(r1)) ' + fis0)[sk1] = refval(r0) ∧ ((sk3 × refval(r1)) ' + fis0)[sk2] = refval(r0) ∧ (sk1 = sk3 ∨ sk1 ∈ fis0) ∧ (sk2 = sk3 ∨ sk2 ∈ fis0)

User applied rule exists right

And obtained the following new goals

Rule argument

[EXRARG : pos: right 5, [extquanttermlist: (r1, sk3, sk) abort:F , discard:F , 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], [CSIMPNAMEDSEQ : assoclist, fieldinit, assoc-no, ⊦ sk ≠ sk0 → ((sk0 × val) ' + fis)[sk] = fis[sk]])], [EXTRATERMS : (r1, sk3, sk)]]