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Proof step 16 from proof nonnullrefs-dups-key (previous step)

fis0[sk] = refval(r), fis0[sk0] = refval(r), sk ∈ fis0, sk0 ∈ fis0 ⊦ r1 = jvmref, r = jvmref, sk = sk0, dups(fis0 .keys), sk3 ∈ fis0 .keys, ∃ 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

Proof step 15:
fis0[sk] = refval(r), fis0[sk0] = refval(r), sk ∈ fis0, sk0 ∈ fis0 ⊦ r1 = jvmref, r = jvmref, sk = sk0, dups(fis0 .keys), sk3 ∈ fis0 .keys, sk ≠ sk0 ∧ r ≠ jvmref ∧ ((sk3 × refval(r1)) ' + fis0)[sk] = refval(r) ∧ ((sk3 × refval(r1)) ' + fis0)[sk0] = refval(r) ∧ (sk = sk3 ∨ sk ∈ fis0) ∧ (sk0 = sk3 ∨ sk0 ∈ fis0), ∃ 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)

Rule argument

[EXRARG : pos: right 6, [extquanttermlist: (r, sk, sk0) abort:F , discard:F , computed:T ]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (r, sk, sk0)]]