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Proof step 25 from proof nonnullrefs-in-key (previous step)

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)

User applied rule exists right

And obtained the following new goals

Proof step 24:
fis0[sk] = refval(r), sk ∈ fis0, ¬ is_referencevalue(val) ∨ val .refval = jvmref ⊦ r = jvmref, dups(fis0 .keys), sk0 ∈ fis0 .keys, ((sk0 × val) ' + fis0)[sk] = refval(r) ∧ (sk = sk0 ∨ sk ∈ fis0), ∃ sk. ((sk0 × val) ' + fis0)[sk] = refval(r) ∧ (sk = sk0 ∨ sk ∈ fis0)

Rule argument

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

Additional information

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