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Proof step 9 from proof nonnullrefs-in (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 4, [QUANTTERMLIST : (sk, r)]]

Additional information

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