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Proof step 14 from proof stsubset-new (previous step)

rk ∈ st, ∀ rk. rk ∈ st → rk .ref ≠ r, ∀ rk. rk ∈ st → rk ∈ st0 ∧ st[rk] = st0[rk] ⊦ rk = r -- sk, r = jvmref, st[rk] = st0[rk]

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 3, [QUANTTERMLIST : (rk)]]

Additional information

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