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Proof step 5 from proof subset-notin (previous step)

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

Heuristic quantifier closing applied rule all left

And obtained the following new goals

Rule argument

[EXRARG : pos: left 3, [QUANTTERMLIST : (r -- sk)]]

Additional information

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