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Proof step 3 from proof hasKeys-all (previous step)

r -- sk ∈ st, hasKeys(r, sks0, st), sk0 ∈ sks0, r ∈ st, ∀ r, st. hasKeys(r, sks0, st) ↔ (∀ sk. sk ∈ sks0 → r -- sk ∈ st), ∀ sk. sk ∈ sks0 → r -- sk ∈ st ⊦ r -- sk0 ∈ st

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 6, [QUANTTERMLIST : (sk0)]]

Additional information

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