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Proof step 7 from proof in-del-01 (previous step)

r -- sk ∈ st, r ∈ st, ∀ r1, sk0. r0 ≠ r1 → (r1 -- sk0 ∈ st ↔ r1 -- sk0 ∈ st -- r0) ⊦ r = r0, r -- sk ∈ st -- r0, r0 ∈ st -- r0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

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

Additional information

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