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Proof step 7 from proof union-subset-left (previous step)

∀ r. ¬ r ∈ st .refs ∨ ¬ r ∈ st0 .refs, r -- sk ∈ st0, r -- sk ∈ st, r ∈ st, r ∈ st0 ⊦ st[r -- sk] = st0[r -- sk]

Heuristic quantifier closing applied rule all left

And obtained the following new goals

Proof step 6:
⊦ ¬ r ∈ st .refs → ¬ r ∈ st

Rule argument

[EXRARG : pos: left 1, [QUANTTERMLIST : (r)]]

Additional information

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