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Proof step 7 from proof in-butmin (previous step)

st .min = r0 -- sk, r -- sk0 ∈ st, r ∈ st ⊦ r = r0, ∃ sk. r -- sk ∈ st .butmin

User applied rule exists right

And obtained the following new goals

Proof step 6:
st .min = r0 -- sk, r -- sk0 ∈ st, r ∈ st ⊦ r = r0, r -- sk0 ∈ st .butmin

Rule argument

[EXRARG : pos: right 2, [extquanttermlist: (sk0)abort:F, discard:T, computed:F]]

Additional information

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