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Proof step 15 from proof union-in-01 (previous step)

r -- sk ∈ st ∨ r -- sk0 ∈ st0 ⊦ ∃ sk1. r -- sk1 ∈ st ∨ r -- sk1 ∈ st0

User applied rule exists right

And obtained the following new goals

Proof step 14:
r -- sk ∈ st ∨ r -- sk0 ∈ st0 ⊦ r -- sk ∈ st ∨ r -- sk ∈ st0, ∃ sk1. r -- sk1 ∈ st ∨ r -- sk1 ∈ st0

Rule argument

[EXRARG : pos: right 1, [extquanttermlist: (sk)abort:F, discard:F, computed:T]]

Additional information

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