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Proof step 9 from proof in-put-rkvs (previous step)

∀ rk, st. rk ∈ st[rkvs0] ↔ rk ∈ st ∨ rk ∈ rkvs0 ⊦ rk ∈ st[rkvs0] ↔ rk ∈ st ∨ rk ∈ rkvs0, rk = rk0, rk0 ∈ rkvs0

User applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 1, [extquanttermlist: (rk, st)abort:F, discard:T, computed:T]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (rk, st)]]