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Proof step 3 from proof indel (previous step)

r -- sk0 ∈ st, r -- sk1 ∈ st ⊦ r = r0, ∃ sk0. r -- sk0 ∈ st -- (r0 -- sk)

User applied rule exists right

And closed the goal!

Rule argument

[EXRARG : pos: right 2, [extquanttermlist: (sk1)abort:F, discard:T, computed:F]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r -- sk = r0 -- sk0 ↔ r = r0 ∧ sk = sk0], [CSIMPSEQ : ⊦ rk ∈ st -- rk0 ↔ rk ≠ rk0 ∧ rk ∈ st])], [EXTRATERMS : (sk1)]]