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Proof step 5 from proof in-1dim-06 (previous step)

r -- sk ∈ st, r ∈ st ⊦ ∃ sk. r -- sk ∈ addarray(r0, ty0, i, st)

User applied rule exists right

And obtained the following new goals

Proof step 4:
r -- sk ∈ st, r ∈ st ⊦ r -- sk ∈ addarray(r0, ty0, i, st)

Rule argument

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

Additional information

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