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Proof step 2 from proof K-reduce (previous step)

value(v0), value(v) ⊦ ∃ v2. K _ v _ v0 = K _ v _ v2 ∧ value(v2), ∃ v1, v2, v3. K _ (v1 _ v3 _ (v2 _ v3)) _ v0 = S _ v1 _ v2 _ v3 ∧ v = v1 _ v3 _ (v2 _ v3) ∧ value(v1) ∧ value(v2) ∧ value(v3)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

[EXRARG : pos: right 1, [QUANTTERMLIST : (v0)]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (v0)]]