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Proof step 3 from proof K-top-reduce (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
value(v), value(v0) ⊦ ∃ 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)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]