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

value(v3), value(v2), value(v1) ⊦ (∃ v, v0. S _ v1 _ v2 _ v3 = K _ v _ v0 ∧ v1 _ v3 _ (v2 _ v3) = v ∧ value(v) ∧ value(v0)) ∨ (∃ v, v0, v4. S _ v1 _ v2 _ v3 = S _ v _ v0 _ v4 ∧ v1 _ v3 _ (v2 _ v3) = v _ v4 _ (v0 _ v4) ∧ value(v) ∧ value(v0) ∧ value(v4))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
value(v3), value(v2), value(v1) ⊦ ∃ v, v0, v4. S _ v1 _ v2 _ v3 = S _ v _ v0 _ v4 ∧ v1 _ v3 _ (v2 _ v3) = v _ v4 _ (v0 _ v4) ∧ value(v) ∧ value(v0) ∧ value(v4), ∃ v0. S _ v1 _ v2 _ v3 = K _ (v1 _ v3 _ (v2 _ v3)) _ v0 ∧ value(v1 _ v3 _ (v2 _ v3)) ∧ value(v0)

Rule argument

[emptyarg]

Additional information

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