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Proof step 6 from proof S-reduce (conclusion)

⊦ value(v1) ∧ value(v2) ∧ value(v3) → S _ v1 _ v2 _ v3 => v1 _ v3 _ (v2 _ v3)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
value(v3), value(v2), value(v1) ⊦ S _ v1 _ v2 _ v3 => v1 _ v3 _ (v2 _ v3)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]