MAYBE Problem: f(X) -> f(a()) b() -> a() Proof: DP Processor: DPs: f#(X) -> f#(a()) TRS: f(X) -> f(a()) b() -> a() SCC Processor: #sccs: 1 #rules: 1 #arcs: 1/1 DPs: f#(X) -> f#(a()) TRS: f(X) -> f(a()) b() -> a() Open