YES Problem: a(x1) -> x1 a(x1) -> b(x1) a(c(x1)) -> c(c(a(a(b(x1))))) b(b(x1)) -> x1 Proof: String Reversal Processor: a(x1) -> x1 a(x1) -> b(x1) c(a(x1)) -> b(a(a(c(c(x1))))) b(b(x1)) -> x1 DP Processor: DPs: a#(x1) -> b#(x1) c#(a(x1)) -> c#(x1) c#(a(x1)) -> c#(c(x1)) c#(a(x1)) -> a#(c(c(x1))) c#(a(x1)) -> a#(a(c(c(x1)))) c#(a(x1)) -> b#(a(a(c(c(x1))))) TRS: a(x1) -> x1 a(x1) -> b(x1) c(a(x1)) -> b(a(a(c(c(x1))))) b(b(x1)) -> x1 TDG Processor: DPs: a#(x1) -> b#(x1) c#(a(x1)) -> c#(x1) c#(a(x1)) -> c#(c(x1)) c#(a(x1)) -> a#(c(c(x1))) c#(a(x1)) -> a#(a(c(c(x1)))) c#(a(x1)) -> b#(a(a(c(c(x1))))) TRS: a(x1) -> x1 a(x1) -> b(x1) c(a(x1)) -> b(a(a(c(c(x1))))) b(b(x1)) -> x1 graph: c#(a(x1)) -> c#(c(x1)) -> c#(a(x1)) -> b#(a(a(c(c(x1))))) c#(a(x1)) -> c#(c(x1)) -> c#(a(x1)) -> a#(a(c(c(x1)))) c#(a(x1)) -> c#(c(x1)) -> c#(a(x1)) -> a#(c(c(x1))) c#(a(x1)) -> c#(c(x1)) -> c#(a(x1)) -> c#(c(x1)) c#(a(x1)) -> c#(c(x1)) -> c#(a(x1)) -> c#(x1) c#(a(x1)) -> c#(x1) -> c#(a(x1)) -> b#(a(a(c(c(x1))))) c#(a(x1)) -> c#(x1) -> c#(a(x1)) -> a#(a(c(c(x1)))) c#(a(x1)) -> c#(x1) -> c#(a(x1)) -> a#(c(c(x1))) c#(a(x1)) -> c#(x1) -> c#(a(x1)) -> c#(c(x1)) c#(a(x1)) -> c#(x1) -> c#(a(x1)) -> c#(x1) c#(a(x1)) -> a#(c(c(x1))) -> a#(x1) -> b#(x1) c#(a(x1)) -> a#(a(c(c(x1)))) -> a#(x1) -> b#(x1) SCC Processor: #sccs: 1 #rules: 2 #arcs: 12/36 DPs: c#(a(x1)) -> c#(c(x1)) c#(a(x1)) -> c#(x1) TRS: a(x1) -> x1 a(x1) -> b(x1) c(a(x1)) -> b(a(a(c(c(x1))))) b(b(x1)) -> x1 Arctic Interpretation Processor: dimension: 2 usable rules: a(x1) -> x1 a(x1) -> b(x1) c(a(x1)) -> b(a(a(c(c(x1))))) b(b(x1)) -> x1 interpretation: [-& 0 ] [0 ] [b](x0) = [0 0 ]x0 + [-&], [c#](x0) = [1 0]x0 + [0], [1 0] [3] [a](x0) = [0 0]x0 + [0], [0 -&] [0] [c](x0) = [1 0 ]x0 + [0] orientation: c#(a(x1)) = [2 1]x1 + [4] >= [1 0]x1 + [1] = c#(c(x1)) c#(a(x1)) = [2 1]x1 + [4] >= [1 0]x1 + [0] = c#(x1) [1 0] [3] a(x1) = [0 0]x1 + [0] >= x1 = x1 [1 0] [3] [-& 0 ] [0 ] a(x1) = [0 0]x1 + [0] >= [0 0 ]x1 + [-&] = b(x1) [1 0] [3] [1 0] [3] c(a(x1)) = [2 1]x1 + [4] >= [2 1]x1 + [4] = b(a(a(c(c(x1))))) [0 0] [0] b(b(x1)) = [0 0]x1 + [0] >= x1 = x1 problem: DPs: TRS: a(x1) -> x1 a(x1) -> b(x1) c(a(x1)) -> b(a(a(c(c(x1))))) b(b(x1)) -> x1 Qed