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