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: [c#](x0) = [0 0]x0 + [0], [0 -&] [-&] [c](x0) = [0 -&]x0 + [1 ], [0 1 ] [0 ] [b](x0) = [-& 0 ]x0 + [-&], [0 1] [0] [a](x0) = [0 0]x0 + [3] orientation: c#(b(a(x1))) = [1 1]x1 + [4] >= [0 -&]x1 + [1] = c#(c(x1)) c#(b(a(x1))) = [1 1]x1 + [4] >= [0 0]x1 + [0] = c#(x1) [0 1] [0] a(x1) = [0 0]x1 + [3] >= x1 = x1 [0 1] [0] [0 1 ] [0 ] a(x1) = [0 0]x1 + [3] >= [-& 0 ]x1 + [-&] = b(x1) [1 1] [4] [1 -&] [4] c(b(a(x1))) = [1 1]x1 + [4] >= [1 -&]x1 + [4] = a(b(a(c(c(x1))))) [0 1 ] [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