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: 3 usable rules: a(x1) -> x1 a(x1) -> b(x1) c(a(x1)) -> b(a(a(c(c(x1))))) b(b(x1)) -> x1 interpretation: [c#](x0) = [0 0 -&]x0 + [0], [-& 0 -&] [0] [c](x0) = [-& 0 -&]x0 + [0] [1 1 0 ] [1], [-& -& 0 ] [0 ] [b](x0) = [-& -& 0 ]x0 + [-&] [0 0 0 ] [0 ], [0 1 0] [0] [a](x0) = [1 1 0]x0 + [1] [0 0 0] [0] orientation: c#(a(x1)) = [1 1 0]x1 + [1] >= [-& 0 -&]x1 + [0] = c#(c(x1)) c#(a(x1)) = [1 1 0]x1 + [1] >= [0 0 -&]x1 + [0] = c#(x1) [0 1 0] [0] a(x1) = [1 1 0]x1 + [1] >= x1 = x1 [0 0 0] [0] [0 1 0] [0] [-& -& 0 ] [0 ] a(x1) = [1 1 0]x1 + [1] >= [-& -& 0 ]x1 + [-&] = b(x1) [0 0 0] [0] [0 0 0 ] [0 ] [1 1 0] [1] [1 1 0] [1] c(a(x1)) = [1 1 0]x1 + [1] >= [1 1 0]x1 + [1] = b(a(a(c(c(x1))))) [2 2 1] [2] [2 2 1] [2] [0 0 0] [0] b(b(x1)) = [0 0 0]x1 + [0] >= x1 = x1 [0 0 0] [0] problem: DPs: TRS: a(x1) -> x1 a(x1) -> b(x1) c(a(x1)) -> b(a(a(c(c(x1))))) b(b(x1)) -> x1 Qed