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