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