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