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