YES Problem: a(x1) -> x1 a(b(b(x1))) -> b(b(b(c(x1)))) b(c(x1)) -> a(a(x1)) Proof: String Reversal Processor: a(x1) -> x1 b(b(a(x1))) -> c(b(b(b(x1)))) c(b(x1)) -> a(a(x1)) DP Processor: DPs: b#(b(a(x1))) -> b#(x1) b#(b(a(x1))) -> b#(b(x1)) b#(b(a(x1))) -> b#(b(b(x1))) b#(b(a(x1))) -> c#(b(b(b(x1)))) c#(b(x1)) -> a#(x1) c#(b(x1)) -> a#(a(x1)) TRS: a(x1) -> x1 b(b(a(x1))) -> c(b(b(b(x1)))) c(b(x1)) -> a(a(x1)) TDG Processor: DPs: b#(b(a(x1))) -> b#(x1) b#(b(a(x1))) -> b#(b(x1)) b#(b(a(x1))) -> b#(b(b(x1))) b#(b(a(x1))) -> c#(b(b(b(x1)))) c#(b(x1)) -> a#(x1) c#(b(x1)) -> a#(a(x1)) TRS: a(x1) -> x1 b(b(a(x1))) -> c(b(b(b(x1)))) c(b(x1)) -> a(a(x1)) graph: b#(b(a(x1))) -> c#(b(b(b(x1)))) -> c#(b(x1)) -> a#(a(x1)) b#(b(a(x1))) -> c#(b(b(b(x1)))) -> c#(b(x1)) -> a#(x1) b#(b(a(x1))) -> b#(b(b(x1))) -> b#(b(a(x1))) -> c#(b(b(b(x1)))) b#(b(a(x1))) -> b#(b(b(x1))) -> b#(b(a(x1))) -> b#(b(b(x1))) b#(b(a(x1))) -> b#(b(b(x1))) -> b#(b(a(x1))) -> b#(b(x1)) b#(b(a(x1))) -> b#(b(b(x1))) -> b#(b(a(x1))) -> b#(x1) b#(b(a(x1))) -> b#(b(x1)) -> b#(b(a(x1))) -> c#(b(b(b(x1)))) b#(b(a(x1))) -> b#(b(x1)) -> b#(b(a(x1))) -> b#(b(b(x1))) b#(b(a(x1))) -> b#(b(x1)) -> b#(b(a(x1))) -> b#(b(x1)) b#(b(a(x1))) -> b#(b(x1)) -> b#(b(a(x1))) -> b#(x1) b#(b(a(x1))) -> b#(x1) -> b#(b(a(x1))) -> c#(b(b(b(x1)))) b#(b(a(x1))) -> b#(x1) -> b#(b(a(x1))) -> b#(b(b(x1))) b#(b(a(x1))) -> b#(x1) -> b#(b(a(x1))) -> b#(b(x1)) b#(b(a(x1))) -> b#(x1) -> b#(b(a(x1))) -> b#(x1) SCC Processor: #sccs: 1 #rules: 3 #arcs: 14/36 DPs: b#(b(a(x1))) -> b#(b(b(x1))) b#(b(a(x1))) -> b#(x1) b#(b(a(x1))) -> b#(b(x1)) TRS: a(x1) -> x1 b(b(a(x1))) -> c(b(b(b(x1)))) c(b(x1)) -> a(a(x1)) Arctic Interpretation Processor: dimension: 2 interpretation: [b#](x0) = [0 -&]x0, [-& 0 ] [0] [c](x0) = [0 1 ]x0 + [0], [-& 0 ] [0] [b](x0) = [0 -&]x0 + [1], [0 -&] [0] [a](x0) = [1 0 ]x0 + [2] orientation: b#(b(a(x1))) = [1 0]x1 + [2] >= [0 -&]x1 + [1] = b#(b(b(x1))) b#(b(a(x1))) = [1 0]x1 + [2] >= [0 -&]x1 = b#(x1) b#(b(a(x1))) = [1 0]x1 + [2] >= [-& 0 ]x1 + [0] = b#(b(x1)) [0 -&] [0] a(x1) = [1 0 ]x1 + [2] >= x1 = x1 [0 -&] [1] [0 -&] [1] b(b(a(x1))) = [1 0 ]x1 + [2] >= [1 0 ]x1 + [2] = c(b(b(b(x1)))) [0 -&] [1] [0 -&] [0] c(b(x1)) = [1 0 ]x1 + [2] >= [1 0 ]x1 + [2] = a(a(x1)) problem: DPs: b#(b(a(x1))) -> b#(b(x1)) TRS: a(x1) -> x1 b(b(a(x1))) -> c(b(b(b(x1)))) c(b(x1)) -> a(a(x1)) EDG Processor: DPs: b#(b(a(x1))) -> b#(b(x1)) TRS: a(x1) -> x1 b(b(a(x1))) -> c(b(b(b(x1)))) c(b(x1)) -> a(a(x1)) graph: b#(b(a(x1))) -> b#(b(x1)) -> b#(b(a(x1))) -> b#(b(x1)) Matrix Interpretation Processor: dim=1 interpretation: [b#](x0) = 1/2x0 + 13/2, [c](x0) = 1/2x0 + 1/2, [b](x0) = 2x0 + 3, [a](x0) = x0 + 1 orientation: b#(b(a(x1))) = x1 + 9 >= x1 + 8 = b#(b(x1)) a(x1) = x1 + 1 >= x1 = x1 b(b(a(x1))) = 4x1 + 13 >= 4x1 + 11 = c(b(b(b(x1)))) c(b(x1)) = x1 + 2 >= x1 + 2 = a(a(x1)) problem: DPs: TRS: a(x1) -> x1 b(b(a(x1))) -> c(b(b(b(x1)))) c(b(x1)) -> a(a(x1)) Qed