YES Problem: a(x1) -> x1 a(b(x1)) -> b(c(x1)) b(x1) -> x1 c(c(x1)) -> a(c(b(a(x1)))) Proof: String Reversal Processor: a(x1) -> x1 b(a(x1)) -> c(b(x1)) b(x1) -> x1 c(c(x1)) -> a(b(c(a(x1)))) DP Processor: DPs: b#(a(x1)) -> b#(x1) b#(a(x1)) -> c#(b(x1)) c#(c(x1)) -> a#(x1) c#(c(x1)) -> c#(a(x1)) c#(c(x1)) -> b#(c(a(x1))) c#(c(x1)) -> a#(b(c(a(x1)))) TRS: a(x1) -> x1 b(a(x1)) -> c(b(x1)) b(x1) -> x1 c(c(x1)) -> a(b(c(a(x1)))) TDG Processor: DPs: b#(a(x1)) -> b#(x1) b#(a(x1)) -> c#(b(x1)) c#(c(x1)) -> a#(x1) c#(c(x1)) -> c#(a(x1)) c#(c(x1)) -> b#(c(a(x1))) c#(c(x1)) -> a#(b(c(a(x1)))) TRS: a(x1) -> x1 b(a(x1)) -> c(b(x1)) b(x1) -> x1 c(c(x1)) -> a(b(c(a(x1)))) graph: c#(c(x1)) -> c#(a(x1)) -> c#(c(x1)) -> a#(b(c(a(x1)))) c#(c(x1)) -> c#(a(x1)) -> c#(c(x1)) -> b#(c(a(x1))) c#(c(x1)) -> c#(a(x1)) -> c#(c(x1)) -> c#(a(x1)) c#(c(x1)) -> c#(a(x1)) -> c#(c(x1)) -> a#(x1) c#(c(x1)) -> b#(c(a(x1))) -> b#(a(x1)) -> c#(b(x1)) c#(c(x1)) -> b#(c(a(x1))) -> b#(a(x1)) -> b#(x1) b#(a(x1)) -> c#(b(x1)) -> c#(c(x1)) -> a#(b(c(a(x1)))) b#(a(x1)) -> c#(b(x1)) -> c#(c(x1)) -> b#(c(a(x1))) b#(a(x1)) -> c#(b(x1)) -> c#(c(x1)) -> c#(a(x1)) b#(a(x1)) -> c#(b(x1)) -> c#(c(x1)) -> a#(x1) b#(a(x1)) -> b#(x1) -> b#(a(x1)) -> c#(b(x1)) b#(a(x1)) -> b#(x1) -> b#(a(x1)) -> b#(x1) SCC Processor: #sccs: 1 #rules: 4 #arcs: 12/36 DPs: c#(c(x1)) -> c#(a(x1)) c#(c(x1)) -> b#(c(a(x1))) b#(a(x1)) -> b#(x1) b#(a(x1)) -> c#(b(x1)) TRS: a(x1) -> x1 b(a(x1)) -> c(b(x1)) b(x1) -> x1 c(c(x1)) -> a(b(c(a(x1)))) Arctic Interpretation Processor: dimension: 2 interpretation: [c#](x0) = [-& 0 ]x0 + [0], [b#](x0) = [0 -&]x0 + [0], [-& 0 ] [1] [c](x0) = [-& 2 ]x0 + [3], [0 -&] [0] [b](x0) = [2 0 ]x0 + [0], [2 0 ] [1] [a](x0) = [-& 0 ]x0 + [1] orientation: c#(c(x1)) = [-& 2 ]x1 + [3] >= [-& 0 ]x1 + [1] = c#(a(x1)) c#(c(x1)) = [-& 2 ]x1 + [3] >= [-& 0 ]x1 + [1] = b#(c(a(x1))) b#(a(x1)) = [2 0]x1 + [1] >= [0 -&]x1 + [0] = b#(x1) b#(a(x1)) = [2 0]x1 + [1] >= [2 0]x1 + [0] = c#(b(x1)) [2 0 ] [1] a(x1) = [-& 0 ]x1 + [1] >= x1 = x1 [2 0] [1] [2 0] [1] b(a(x1)) = [4 2]x1 + [3] >= [4 2]x1 + [3] = c(b(x1)) [0 -&] [0] b(x1) = [2 0 ]x1 + [0] >= x1 = x1 [-& 2 ] [3] [-& 2 ] [3] c(c(x1)) = [-& 4 ]x1 + [5] >= [-& 2 ]x1 + [3] = a(b(c(a(x1)))) problem: DPs: b#(a(x1)) -> c#(b(x1)) TRS: a(x1) -> x1 b(a(x1)) -> c(b(x1)) b(x1) -> x1 c(c(x1)) -> a(b(c(a(x1)))) EDG Processor: DPs: b#(a(x1)) -> c#(b(x1)) TRS: a(x1) -> x1 b(a(x1)) -> c(b(x1)) b(x1) -> x1 c(c(x1)) -> a(b(c(a(x1)))) graph: Qed