YES Problem: a(x1) -> b(x1) a(b(x1)) -> c(a(c(a(x1)))) c(b(c(x1))) -> b(x1) Proof: DP Processor: DPs: a#(b(x1)) -> a#(x1) a#(b(x1)) -> c#(a(x1)) a#(b(x1)) -> a#(c(a(x1))) a#(b(x1)) -> c#(a(c(a(x1)))) TRS: a(x1) -> b(x1) a(b(x1)) -> c(a(c(a(x1)))) c(b(c(x1))) -> b(x1) TDG Processor: DPs: a#(b(x1)) -> a#(x1) a#(b(x1)) -> c#(a(x1)) a#(b(x1)) -> a#(c(a(x1))) a#(b(x1)) -> c#(a(c(a(x1)))) TRS: a(x1) -> b(x1) a(b(x1)) -> c(a(c(a(x1)))) c(b(c(x1))) -> b(x1) graph: a#(b(x1)) -> a#(c(a(x1))) -> a#(b(x1)) -> c#(a(c(a(x1)))) a#(b(x1)) -> a#(c(a(x1))) -> a#(b(x1)) -> a#(c(a(x1))) a#(b(x1)) -> a#(c(a(x1))) -> a#(b(x1)) -> c#(a(x1)) a#(b(x1)) -> a#(c(a(x1))) -> a#(b(x1)) -> a#(x1) a#(b(x1)) -> a#(x1) -> a#(b(x1)) -> c#(a(c(a(x1)))) a#(b(x1)) -> a#(x1) -> a#(b(x1)) -> a#(c(a(x1))) a#(b(x1)) -> a#(x1) -> a#(b(x1)) -> c#(a(x1)) a#(b(x1)) -> a#(x1) -> a#(b(x1)) -> a#(x1) SCC Processor: #sccs: 1 #rules: 2 #arcs: 8/16 DPs: a#(b(x1)) -> a#(c(a(x1))) a#(b(x1)) -> a#(x1) TRS: a(x1) -> b(x1) a(b(x1)) -> c(a(c(a(x1)))) c(b(c(x1))) -> b(x1) Arctic Interpretation Processor: dimension: 1 interpretation: [a#](x0) = x0, [c](x0) = x0, [b](x0) = 10x0, [a](x0) = 10x0 orientation: a#(b(x1)) = 10x1 >= 10x1 = a#(c(a(x1))) a#(b(x1)) = 10x1 >= x1 = a#(x1) a(x1) = 10x1 >= 10x1 = b(x1) a(b(x1)) = 20x1 >= 20x1 = c(a(c(a(x1)))) c(b(c(x1))) = 10x1 >= 10x1 = b(x1) problem: DPs: a#(b(x1)) -> a#(c(a(x1))) TRS: a(x1) -> b(x1) a(b(x1)) -> c(a(c(a(x1)))) c(b(c(x1))) -> b(x1) EDG Processor: DPs: a#(b(x1)) -> a#(c(a(x1))) TRS: a(x1) -> b(x1) a(b(x1)) -> c(a(c(a(x1)))) c(b(c(x1))) -> b(x1) graph: a#(b(x1)) -> a#(c(a(x1))) -> a#(b(x1)) -> a#(c(a(x1))) Arctic Interpretation Processor: dimension: 2 interpretation: [a#](x0) = [-& 0 ]x0 + [0], [1 -&] [2] [c](x0) = [-& -1]x0 + [1], [-& 0 ] [0] [b](x0) = [1 -&]x0 + [2], [-& 0 ] [1] [a](x0) = [1 -&]x0 + [2] orientation: a#(b(x1)) = [1 -&]x1 + [2] >= [0 -&]x1 + [1] = a#(c(a(x1))) [-& 0 ] [1] [-& 0 ] [0] a(x1) = [1 -&]x1 + [2] >= [1 -&]x1 + [2] = b(x1) [1 -&] [2] [1 -&] [2] a(b(x1)) = [-& 1 ]x1 + [2] >= [-& 1 ]x1 + [2] = c(a(c(a(x1)))) [-& 0 ] [2] [-& 0 ] [0] c(b(c(x1))) = [1 -&]x1 + [2] >= [1 -&]x1 + [2] = b(x1) problem: DPs: TRS: a(x1) -> b(x1) a(b(x1)) -> c(a(c(a(x1)))) c(b(c(x1))) -> b(x1) Qed