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