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