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