YES Problem: a(x1) -> b(x1) a(c(x1)) -> c(b(c(b(a(a(x1)))))) b(b(x1)) -> x1 Proof: String Reversal Processor: a(x1) -> b(x1) c(a(x1)) -> a(a(b(c(b(c(x1)))))) b(b(x1)) -> x1 DP Processor: DPs: a#(x1) -> b#(x1) c#(a(x1)) -> c#(x1) c#(a(x1)) -> b#(c(x1)) c#(a(x1)) -> c#(b(c(x1))) c#(a(x1)) -> b#(c(b(c(x1)))) c#(a(x1)) -> a#(b(c(b(c(x1))))) c#(a(x1)) -> a#(a(b(c(b(c(x1)))))) TRS: a(x1) -> b(x1) c(a(x1)) -> a(a(b(c(b(c(x1)))))) b(b(x1)) -> x1 TDG Processor: DPs: a#(x1) -> b#(x1) c#(a(x1)) -> c#(x1) c#(a(x1)) -> b#(c(x1)) c#(a(x1)) -> c#(b(c(x1))) c#(a(x1)) -> b#(c(b(c(x1)))) c#(a(x1)) -> a#(b(c(b(c(x1))))) c#(a(x1)) -> a#(a(b(c(b(c(x1)))))) TRS: a(x1) -> b(x1) c(a(x1)) -> a(a(b(c(b(c(x1)))))) b(b(x1)) -> x1 graph: c#(a(x1)) -> c#(b(c(x1))) -> c#(a(x1)) -> a#(a(b(c(b(c(x1)))))) c#(a(x1)) -> c#(b(c(x1))) -> c#(a(x1)) -> a#(b(c(b(c(x1))))) c#(a(x1)) -> c#(b(c(x1))) -> c#(a(x1)) -> b#(c(b(c(x1)))) c#(a(x1)) -> c#(b(c(x1))) -> c#(a(x1)) -> c#(b(c(x1))) c#(a(x1)) -> c#(b(c(x1))) -> c#(a(x1)) -> b#(c(x1)) c#(a(x1)) -> c#(b(c(x1))) -> c#(a(x1)) -> c#(x1) c#(a(x1)) -> c#(x1) -> c#(a(x1)) -> a#(a(b(c(b(c(x1)))))) c#(a(x1)) -> c#(x1) -> c#(a(x1)) -> a#(b(c(b(c(x1))))) c#(a(x1)) -> c#(x1) -> c#(a(x1)) -> b#(c(b(c(x1)))) c#(a(x1)) -> c#(x1) -> c#(a(x1)) -> c#(b(c(x1))) c#(a(x1)) -> c#(x1) -> c#(a(x1)) -> b#(c(x1)) c#(a(x1)) -> c#(x1) -> c#(a(x1)) -> c#(x1) c#(a(x1)) -> a#(b(c(b(c(x1))))) -> a#(x1) -> b#(x1) c#(a(x1)) -> a#(a(b(c(b(c(x1)))))) -> a#(x1) -> b#(x1) SCC Processor: #sccs: 1 #rules: 2 #arcs: 14/49 DPs: c#(a(x1)) -> c#(b(c(x1))) c#(a(x1)) -> c#(x1) TRS: a(x1) -> b(x1) c(a(x1)) -> a(a(b(c(b(c(x1)))))) b(b(x1)) -> x1 Arctic Interpretation Processor: dimension: 2 usable rules: a(x1) -> b(x1) c(a(x1)) -> a(a(b(c(b(c(x1)))))) b(b(x1)) -> x1 interpretation: [c#](x0) = [0 2]x0 + [3], [-& 0 ] [0] [c](x0) = [0 1 ]x0 + [2], [0 0 ] [0] [b](x0) = [0 -&]x0 + [0], [0 0] [0] [a](x0) = [0 1]x0 + [2] orientation: c#(a(x1)) = [2 3]x1 + [4] >= [0 2]x1 + [3] = c#(b(c(x1))) c#(a(x1)) = [2 3]x1 + [4] >= [0 2]x1 + [3] = c#(x1) [0 0] [0] [0 0 ] [0] a(x1) = [0 1]x1 + [2] >= [0 -&]x1 + [0] = b(x1) [0 1] [2] [0 1] [2] c(a(x1)) = [1 2]x1 + [3] >= [1 2]x1 + [3] = a(a(b(c(b(c(x1)))))) [0 0] [0] b(b(x1)) = [0 0]x1 + [0] >= x1 = x1 problem: DPs: TRS: a(x1) -> b(x1) c(a(x1)) -> a(a(b(c(b(c(x1)))))) b(b(x1)) -> x1 Qed