YES Problem: a(a(b(x1))) -> c(b(a(a(a(x1))))) a(c(x1)) -> b(a(x1)) Proof: DP Processor: DPs: a#(a(b(x1))) -> a#(x1) a#(a(b(x1))) -> a#(a(x1)) a#(a(b(x1))) -> a#(a(a(x1))) a#(c(x1)) -> a#(x1) TRS: a(a(b(x1))) -> c(b(a(a(a(x1))))) a(c(x1)) -> b(a(x1)) Matrix Interpretation Processor: dim=3 interpretation: [a#](x0) = [0 2 1]x0, [0 0 0] [0] [c](x0) = [0 0 0]x0 + [0] [1 2 1] [2], [0 0 1] [a](x0) = [1 0 0]x0 [0 2 0] , [1 1 1] [2] [b](x0) = [0 0 0]x0 + [0] [0 0 0] [0] orientation: a#(a(b(x1))) = [2 2 2]x1 + [4] >= [0 2 1]x1 = a#(x1) a#(a(b(x1))) = [2 2 2]x1 + [4] >= [2 2 0]x1 = a#(a(x1)) a#(a(b(x1))) = [2 2 2]x1 + [4] >= [2 0 2]x1 = a#(a(a(x1))) a#(c(x1)) = [1 2 1]x1 + [2] >= [0 2 1]x1 = a#(x1) [0 0 0] [0] [0 0 0] [0] a(a(b(x1))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = c(b(a(a(a(x1))))) [2 2 2] [4] [2 2 2] [4] [1 2 1] [2] [1 2 1] [2] a(c(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = b(a(x1)) [0 0 0] [0] [0 0 0] [0] problem: DPs: TRS: a(a(b(x1))) -> c(b(a(a(a(x1))))) a(c(x1)) -> b(a(x1)) Qed