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