YES Problem: 0(1(2(3(4(5(1(x1))))))) -> 1(2(3(4(5(1(1(0(1(2(3(4(5(0(1(2(3(4(5(x1))))))))))))))))))) 0(1(2(3(4(5(1(x1))))))) -> 1(2(3(4(5(1(1(0(1(2(3(4(5(0(1(2(3(4(5(0(1(2(3(4(5(x1))))))))))))))))))))))))) Proof: DP Processor: DPs: 0#(1(2(3(4(5(1(x1))))))) -> 0#(1(2(3(4(5(x1)))))) 0#(1(2(3(4(5(1(x1))))))) -> 0#(1(2(3(4(5(0(1(2(3(4(5(x1)))))))))))) 0#(1(2(3(4(5(1(x1))))))) -> 0#(1(2(3(4(5(0(1(2(3(4(5(0(1(2(3(4(5(x1)))))))))))))))))) TRS: 0(1(2(3(4(5(1(x1))))))) -> 1(2(3(4(5(1(1(0(1(2(3(4(5(0(1(2(3(4(5(x1))))))))))))))))))) 0(1(2(3(4(5(1(x1))))))) -> 1(2(3(4(5(1(1(0(1(2(3(4(5(0(1(2(3(4(5(0(1(2(3(4(5(x1))))))))))))))))))))))))) Matrix Interpretation Processor: dim=2 interpretation: [0#](x0) = [2 2]x0, [2 0] [3] [0](x0) = [1 0]x0 + [0], [1 0] [2](x0) = [0 0]x0, [1 0] [3](x0) = [1 1]x0, [4](x0) = x0, [0 1] [5](x0) = [1 0]x0, [1 0] [0] [1](x0) = [0 2]x0 + [1] orientation: 0#(1(2(3(4(5(1(x1))))))) = [0 4]x1 + [4] >= [0 2]x1 + [2] = 0#(1(2(3(4(5(x1)))))) 0#(1(2(3(4(5(1(x1))))))) = [0 4]x1 + [4] >= [0 2]x1 + [2] = 0#(1(2(3(4(5(0(1(2(3(4(5(x1)))))))))))) 0#(1(2(3(4(5(1(x1))))))) = [0 4]x1 + [4] >= [0 2]x1 + [2] = 0#(1(2(3(4(5(0(1(2(3(4(5(0(1(2(3(4(5(x1)))))))))))))))))) [0 4] [5] [0 4] [3] 0(1(2(3(4(5(1(x1))))))) = [0 2]x1 + [1] >= [0 0]x1 + [1] = 1(2(3(4(5(1(1(0(1(2(3(4(5(0(1(2(3(4(5(x1))))))))))))))))))) [0 4] [5] [0 4] [3] 0(1(2(3(4(5(1(x1))))))) = [0 2]x1 + [1] >= [0 0]x1 + [1] = 1(2(3(4(5(1(1(0(1(2(3(4(5(0(1(2(3(4(5(0(1(2(3(4(5(x1))))))))))))))))))))))))) problem: DPs: TRS: 0(1(2(3(4(5(1(x1))))))) -> 1(2(3(4(5(1(1(0(1(2(3(4(5(0(1(2(3(4(5(x1))))))))))))))))))) 0(1(2(3(4(5(1(x1))))))) -> 1(2(3(4(5(1(1(0(1(2(3(4(5(0(1(2(3(4(5(0(1(2(3(4(5(x1))))))))))))))))))))))))) Qed