YES Problem: p(a(a(x0)),p(x1,p(a(x2),x3))) -> p(x2,p(a(a(b(x1))),p(a(a(x0)),x3))) Proof: DP Processor: DPs: p#(a(a(x0)),p(x1,p(a(x2),x3))) -> p#(a(a(x0)),x3) p#(a(a(x0)),p(x1,p(a(x2),x3))) -> p#(a(a(b(x1))),p(a(a(x0)),x3)) p#(a(a(x0)),p(x1,p(a(x2),x3))) -> p#(x2,p(a(a(b(x1))),p(a(a(x0)),x3))) TRS: p(a(a(x0)),p(x1,p(a(x2),x3))) -> p(x2,p(a(a(b(x1))),p(a(a(x0)),x3))) Matrix Interpretation Processor: dim=4 interpretation: [p#](x0, x1) = [0 0 1 0]x0 + [1 0 0 0]x1, [0 0 0 0] [0 0 0 0] [b](x0) = [0 0 0 0]x0 [0 1 0 0] , [0 1 0 0] [1 0 0 0] [1] [1 0 0 0] [0 1 0 0] [0] [p](x0, x1) = [0 0 0 0]x0 + [0 0 0 0]x1 + [0] [0 0 0 0] [0 0 0 0] [1], [1 0 0 0] [0] [0 1 1 0] [0] [a](x0) = [0 1 1 1]x0 + [0] [0 0 0 0] [1] orientation: p#(a(a(x0)),p(x1,p(a(x2),x3))) = [0 2 2 1]x0 + [0 1 0 0]x1 + [0 1 1 0]x2 + [1 0 0 0]x3 + [3] >= [0 2 2 1]x0 + [1 0 0 0]x3 + [1] = p#(a(a(x0)),x3) p#(a(a(x0)),p(x1,p(a(x2),x3))) = [0 2 2 1]x0 + [0 1 0 0]x1 + [0 1 1 0]x2 + [1 0 0 0]x3 + [3] >= [0 2 2 1]x0 + [0 1 0 0]x1 + [1 0 0 0]x3 + [2] = p#(a(a(b(x1))),p(a(a(x0)),x3)) p#(a(a(x0)),p(x1,p(a(x2),x3))) = [0 2 2 1]x0 + [0 1 0 0]x1 + [0 1 1 0]x2 + [1 0 0 0]x3 + [3] >= [0 2 2 1]x0 + [0 1 0 0]x1 + [0 0 1 0]x2 + [1 0 0 0]x3 + [2] = p#(x2,p(a(a(b(x1))),p(a(a(x0)),x3))) [0 2 2 1] [0 1 0 0] [0 1 1 0] [1 0 0 0] [3] [0 2 2 1] [0 1 0 0] [0 1 0 0] [1 0 0 0] [3] [1 0 0 0] [1 0 0 0] [1 0 0 0] [0 1 0 0] [0] [1 0 0 0] [0 0 0 0] [1 0 0 0] [0 1 0 0] [0] p(a(a(x0)),p(x1,p(a(x2),x3))) = [0 0 0 0]x0 + [0 0 0 0]x1 + [0 0 0 0]x2 + [0 0 0 0]x3 + [0] >= [0 0 0 0]x0 + [0 0 0 0]x1 + [0 0 0 0]x2 + [0 0 0 0]x3 + [0] = p(x2,p(a(a(b(x1))),p(a(a(x0)),x3))) [0 0 0 0] [0 0 0 0] [0 0 0 0] [0 0 0 0] [1] [0 0 0 0] [0 0 0 0] [0 0 0 0] [0 0 0 0] [1] problem: DPs: TRS: p(a(a(x0)),p(x1,p(a(x2),x3))) -> p(x2,p(a(a(b(x1))),p(a(a(x0)),x3))) Qed