YES Problem: a__f(f(a())) -> a__f(g(f(a()))) mark(f(X)) -> a__f(X) mark(a()) -> a() mark(g(X)) -> g(mark(X)) a__f(X) -> f(X) Proof: DP Processor: DPs: a__f#(f(a())) -> a__f#(g(f(a()))) mark#(f(X)) -> a__f#(X) mark#(g(X)) -> mark#(X) TRS: a__f(f(a())) -> a__f(g(f(a()))) mark(f(X)) -> a__f(X) mark(a()) -> a() mark(g(X)) -> g(mark(X)) a__f(X) -> f(X) Matrix Interpretation Processor: dim=3 interpretation: [mark#](x0) = [0 1 1]x0, [a__f#](x0) = [0 0 1]x0, [0 1 1] [0] [mark](x0) = [1 0 1]x0 + [1] [1 1 0] [1], [1 0 0] [1] [g](x0) = [0 0 1]x0 + [1] [0 1 0] [0], [0 0 1] [1] [a__f](x0) = [0 0 1]x0 + [0] [0 0 1] [1], [0 0 1] [0] [f](x0) = [0 0 1]x0 + [0] [0 0 0] [1], [0] [a] = [0] [0] orientation: a__f#(f(a())) = 1 >= 0 = a__f#(g(f(a()))) mark#(f(X)) = [0 0 1]X + [1] >= [0 0 1]X = a__f#(X) mark#(g(X)) = [0 1 1]X + [1] >= [0 1 1]X = mark#(X) [2] [1] a__f(f(a())) = [1] >= [0] = a__f(g(f(a()))) [2] [1] [0 0 1] [1] [0 0 1] [1] mark(f(X)) = [0 0 1]X + [2] >= [0 0 1]X + [0] = a__f(X) [0 0 2] [1] [0 0 1] [1] [0] [0] mark(a()) = [1] >= [0] = a() [1] [0] [0 1 1] [1] [0 1 1] [1] mark(g(X)) = [1 1 0]X + [2] >= [1 1 0]X + [2] = g(mark(X)) [1 0 1] [3] [1 0 1] [1] [0 0 1] [1] [0 0 1] [0] a__f(X) = [0 0 1]X + [0] >= [0 0 1]X + [0] = f(X) [0 0 1] [1] [0 0 0] [1] problem: DPs: TRS: a__f(f(a())) -> a__f(g(f(a()))) mark(f(X)) -> a__f(X) mark(a()) -> a() mark(g(X)) -> g(mark(X)) a__f(X) -> f(X) Qed