YES Problem: active(f(f(a()))) -> mark(f(g(f(a())))) mark(f(X)) -> active(f(X)) mark(a()) -> active(a()) mark(g(X)) -> active(g(mark(X))) f(mark(X)) -> f(X) f(active(X)) -> f(X) g(mark(X)) -> g(X) g(active(X)) -> g(X) Proof: Matrix Interpretation Processor: dim=3 interpretation: [1 0 0] [mark](x0) = [0 0 0]x0 [0 1 1] , [1 0 0] [g](x0) = [0 1 1]x0 [0 0 0] , [1 0 0] [active](x0) = [0 0 0]x0 [0 1 1] , [1 0 1] [0] [f](x0) = [0 0 0]x0 + [0] [0 0 0] [1], [0] [a] = [0] [0] orientation: [1] [0] active(f(f(a()))) = [0] >= [0] = mark(f(g(f(a())))) [1] [1] [1 0 1] [0] [1 0 1] [0] mark(f(X)) = [0 0 0]X + [0] >= [0 0 0]X + [0] = active(f(X)) [0 0 0] [1] [0 0 0] [1] [0] [0] mark(a()) = [0] >= [0] = active(a()) [0] [0] [1 0 0] [1 0 0] mark(g(X)) = [0 0 0]X >= [0 0 0]X = active(g(mark(X))) [0 1 1] [0 1 1] [1 1 1] [0] [1 0 1] [0] f(mark(X)) = [0 0 0]X + [0] >= [0 0 0]X + [0] = f(X) [0 0 0] [1] [0 0 0] [1] [1 1 1] [0] [1 0 1] [0] f(active(X)) = [0 0 0]X + [0] >= [0 0 0]X + [0] = f(X) [0 0 0] [1] [0 0 0] [1] [1 0 0] [1 0 0] g(mark(X)) = [0 1 1]X >= [0 1 1]X = g(X) [0 0 0] [0 0 0] [1 0 0] [1 0 0] g(active(X)) = [0 1 1]X >= [0 1 1]X = g(X) [0 0 0] [0 0 0] problem: mark(f(X)) -> active(f(X)) mark(a()) -> active(a()) mark(g(X)) -> active(g(mark(X))) f(mark(X)) -> f(X) f(active(X)) -> f(X) g(mark(X)) -> g(X) g(active(X)) -> g(X) Matrix Interpretation Processor: dim=1 interpretation: [mark](x0) = 2x0, [g](x0) = x0 + 2, [active](x0) = x0, [f](x0) = 4x0, [a] = 5 orientation: mark(f(X)) = 8X >= 4X = active(f(X)) mark(a()) = 10 >= 5 = active(a()) mark(g(X)) = 2X + 4 >= 2X + 2 = active(g(mark(X))) f(mark(X)) = 8X >= 4X = f(X) f(active(X)) = 4X >= 4X = f(X) g(mark(X)) = 2X + 2 >= X + 2 = g(X) g(active(X)) = X + 2 >= X + 2 = g(X) problem: mark(f(X)) -> active(f(X)) f(mark(X)) -> f(X) f(active(X)) -> f(X) g(mark(X)) -> g(X) g(active(X)) -> g(X) Matrix Interpretation Processor: dim=1 interpretation: [mark](x0) = 3x0 + 6, [g](x0) = 4x0 + 6, [active](x0) = 3x0 + 2, [f](x0) = x0 + 4 orientation: mark(f(X)) = 3X + 18 >= 3X + 14 = active(f(X)) f(mark(X)) = 3X + 10 >= X + 4 = f(X) f(active(X)) = 3X + 6 >= X + 4 = f(X) g(mark(X)) = 12X + 30 >= 4X + 6 = g(X) g(active(X)) = 12X + 14 >= 4X + 6 = g(X) problem: Qed