YES Problem: ++(nil(),y) -> y ++(x,nil()) -> x ++(.(x,y),z) -> .(x,++(y,z)) ++(++(x,y),z) -> ++(x,++(y,z)) Proof: Matrix Interpretation Processor: dim=1 interpretation: [.](x0, x1) = x0 + x1 + 2, [++](x0, x1) = x0 + x1 + 1, [nil] = 6 orientation: ++(nil(),y) = y + 7 >= y = y ++(x,nil()) = x + 7 >= x = x ++(.(x,y),z) = x + y + z + 3 >= x + y + z + 3 = .(x,++(y,z)) ++(++(x,y),z) = x + y + z + 2 >= x + y + z + 2 = ++(x,++(y,z)) problem: ++(.(x,y),z) -> .(x,++(y,z)) ++(++(x,y),z) -> ++(x,++(y,z)) Matrix Interpretation Processor: dim=1 interpretation: [.](x0, x1) = 2x0 + x1 + 4, [++](x0, x1) = 2x0 + x1 + 1 orientation: ++(.(x,y),z) = 4x + 2y + z + 9 >= 2x + 2y + z + 5 = .(x,++(y,z)) ++(++(x,y),z) = 4x + 2y + z + 3 >= 2x + 2y + z + 2 = ++(x,++(y,z)) problem: Qed