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=3 interpretation: [1 0 0] [1 0 0] [0] [.](x0, x1) = [0 0 0]x0 + [0 0 0]x1 + [1] [0 0 0] [0 0 1] [1], [++](x0, x1) = x0 + x1 , [1] [nil] = [0] [0] orientation: [1] ++(nil(),y) = y + [0] >= y = y [0] [1] ++(x,nil()) = x + [0] >= x = x [0] [1 0 0] [1 0 0] [0] [1 0 0] [1 0 0] [1 0 0] [0] ++(.(x,y),z) = [0 0 0]x + [0 0 0]y + z + [1] >= [0 0 0]x + [0 0 0]y + [0 0 0]z + [1] = .(x,++(y,z)) [0 0 0] [0 0 1] [1] [0 0 0] [0 0 1] [0 0 1] [1] ++(++(x,y),z) = x + y + z >= x + y + z = ++(x,++(y,z)) problem: ++(.(x,y),z) -> .(x,++(y,z)) ++(++(x,y),z) -> ++(x,++(y,z)) Matrix Interpretation Processor: dim=3 interpretation: [1 0 0] [0] [.](x0, x1) = [0 0 0]x0 + x1 + [1] [0 0 0] [1], [1 1 0] [++](x0, x1) = [0 0 1]x0 + x1 [0 0 1] orientation: [1 0 0] [1 1 0] [1] [1 0 0] [1 1 0] [0] ++(.(x,y),z) = [0 0 0]x + [0 0 1]y + z + [1] >= [0 0 0]x + [0 0 1]y + z + [1] = .(x,++(y,z)) [0 0 0] [0 0 1] [1] [0 0 0] [0 0 1] [1] [1 1 1] [1 1 0] [1 1 0] [1 1 0] ++(++(x,y),z) = [0 0 1]x + [0 0 1]y + z >= [0 0 1]x + [0 0 1]y + z = ++(x,++(y,z)) [0 0 1] [0 0 1] [0 0 1] [0 0 1] problem: ++(++(x,y),z) -> ++(x,++(y,z)) Matrix Interpretation Processor: dim=1 interpretation: [++](x0, x1) = 2x0 + x1 + 1 orientation: ++(++(x,y),z) = 4x + 2y + z + 3 >= 2x + 2y + z + 2 = ++(x,++(y,z)) problem: Qed