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