YES Problem: d(0()) -> 0() d(s(x)) -> s(s(d(x))) e(0()) -> s(0()) e(r(x)) -> d(e(x)) Proof: Matrix Interpretation Processor: dim=3 interpretation: [1 1 1] [1] [r](x0) = [1 1 1]x0 + [1] [1 1 1] [0], [1 0 0] [0] [e](x0) = [0 0 1]x0 + [1] [0 1 0] [0], [1 0 0] [0] [s](x0) = [0 0 1]x0 + [1] [0 1 0] [0], [1 1 1] [d](x0) = [1 1 1]x0 [1 1 1] , [1] [0] = [0] [0] orientation: [1] [1] d(0()) = [1] >= [0] = 0() [1] [0] [1 1 1] [1] [1 1 1] [0] d(s(x)) = [1 1 1]x + [1] >= [1 1 1]x + [1] = s(s(d(x))) [1 1 1] [1] [1 1 1] [1] [1] [1] e(0()) = [1] >= [1] = s(0()) [0] [0] [1 1 1] [1] [1 1 1] [1] e(r(x)) = [1 1 1]x + [1] >= [1 1 1]x + [1] = d(e(x)) [1 1 1] [1] [1 1 1] [1] problem: d(0()) -> 0() e(0()) -> s(0()) e(r(x)) -> d(e(x)) Matrix Interpretation Processor: dim=3 interpretation: [1 0 1] [0] [r](x0) = [0 0 0]x0 + [0] [0 0 0] [1], [1 0 1] [e](x0) = [0 0 0]x0 [0 0 0] , [1 0 0] [s](x0) = [0 0 0]x0 [0 0 0] , [1 0 1] [d](x0) = [0 0 0]x0 [0 0 1] , [0] [0] = [0] [1] orientation: [1] [0] d(0()) = [0] >= [0] = 0() [1] [1] [1] [0] e(0()) = [0] >= [0] = s(0()) [0] [0] [1 0 1] [1] [1 0 1] e(r(x)) = [0 0 0]x + [0] >= [0 0 0]x = d(e(x)) [0 0 0] [0] [0 0 0] problem: Qed