YES Problem: main(S(x1)) -> x1 main(0()) -> 0() Proof: Matrix Interpretation Processor: dim=3 interpretation: [0] [0] = [0] [0], [1 0 0] [1] [main](x0) = [0 0 1]x0 + [0] [0 1 0] [0], [1 0 0] [S](x0) = [0 0 1]x0 [0 1 0] orientation: [1] main(S(x1)) = x1 + [0] >= x1 = x1 [0] [1] [0] main(0()) = [0] >= [0] = 0() [0] [0] problem: Qed