YES Problem: f(a(),f(a(),x)) -> f(a(),f(f(a(),a()),f(a(),x))) Proof: Uncurry Processor: a2(a1(x),x3) -> a2(a2(a(),a1(x)),x3) a1(a1(x)) -> a1(a2(a(),a1(x))) f(a1(x1),x2) -> a2(x1,x2) f(a(),x2) -> a1(x2) Matrix Interpretation Processor: dim=3 interpretation: [1 0 1] [1 0 0] [a2](x0, x1) = [0 0 0]x0 + [0 0 0]x1 [0 0 0] [0 0 0] , [1 0 1] [0] [a1](x0) = [0 0 0]x0 + [0] [0 0 0] [1], [1 0 0] [1 0 1] [0] [f](x0, x1) = [0 0 0]x0 + [1 0 0]x1 + [0] [0 0 0] [0 0 0] [1], [0] [a] = [0] [0] orientation: [1 0 1] [1 0 0] [1] [1 0 1] [1 0 0] a2(a1(x),x3) = [0 0 0]x + [0 0 0]x3 + [0] >= [0 0 0]x + [0 0 0]x3 = a2(a2(a(),a1(x)),x3) [0 0 0] [0 0 0] [0] [0 0 0] [0 0 0] [1 0 1] [1] [1 0 1] [0] a1(a1(x)) = [0 0 0]x + [0] >= [0 0 0]x + [0] = a1(a2(a(),a1(x))) [0 0 0] [1] [0 0 0] [1] [1 0 1] [1 0 1] [0] [1 0 1] [1 0 0] f(a1(x1),x2) = [0 0 0]x1 + [1 0 0]x2 + [0] >= [0 0 0]x1 + [0 0 0]x2 = a2(x1,x2) [0 0 0] [0 0 0] [1] [0 0 0] [0 0 0] [1 0 1] [0] [1 0 1] [0] f(a(),x2) = [1 0 0]x2 + [0] >= [0 0 0]x2 + [0] = a1(x2) [0 0 0] [1] [0 0 0] [1] problem: f(a1(x1),x2) -> a2(x1,x2) f(a(),x2) -> a1(x2) Matrix Interpretation Processor: dim=3 interpretation: [1 0 0] [1 0 0] [a2](x0, x1) = [0 0 0]x0 + [0 0 0]x1 [0 0 0] [0 0 0] , [1 0 0] [a1](x0) = [0 0 0]x0 [0 0 0] , [1 0 0] [1 0 0] [1] [f](x0, x1) = [0 0 0]x0 + [0 0 0]x1 + [0] [0 0 0] [0 0 0] [0], [0] [a] = [0] [0] orientation: [1 0 0] [1 0 0] [1] [1 0 0] [1 0 0] f(a1(x1),x2) = [0 0 0]x1 + [0 0 0]x2 + [0] >= [0 0 0]x1 + [0 0 0]x2 = a2(x1,x2) [0 0 0] [0 0 0] [0] [0 0 0] [0 0 0] [1 0 0] [1] [1 0 0] f(a(),x2) = [0 0 0]x2 + [0] >= [0 0 0]x2 = a1(x2) [0 0 0] [0] [0 0 0] problem: Qed