YES(?,O(n^2)) Problem: q0(0(x1)) -> 0'(q1(x1)) q1(0(x1)) -> 0(q1(x1)) q1(1'(x1)) -> 1'(q1(x1)) 0(q1(1(x1))) -> q2(0(1'(x1))) 0'(q1(1(x1))) -> q2(0'(1'(x1))) 1'(q1(1(x1))) -> q2(1'(1'(x1))) 0(q2(0(x1))) -> q2(0(0(x1))) 0'(q2(0(x1))) -> q2(0'(0(x1))) 1'(q2(0(x1))) -> q2(1'(0(x1))) 0(q2(1'(x1))) -> q2(0(1'(x1))) 0'(q2(1'(x1))) -> q2(0'(1'(x1))) 1'(q2(1'(x1))) -> q2(1'(1'(x1))) q2(0'(x1)) -> 0'(q0(x1)) q0(1'(x1)) -> 1'(q3(x1)) q3(1'(x1)) -> 1'(q3(x1)) q3(b(x1)) -> b(q4(x1)) Proof: Matrix Interpretation Processor: dimension: 2 interpretation: [1 2] [2] [q4](x0) = [0 0]x0 + [0], [1 1] [4] [b](x0) = [0 1]x0 + [3], [1 1] [q3](x0) = [0 1]x0, [1 2] [1] [q2](x0) = [0 1]x0 + [3], [1 3] [2] [1](x0) = [0 1]x0 + [5], [1 2] [0] [1'](x0) = [0 1]x0 + [2], [1 1] [0'](x0) = [0 1]x0, [1 1] [1] [q1](x0) = [0 1]x0 + [0], [1 1] [q0](x0) = [0 1]x0, [1 3] [0] [0](x0) = [0 1]x0 + [4] orientation: [1 4] [4] [1 2] [1] q0(0(x1)) = [0 1]x1 + [4] >= [0 1]x1 + [0] = 0'(q1(x1)) [1 4] [5] [1 4] [1] q1(0(x1)) = [0 1]x1 + [4] >= [0 1]x1 + [4] = 0(q1(x1)) [1 3] [3] [1 3] [1] q1(1'(x1)) = [0 1]x1 + [2] >= [0 1]x1 + [2] = 1'(q1(x1)) [1 7] [23] [1 7] [19] 0(q1(1(x1))) = [0 1]x1 + [9 ] >= [0 1]x1 + [9 ] = q2(0(1'(x1))) [1 5] [13] [1 5] [7] 0'(q1(1(x1))) = [0 1]x1 + [5 ] >= [0 1]x1 + [5] = q2(0'(1'(x1))) [1 6] [18] [1 6] [13] 1'(q1(1(x1))) = [0 1]x1 + [7 ] >= [0 1]x1 + [7 ] = q2(1'(1'(x1))) [1 8] [30] [1 8] [29] 0(q2(0(x1))) = [0 1]x1 + [11] >= [0 1]x1 + [11] = q2(0(0(x1))) [1 6] [16] [1 6] [13] 0'(q2(0(x1))) = [0 1]x1 + [7 ] >= [0 1]x1 + [7 ] = q2(0'(0(x1))) [1 7] [23] [1 7] [21] 1'(q2(0(x1))) = [0 1]x1 + [9 ] >= [0 1]x1 + [9 ] = q2(1'(0(x1))) [1 7] [20] [1 7] [19] 0(q2(1'(x1))) = [0 1]x1 + [9 ] >= [0 1]x1 + [9 ] = q2(0(1'(x1))) [1 5] [10] [1 5] [7] 0'(q2(1'(x1))) = [0 1]x1 + [5 ] >= [0 1]x1 + [5] = q2(0'(1'(x1))) [1 6] [15] [1 6] [13] 1'(q2(1'(x1))) = [0 1]x1 + [7 ] >= [0 1]x1 + [7 ] = q2(1'(1'(x1))) [1 3] [1] [1 2] q2(0'(x1)) = [0 1]x1 + [3] >= [0 1]x1 = 0'(q0(x1)) [1 3] [2] [1 3] [0] q0(1'(x1)) = [0 1]x1 + [2] >= [0 1]x1 + [2] = 1'(q3(x1)) [1 3] [2] [1 3] [0] q3(1'(x1)) = [0 1]x1 + [2] >= [0 1]x1 + [2] = 1'(q3(x1)) [1 2] [7] [1 2] [6] q3(b(x1)) = [0 1]x1 + [3] >= [0 0]x1 + [3] = b(q4(x1)) problem: Qed