YES(?,O(n^1))
0.16/0.23	YES(?,O(n^1))
0.16/0.23	
0.16/0.23	Problem:
0.16/0.23	 double(0()) -> 0()
0.16/0.23	 double(s(x)) -> s(s(double(x)))
0.16/0.23	 half(0()) -> 0()
0.16/0.23	 half(s(0())) -> 0()
0.16/0.23	 half(s(s(x))) -> s(half(x))
0.16/0.23	 -(x,0()) -> x
0.16/0.23	 -(s(x),s(y)) -> -(x,y)
0.16/0.23	 if(0(),y,z) -> y
0.16/0.23	 if(s(x),y,z) -> z
0.16/0.23	 half(double(x)) -> x
0.16/0.23	
0.16/0.23	Proof:
0.16/0.23	 Bounds Processor:
0.16/0.23	  bound: 1
0.16/0.23	  enrichment: match
0.16/0.23	  automaton:
0.16/0.23	   final states: {6,5,4,3}
0.16/0.23	   transitions:
0.16/0.23	    -1(1,2) -> 5*
0.16/0.23	    -1(2,1) -> 5*
0.16/0.23	    -1(1,1) -> 5*
0.16/0.23	    -1(2,2) -> 5*
0.16/0.23	    s1(12) -> 12,4
0.16/0.23	    s1(7) -> 8*
0.16/0.23	    s1(8) -> 7,3
0.16/0.23	    half1(2) -> 12*
0.16/0.23	    half1(1) -> 12*
0.16/0.23	    01() -> 12,7,4,3
0.16/0.23	    double1(2) -> 7*
0.16/0.23	    double1(1) -> 7*
0.16/0.23	    double0(2) -> 3*
0.16/0.23	    double0(1) -> 3*
0.16/0.23	    00() -> 1*
0.16/0.23	    s0(2) -> 2*
0.16/0.23	    s0(1) -> 2*
0.16/0.23	    half0(2) -> 4*
0.16/0.23	    half0(1) -> 4*
0.16/0.23	    -0(1,2) -> 5*
0.16/0.23	    -0(2,1) -> 5*
0.16/0.23	    -0(1,1) -> 5*
0.16/0.23	    -0(2,2) -> 5*
0.16/0.23	    if0(1,1,1) -> 6*
0.16/0.23	    if0(2,2,1) -> 6*
0.16/0.23	    if0(1,1,2) -> 6*
0.16/0.23	    if0(2,2,2) -> 6*
0.16/0.23	    if0(1,2,1) -> 6*
0.16/0.23	    if0(2,1,1) -> 6*
0.16/0.23	    if0(1,2,2) -> 6*
0.16/0.23	    if0(2,1,2) -> 6*
0.16/0.23	    1 -> 6,5
0.16/0.23	    2 -> 6,5
0.16/0.23	  problem:
0.16/0.23	   
0.16/0.23	  Qed
0.16/0.24	EOF