0.00/0.11	YES
0.00/0.11	
0.00/0.11	DP problem for innermost termination.
0.00/0.11	P =
0.00/0.11	  cd#(true, x) -> cd#(x > 0, x - 1) 
0.00/0.11	R = 
0.00/0.11	  cd(true, x) -> cd(x > 0, x - 1) 
0.00/0.11	
0.00/0.11	This problem is converted using chaining, where edges between chained DPs are removed.
0.00/0.11	
0.00/0.11	DP problem for innermost termination.
0.00/0.11	P =
0.00/0.11	  cd#(true, x) -> cd_1#(x) 
0.00/0.11	  cd_1#(x) -> cd#(x > 0, x - 1) 
0.00/0.11	  cd_1#(x) -> cd_1#(x - 1) [x > 0]
0.00/0.11	R = 
0.00/0.11	  cd(true, x) -> cd(x > 0, x - 1) 
0.00/0.11	
0.00/0.11	The dependency graph for this problem is:
0.00/0.11	  0 -> 2, 1
0.00/0.11	  1 -> 
0.00/0.11	  2 -> 2, 1
0.00/0.11	Where:
0.00/0.11	  0) cd#(true, x) -> cd_1#(x) 
0.00/0.11	  1) cd_1#(x) -> cd#(x > 0, x - 1) 
0.00/0.11	  2) cd_1#(x) -> cd_1#(x - 1) [x > 0]
0.00/0.11	
0.00/0.11	We have the following SCCs.
0.00/0.11	  { 2 }
0.00/0.11	
0.00/0.11	DP problem for innermost termination.
0.00/0.11	P =
0.00/0.11	  cd_1#(x) -> cd_1#(x - 1) [x > 0]
0.00/0.11	R = 
0.00/0.11	  cd(true, x) -> cd(x > 0, x - 1) 
0.00/0.11	
0.00/0.11	We use the reverse value criterion with the projection function NU:
0.00/0.11	  NU[cd_1#(z1)] = z1
0.00/0.11	
0.00/0.11	This gives the following inequalities:
0.00/0.11	  x > 0  ==>  x > x - 1  with  x >= 0
0.00/0.11	
0.00/0.11	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.09	EOF