0.00/0.13	YES
0.00/0.13	
0.00/0.13	DP problem for innermost termination.
0.00/0.13	P =
0.00/0.13	  rand#(I1, y) -> rand#(I1 - 1, id_inc(y)) [I1 > 0]
0.00/0.13	  rand#(I1, y) -> id_inc#(y) [I1 > 0]
0.00/0.13	  random#(I3) -> rand#(I3, w(0)) [I3 >= 0]
0.00/0.13	R = 
0.00/0.13	  id_inc(w(x)) -> w(x + 1) 
0.00/0.13	  id_inc(w(I0)) -> w(I0) 
0.00/0.13	  rand(I1, y) -> rand(I1 - 1, id_inc(y)) [I1 > 0]
0.00/0.13	  rand(I2, B0) -> B0 [I2 = 0]
0.00/0.13	  random(I3) -> rand(I3, w(0)) [I3 >= 0]
0.00/0.13	
0.00/0.13	The dependency graph for this problem is:
0.00/0.13	  0 -> 0, 1
0.00/0.13	  1 -> 
0.00/0.13	  2 -> 0, 1
0.00/0.13	Where:
0.00/0.13	  0) rand#(I1, y) -> rand#(I1 - 1, id_inc(y)) [I1 > 0]
0.00/0.13	  1) rand#(I1, y) -> id_inc#(y) [I1 > 0]
0.00/0.13	  2) random#(I3) -> rand#(I3, w(0)) [I3 >= 0]
0.00/0.13	
0.00/0.13	We have the following SCCs.
0.00/0.13	  { 0 }
0.00/0.13	
0.00/0.13	DP problem for innermost termination.
0.00/0.13	P =
0.00/0.13	  rand#(I1, y) -> rand#(I1 - 1, id_inc(y)) [I1 > 0]
0.00/0.13	R = 
0.00/0.13	  id_inc(w(x)) -> w(x + 1) 
0.00/0.13	  id_inc(w(I0)) -> w(I0) 
0.00/0.13	  rand(I1, y) -> rand(I1 - 1, id_inc(y)) [I1 > 0]
0.00/0.13	  rand(I2, B0) -> B0 [I2 = 0]
0.00/0.13	  random(I3) -> rand(I3, w(0)) [I3 >= 0]
0.00/0.13	
0.00/0.13	We use the reverse value criterion with the projection function NU:
0.00/0.13	  NU[rand#(z1,z2)] = z1
0.00/0.13	
0.00/0.13	This gives the following inequalities:
0.00/0.13	  I1 > 0  ==>  I1 > I1 - 1  with  I1 >= 0
0.00/0.13	
0.00/0.13	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.11	EOF