0.00/0.14	YES
0.00/0.14	
0.00/0.14	DP problem for innermost termination.
0.00/0.14	P =
0.00/0.14	  rand#(I1, y) -> rand#(I1 - 1, id_inc(y)) [I1 > 0]
0.00/0.14	  rand#(I1, y) -> id_inc#(y) [I1 > 0]
0.00/0.14	  random#(I4) -> rand#(I4, 0) [I4 >= 0]
0.00/0.14	R = 
0.00/0.14	  id_inc(x) -> x + 1 
0.00/0.14	  id_inc(I0) -> I0 
0.00/0.14	  rand(I1, y) -> rand(I1 - 1, id_inc(y)) [I1 > 0]
0.00/0.14	  rand(I2, I3) -> I3 [I2 = 0]
0.00/0.14	  random(I4) -> rand(I4, 0) [I4 >= 0]
0.00/0.14	
0.00/0.14	The dependency graph for this problem is:
0.00/0.14	  0 -> 0, 1
0.00/0.14	  1 -> 
0.00/0.14	  2 -> 0, 1
0.00/0.14	Where:
0.00/0.14	  0) rand#(I1, y) -> rand#(I1 - 1, id_inc(y)) [I1 > 0]
0.00/0.14	  1) rand#(I1, y) -> id_inc#(y) [I1 > 0]
0.00/0.14	  2) random#(I4) -> rand#(I4, 0) [I4 >= 0]
0.00/0.14	
0.00/0.14	We have the following SCCs.
0.00/0.14	  { 0 }
0.00/0.14	
0.00/0.14	DP problem for innermost termination.
0.00/0.14	P =
0.00/0.14	  rand#(I1, y) -> rand#(I1 - 1, id_inc(y)) [I1 > 0]
0.00/0.14	R = 
0.00/0.14	  id_inc(x) -> x + 1 
0.00/0.14	  id_inc(I0) -> I0 
0.00/0.14	  rand(I1, y) -> rand(I1 - 1, id_inc(y)) [I1 > 0]
0.00/0.14	  rand(I2, I3) -> I3 [I2 = 0]
0.00/0.14	  random(I4) -> rand(I4, 0) [I4 >= 0]
0.00/0.14	
0.00/0.14	We use the reverse value criterion with the projection function NU:
0.00/0.14	  NU[rand#(z1,z2)] = z1
0.00/0.14	
0.00/0.14	This gives the following inequalities:
0.00/0.14	  I1 > 0  ==>  I1 > I1 - 1  with  I1 >= 0
0.00/0.14	
0.00/0.14	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.12	EOF