0.00/0.34	MAYBE
0.00/0.34	
0.00/0.34	DP problem for innermost termination.
0.00/0.34	P =
0.00/0.34	  rand#(I3, y) -> rand#(0 - I3 - 1, id_dec(y)) [0 > I3]
0.00/0.34	  rand#(I3, y) -> id_dec#(y) [0 > I3]
0.00/0.34	  rand#(I4, I5) -> rand#(0 - I4 + 1, id_inc(I5)) [I4 > 0]
0.00/0.34	  rand#(I4, I5) -> id_inc#(I5) [I4 > 0]
0.00/0.34	  random#(I8) -> rand#(I8, 0) 
0.00/0.34	R = 
0.00/0.34	  id_dec(x) -> x - 1 
0.00/0.34	  id_dec(I0) -> I0 
0.00/0.34	  id_inc(I1) -> I1 + 1 
0.00/0.34	  id_inc(I2) -> I2 
0.00/0.34	  rand(I3, y) -> rand(0 - I3 - 1, id_dec(y)) [0 > I3]
0.00/0.34	  rand(I4, I5) -> rand(0 - I4 + 1, id_inc(I5)) [I4 > 0]
0.00/0.34	  rand(I6, I7) -> I7 [I6 = 0]
0.00/0.34	  random(I8) -> rand(I8, 0) 
0.00/0.34	
0.00/0.34	The dependency graph for this problem is:
0.00/0.34	  0 -> 2, 3
0.00/0.34	  1 -> 
0.00/0.34	  2 -> 0, 1
0.00/0.34	  3 -> 
0.00/0.34	  4 -> 0, 1, 2, 3
0.00/0.34	Where:
0.00/0.34	  0) rand#(I3, y) -> rand#(0 - I3 - 1, id_dec(y)) [0 > I3]
0.00/0.34	  1) rand#(I3, y) -> id_dec#(y) [0 > I3]
0.00/0.34	  2) rand#(I4, I5) -> rand#(0 - I4 + 1, id_inc(I5)) [I4 > 0]
0.00/0.34	  3) rand#(I4, I5) -> id_inc#(I5) [I4 > 0]
0.00/0.34	  4) random#(I8) -> rand#(I8, 0) 
0.00/0.34	
0.00/0.34	We have the following SCCs.
0.00/0.34	  { 0, 2 }
0.00/0.34	
0.00/0.34	DP problem for innermost termination.
0.00/0.34	P =
0.00/0.34	  rand#(I3, y) -> rand#(0 - I3 - 1, id_dec(y)) [0 > I3]
0.00/0.34	  rand#(I4, I5) -> rand#(0 - I4 + 1, id_inc(I5)) [I4 > 0]
0.00/0.35	R = 
0.00/0.35	  id_dec(x) -> x - 1 
0.00/0.35	  id_dec(I0) -> I0 
0.00/0.35	  id_inc(I1) -> I1 + 1 
0.00/0.35	  id_inc(I2) -> I2 
0.00/0.35	  rand(I3, y) -> rand(0 - I3 - 1, id_dec(y)) [0 > I3]
0.00/0.35	  rand(I4, I5) -> rand(0 - I4 + 1, id_inc(I5)) [I4 > 0]
0.00/0.35	  rand(I6, I7) -> I7 [I6 = 0]
0.00/0.35	  random(I8) -> rand(I8, 0) 
0.00/0.35	
0.00/3.33	EOF