0.00/0.28	YES
0.00/0.28	
0.00/0.28	DP problem for innermost termination.
0.00/0.28	P =
0.00/0.28	  rand#(I3, y) -> rand#(I3 + 1, id_dec(y)) [0 > I3]
0.00/0.28	  rand#(I3, y) -> id_dec#(y) [0 > I3]
0.00/0.28	  rand#(I4, I5) -> rand#(I4 - 1, id_inc(I5)) [I4 > 0]
0.00/0.28	  rand#(I4, I5) -> id_inc#(I5) [I4 > 0]
0.00/0.28	  random#(I8) -> rand#(I8, 0) 
0.00/0.28	R = 
0.00/0.28	  id_dec(x) -> x - 1 
0.00/0.28	  id_dec(I0) -> I0 
0.00/0.28	  id_inc(I1) -> I1 + 1 
0.00/0.28	  id_inc(I2) -> I2 
0.00/0.28	  rand(I3, y) -> rand(I3 + 1, id_dec(y)) [0 > I3]
0.00/0.28	  rand(I4, I5) -> rand(I4 - 1, id_inc(I5)) [I4 > 0]
0.00/0.28	  rand(I6, I7) -> I7 [I6 = 0]
0.00/0.28	  random(I8) -> rand(I8, 0) 
0.00/0.28	
0.00/0.28	The dependency graph for this problem is:
0.00/0.28	  0 -> 0, 1
0.00/0.28	  1 -> 
0.00/0.28	  2 -> 2, 3
0.00/0.28	  3 -> 
0.00/0.28	  4 -> 0, 1, 2, 3
0.00/0.28	Where:
0.00/0.28	  0) rand#(I3, y) -> rand#(I3 + 1, id_dec(y)) [0 > I3]
0.00/0.28	  1) rand#(I3, y) -> id_dec#(y) [0 > I3]
0.00/0.28	  2) rand#(I4, I5) -> rand#(I4 - 1, id_inc(I5)) [I4 > 0]
0.00/0.28	  3) rand#(I4, I5) -> id_inc#(I5) [I4 > 0]
0.00/0.28	  4) random#(I8) -> rand#(I8, 0) 
0.00/0.28	
0.00/0.28	We have the following SCCs.
0.00/0.28	  { 2 }
0.00/0.28	  { 0 }
0.00/0.28	
0.00/0.28	DP problem for innermost termination.
0.00/0.28	P =
0.00/0.28	  rand#(I3, y) -> rand#(I3 + 1, id_dec(y)) [0 > I3]
0.00/0.28	R = 
0.00/0.28	  id_dec(x) -> x - 1 
0.00/0.28	  id_dec(I0) -> I0 
0.00/0.28	  id_inc(I1) -> I1 + 1 
0.00/0.28	  id_inc(I2) -> I2 
0.00/0.28	  rand(I3, y) -> rand(I3 + 1, id_dec(y)) [0 > I3]
0.00/0.28	  rand(I4, I5) -> rand(I4 - 1, id_inc(I5)) [I4 > 0]
0.00/0.28	  rand(I6, I7) -> I7 [I6 = 0]
0.00/0.28	  random(I8) -> rand(I8, 0) 
0.00/0.28	
0.00/0.28	We use the reverse value criterion with the projection function NU:
0.00/0.28	  NU[rand#(z1,z2)] = 0 + -1 * z1
0.00/0.28	
0.00/0.28	This gives the following inequalities:
0.00/0.28	  0 > I3  ==>  0 + -1 * I3 > 0 + -1 * (I3 + 1)  with  0 + -1 * I3 >= 0
0.00/0.28	
0.00/0.28	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/0.28	
0.00/0.28	DP problem for innermost termination.
0.00/0.28	P =
0.00/0.28	  rand#(I4, I5) -> rand#(I4 - 1, id_inc(I5)) [I4 > 0]
0.00/0.28	R = 
0.00/0.28	  id_dec(x) -> x - 1 
0.00/0.28	  id_dec(I0) -> I0 
0.00/0.28	  id_inc(I1) -> I1 + 1 
0.00/0.28	  id_inc(I2) -> I2 
0.00/0.28	  rand(I3, y) -> rand(I3 + 1, id_dec(y)) [0 > I3]
0.00/0.28	  rand(I4, I5) -> rand(I4 - 1, id_inc(I5)) [I4 > 0]
0.00/0.28	  rand(I6, I7) -> I7 [I6 = 0]
0.00/0.28	  random(I8) -> rand(I8, 0) 
0.00/0.28	
0.00/0.28	We use the reverse value criterion with the projection function NU:
0.00/0.28	  NU[rand#(z1,z2)] = z1
0.00/0.28	
0.00/0.28	This gives the following inequalities:
0.00/0.28	  I4 > 0  ==>  I4 > I4 - 1  with  I4 >= 0
0.00/0.28	
0.00/0.28	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.26	EOF