0.00/0.30	YES
0.00/0.30	
0.00/0.30	DP problem for innermost termination.
0.00/0.30	P =
0.00/0.30	  rand#(I3, y) -> rand#(I3 + 1, id_dec(y)) [0 > I3]
0.00/0.30	  rand#(I3, y) -> id_dec#(y) [0 > I3]
0.00/0.30	  rand#(I4, B0) -> rand#(I4 - 1, id_inc(B0)) [I4 > 0]
0.00/0.30	  rand#(I4, B0) -> id_inc#(B0) [I4 > 0]
0.00/0.30	  random#(I6) -> rand#(I6, w(0)) 
0.00/0.30	R = 
0.00/0.30	  id_dec(w(x)) -> w(x - 1) 
0.00/0.30	  id_dec(w(I0)) -> w(I0) 
0.00/0.30	  id_inc(w(I1)) -> w(I1 + 1) 
0.00/0.30	  id_inc(w(I2)) -> w(I2) 
0.00/0.30	  rand(I3, y) -> rand(I3 + 1, id_dec(y)) [0 > I3]
0.00/0.30	  rand(I4, B0) -> rand(I4 - 1, id_inc(B0)) [I4 > 0]
0.00/0.30	  rand(I5, B1) -> B1 [I5 = 0]
0.00/0.30	  random(I6) -> rand(I6, w(0)) 
0.00/0.30	
0.00/0.30	The dependency graph for this problem is:
0.00/0.30	  0 -> 0, 1
0.00/0.30	  1 -> 
0.00/0.30	  2 -> 2, 3
0.00/0.30	  3 -> 
0.00/0.30	  4 -> 0, 1, 2, 3
0.00/0.30	Where:
0.00/0.30	  0) rand#(I3, y) -> rand#(I3 + 1, id_dec(y)) [0 > I3]
0.00/0.30	  1) rand#(I3, y) -> id_dec#(y) [0 > I3]
0.00/0.30	  2) rand#(I4, B0) -> rand#(I4 - 1, id_inc(B0)) [I4 > 0]
0.00/0.30	  3) rand#(I4, B0) -> id_inc#(B0) [I4 > 0]
0.00/0.30	  4) random#(I6) -> rand#(I6, w(0)) 
0.00/0.30	
0.00/0.30	We have the following SCCs.
0.00/0.30	  { 2 }
0.00/0.30	  { 0 }
0.00/0.30	
0.00/0.30	DP problem for innermost termination.
0.00/0.30	P =
0.00/0.30	  rand#(I3, y) -> rand#(I3 + 1, id_dec(y)) [0 > I3]
0.00/0.30	R = 
0.00/0.30	  id_dec(w(x)) -> w(x - 1) 
0.00/0.30	  id_dec(w(I0)) -> w(I0) 
0.00/0.30	  id_inc(w(I1)) -> w(I1 + 1) 
0.00/0.30	  id_inc(w(I2)) -> w(I2) 
0.00/0.30	  rand(I3, y) -> rand(I3 + 1, id_dec(y)) [0 > I3]
0.00/0.30	  rand(I4, B0) -> rand(I4 - 1, id_inc(B0)) [I4 > 0]
0.00/0.30	  rand(I5, B1) -> B1 [I5 = 0]
0.00/0.30	  random(I6) -> rand(I6, w(0)) 
0.00/0.30	
0.00/0.30	We use the reverse value criterion with the projection function NU:
0.00/0.30	  NU[rand#(z1,z2)] = 0 + -1 * z1
0.00/0.30	
0.00/0.30	This gives the following inequalities:
0.00/0.30	  0 > I3  ==>  0 + -1 * I3 > 0 + -1 * (I3 + 1)  with  0 + -1 * I3 >= 0
0.00/0.30	
0.00/0.30	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/0.30	
0.00/0.30	DP problem for innermost termination.
0.00/0.30	P =
0.00/0.30	  rand#(I4, B0) -> rand#(I4 - 1, id_inc(B0)) [I4 > 0]
0.00/0.30	R = 
0.00/0.30	  id_dec(w(x)) -> w(x - 1) 
0.00/0.30	  id_dec(w(I0)) -> w(I0) 
0.00/0.30	  id_inc(w(I1)) -> w(I1 + 1) 
0.00/0.30	  id_inc(w(I2)) -> w(I2) 
0.00/0.30	  rand(I3, y) -> rand(I3 + 1, id_dec(y)) [0 > I3]
0.00/0.30	  rand(I4, B0) -> rand(I4 - 1, id_inc(B0)) [I4 > 0]
0.00/0.30	  rand(I5, B1) -> B1 [I5 = 0]
0.00/0.30	  random(I6) -> rand(I6, w(0)) 
0.00/0.30	
0.00/0.30	We use the reverse value criterion with the projection function NU:
0.00/0.30	  NU[rand#(z1,z2)] = z1
0.00/0.30	
0.00/0.30	This gives the following inequalities:
0.00/0.30	  I4 > 0  ==>  I4 > I4 - 1  with  I4 >= 0
0.00/0.30	
0.00/0.30	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.28	EOF