0.00/0.20	YES
0.00/0.20	
0.00/0.20	DP problem for innermost termination.
0.00/0.20	P =
0.00/0.20	  init#(x1) -> f1#(rnd1) 
0.00/0.20	  f3#(I0) -> f3#(I0 - 1) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1]
0.00/0.20	  f2#(I1) -> f2#(I1 - 1) [-1 <= I1 - 1]
0.00/0.20	  f2#(I2) -> f3#(I2) [-1 <= I2 - 1]
0.00/0.20	  f1#(I3) -> f2#(10) 
0.00/0.20	R = 
0.00/0.20	  init(x1) -> f1(rnd1) 
0.00/0.20	  f3(I0) -> f3(I0 - 1) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1]
0.00/0.20	  f2(I1) -> f2(I1 - 1) [-1 <= I1 - 1]
0.00/0.20	  f2(I2) -> f3(I2) [-1 <= I2 - 1]
0.00/0.20	  f1(I3) -> f2(10) 
0.00/0.20	
0.00/0.20	The dependency graph for this problem is:
0.00/0.20	  0 -> 4
0.00/0.20	  1 -> 1
0.00/0.20	  2 -> 2, 3
0.00/0.20	  3 -> 1
0.00/0.20	  4 -> 2, 3
0.00/0.20	Where:
0.00/0.20	  0) init#(x1) -> f1#(rnd1) 
0.00/0.20	  1) f3#(I0) -> f3#(I0 - 1) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1]
0.00/0.20	  2) f2#(I1) -> f2#(I1 - 1) [-1 <= I1 - 1]
0.00/0.20	  3) f2#(I2) -> f3#(I2) [-1 <= I2 - 1]
0.00/0.20	  4) f1#(I3) -> f2#(10) 
0.00/0.20	
0.00/0.20	We have the following SCCs.
0.00/0.20	  { 2 }
0.00/0.20	  { 1 }
0.00/0.20	
0.00/0.20	DP problem for innermost termination.
0.00/0.20	P =
0.00/0.20	  f3#(I0) -> f3#(I0 - 1) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1]
0.00/0.20	R = 
0.00/0.20	  init(x1) -> f1(rnd1) 
0.00/0.20	  f3(I0) -> f3(I0 - 1) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1]
0.00/0.20	  f2(I1) -> f2(I1 - 1) [-1 <= I1 - 1]
0.00/0.20	  f2(I2) -> f3(I2) [-1 <= I2 - 1]
0.00/0.20	  f1(I3) -> f2(10) 
0.00/0.20	
0.00/0.20	We use the basic value criterion with the projection function NU:
0.00/0.20	  NU[f3#(z1)] = z1
0.00/0.20	
0.00/0.20	This gives the following inequalities:
0.00/0.20	  I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1  ==>  I0 >! I0 - 1
0.00/0.20	
0.00/0.20	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/0.20	
0.00/0.20	DP problem for innermost termination.
0.00/0.20	P =
0.00/0.20	  f2#(I1) -> f2#(I1 - 1) [-1 <= I1 - 1]
0.00/0.20	R = 
0.00/0.20	  init(x1) -> f1(rnd1) 
0.00/0.20	  f3(I0) -> f3(I0 - 1) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1]
0.00/0.20	  f2(I1) -> f2(I1 - 1) [-1 <= I1 - 1]
0.00/0.20	  f2(I2) -> f3(I2) [-1 <= I2 - 1]
0.00/0.20	  f1(I3) -> f2(10) 
0.00/0.20	
0.00/0.20	We use the basic value criterion with the projection function NU:
0.00/0.20	  NU[f2#(z1)] = z1
0.00/0.20	
0.00/0.20	This gives the following inequalities:
0.00/0.20	  -1 <= I1 - 1  ==>  I1 >! I1 - 1
0.00/0.20	
0.00/0.20	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.19	EOF