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	  f4#(x1) -> f3#(x1) 
0.00/0.14	  f3#(I0) -> f1#(I0) [1 <= I0]
0.00/0.14	  f2#(I1) -> f1#(I1) 
0.00/0.14	  f1#(I2) -> f2#(-1 + I2) [1 <= -1 + I2]
0.00/0.14	R = 
0.00/0.14	  f4(x1) -> f3(x1) 
0.00/0.14	  f3(I0) -> f1(I0) [1 <= I0]
0.00/0.14	  f2(I1) -> f1(I1) 
0.00/0.14	  f1(I2) -> f2(-1 + I2) [1 <= -1 + I2]
0.00/0.14	
0.00/0.14	The dependency graph for this problem is:
0.00/0.14	  0 -> 1
0.00/0.14	  1 -> 3
0.00/0.14	  2 -> 3
0.00/0.14	  3 -> 2
0.00/0.14	Where:
0.00/0.14	  0) f4#(x1) -> f3#(x1) 
0.00/0.14	  1) f3#(I0) -> f1#(I0) [1 <= I0]
0.00/0.14	  2) f2#(I1) -> f1#(I1) 
0.00/0.14	  3) f1#(I2) -> f2#(-1 + I2) [1 <= -1 + I2]
0.00/0.14	
0.00/0.14	We have the following SCCs.
0.00/0.14	  { 2, 3 }
0.00/0.14	
0.00/0.14	DP problem for innermost termination.
0.00/0.14	P =
0.00/0.14	  f2#(I1) -> f1#(I1) 
0.00/0.14	  f1#(I2) -> f2#(-1 + I2) [1 <= -1 + I2]
0.00/0.14	R = 
0.00/0.14	  f4(x1) -> f3(x1) 
0.00/0.14	  f3(I0) -> f1(I0) [1 <= I0]
0.00/0.14	  f2(I1) -> f1(I1) 
0.00/0.14	  f1(I2) -> f2(-1 + I2) [1 <= -1 + I2]
0.00/0.14	
0.00/0.14	We use the basic value criterion with the projection function NU:
0.00/0.14	  NU[f1#(z1)] = z1
0.00/0.14	  NU[f2#(z1)] = z1
0.00/0.14	
0.00/0.14	This gives the following inequalities:
0.00/0.14	    ==>  I1 (>! \union =) I1
0.00/0.14	  1 <= -1 + I2  ==>  I2 >! -1 + I2
0.00/0.14	
0.00/0.14	We remove all the strictly oriented dependency pairs.
0.00/0.14	
0.00/0.14	DP problem for innermost termination.
0.00/0.14	P =
0.00/0.14	  f2#(I1) -> f1#(I1) 
0.00/0.14	R = 
0.00/0.14	  f4(x1) -> f3(x1) 
0.00/0.14	  f3(I0) -> f1(I0) [1 <= I0]
0.00/0.14	  f2(I1) -> f1(I1) 
0.00/0.14	  f1(I2) -> f2(-1 + I2) [1 <= -1 + I2]
0.00/0.14	
0.00/0.14	The dependency graph for this problem is:
0.00/0.14	  2 -> 
0.00/0.14	Where:
0.00/0.14	  2) f2#(I1) -> f1#(I1) 
0.00/0.14	
0.00/0.14	We have the following SCCs.
0.00/0.14	  
0.00/3.12	EOF