0.00/0.33	YES
0.00/0.33	
0.00/0.33	DP problem for innermost termination.
0.00/0.33	P =
0.00/0.33	  f5#(x1, x2) -> f4#(x1, x2) 
0.00/0.33	  f4#(I0, I1) -> f3#(0, I1) 
0.00/0.33	  f3#(I2, I3) -> f1#(I2, I3) 
0.00/0.33	  f1#(I4, I5) -> f3#(I4, -1 + I5) [1 <= I5]
0.00/0.33	R = 
0.00/0.33	  f5(x1, x2) -> f4(x1, x2) 
0.00/0.33	  f4(I0, I1) -> f3(0, I1) 
0.00/0.33	  f3(I2, I3) -> f1(I2, I3) 
0.00/0.33	  f1(I4, I5) -> f3(I4, -1 + I5) [1 <= I5]
0.00/0.33	  f1(I6, I7) -> f2(1, I7) [I7 <= 0]
0.00/0.33	
0.00/0.33	The dependency graph for this problem is:
0.00/0.33	  0 -> 1
0.00/0.33	  1 -> 2
0.00/0.33	  2 -> 3
0.00/0.33	  3 -> 2
0.00/0.33	Where:
0.00/0.33	  0) f5#(x1, x2) -> f4#(x1, x2) 
0.00/0.33	  1) f4#(I0, I1) -> f3#(0, I1) 
0.00/0.33	  2) f3#(I2, I3) -> f1#(I2, I3) 
0.00/0.33	  3) f1#(I4, I5) -> f3#(I4, -1 + I5) [1 <= I5]
0.00/0.33	
0.00/0.33	We have the following SCCs.
0.00/0.33	  { 2, 3 }
0.00/0.33	
0.00/0.33	DP problem for innermost termination.
0.00/0.33	P =
0.00/0.33	  f3#(I2, I3) -> f1#(I2, I3) 
0.00/0.33	  f1#(I4, I5) -> f3#(I4, -1 + I5) [1 <= I5]
0.00/0.33	R = 
0.00/0.33	  f5(x1, x2) -> f4(x1, x2) 
0.00/0.33	  f4(I0, I1) -> f3(0, I1) 
0.00/0.33	  f3(I2, I3) -> f1(I2, I3) 
0.00/0.33	  f1(I4, I5) -> f3(I4, -1 + I5) [1 <= I5]
0.00/0.33	  f1(I6, I7) -> f2(1, I7) [I7 <= 0]
0.00/0.33	
0.00/0.33	We use the basic value criterion with the projection function NU:
0.00/0.33	  NU[f1#(z1,z2)] = z2
0.00/0.33	  NU[f3#(z1,z2)] = z2
0.00/0.33	
0.00/0.33	This gives the following inequalities:
0.00/0.33	    ==>  I3 (>! \union =) I3
0.00/0.33	  1 <= I5  ==>  I5 >! -1 + I5
0.00/0.33	
0.00/0.33	We remove all the strictly oriented dependency pairs.
0.00/0.33	
0.00/0.33	DP problem for innermost termination.
0.00/0.33	P =
0.00/0.33	  f3#(I2, I3) -> f1#(I2, I3) 
0.00/0.33	R = 
0.00/0.33	  f5(x1, x2) -> f4(x1, x2) 
0.00/0.33	  f4(I0, I1) -> f3(0, I1) 
0.00/0.33	  f3(I2, I3) -> f1(I2, I3) 
0.00/0.33	  f1(I4, I5) -> f3(I4, -1 + I5) [1 <= I5]
0.00/0.33	  f1(I6, I7) -> f2(1, I7) [I7 <= 0]
0.00/0.33	
0.00/0.33	The dependency graph for this problem is:
0.00/0.33	  2 -> 
0.00/0.33	Where:
0.00/0.33	  2) f3#(I2, I3) -> f1#(I2, I3) 
0.00/0.33	
0.00/0.33	We have the following SCCs.
0.00/0.33	  
0.00/3.31	EOF