0.95/1.15	YES
0.95/1.15	
0.95/1.15	DP problem for innermost termination.
0.95/1.15	P =
0.95/1.15	  f5#(x1, x2) -> f4#(x1, x2) 
0.95/1.15	  f4#(I0, I1) -> f3#(1, I1) 
0.95/1.15	  f3#(I2, I3) -> f1#(I2, I3) 
0.95/1.15	  f1#(I4, I5) -> f3#(1 + I4, 10 - I4) [I4 <= 10]
0.95/1.15	R = 
0.95/1.15	  f5(x1, x2) -> f4(x1, x2) 
0.95/1.15	  f4(I0, I1) -> f3(1, I1) 
0.95/1.15	  f3(I2, I3) -> f1(I2, I3) 
0.95/1.15	  f1(I4, I5) -> f3(1 + I4, 10 - I4) [I4 <= 10]
0.95/1.15	  f1(I6, I7) -> f2(I6, I7) [11 <= I6]
0.95/1.15	
0.95/1.15	The dependency graph for this problem is:
0.95/1.15	  0 -> 1
0.95/1.15	  1 -> 2
0.95/1.15	  2 -> 3
0.95/1.15	  3 -> 2
0.95/1.15	Where:
0.95/1.15	  0) f5#(x1, x2) -> f4#(x1, x2) 
0.95/1.15	  1) f4#(I0, I1) -> f3#(1, I1) 
0.95/1.15	  2) f3#(I2, I3) -> f1#(I2, I3) 
0.95/1.15	  3) f1#(I4, I5) -> f3#(1 + I4, 10 - I4) [I4 <= 10]
0.95/1.15	
0.95/1.15	We have the following SCCs.
0.95/1.15	  { 2, 3 }
0.95/1.15	
0.95/1.15	DP problem for innermost termination.
0.95/1.15	P =
0.95/1.15	  f3#(I2, I3) -> f1#(I2, I3) 
0.95/1.15	  f1#(I4, I5) -> f3#(1 + I4, 10 - I4) [I4 <= 10]
0.95/1.15	R = 
0.95/1.15	  f5(x1, x2) -> f4(x1, x2) 
0.95/1.15	  f4(I0, I1) -> f3(1, I1) 
0.95/1.15	  f3(I2, I3) -> f1(I2, I3) 
0.95/1.15	  f1(I4, I5) -> f3(1 + I4, 10 - I4) [I4 <= 10]
0.95/1.15	  f1(I6, I7) -> f2(I6, I7) [11 <= I6]
0.95/1.15	
0.95/1.15	We use the reverse value criterion with the projection function NU:
0.95/1.15	  NU[f1#(z1,z2)] = 10 + -1 * z1
0.95/1.15	  NU[f3#(z1,z2)] = 10 + -1 * z1
0.95/1.15	
0.95/1.15	This gives the following inequalities:
0.95/1.15	    ==>  10 + -1 * I2 >= 10 + -1 * I2
0.95/1.15	  I4 <= 10  ==>  10 + -1 * I4 > 10 + -1 * (1 + I4)  with  10 + -1 * I4 >= 0
0.95/1.15	
0.95/1.15	We remove all the strictly oriented dependency pairs.
0.95/1.15	
0.95/1.15	DP problem for innermost termination.
0.95/1.15	P =
0.95/1.15	  f3#(I2, I3) -> f1#(I2, I3) 
0.95/1.15	R = 
0.95/1.15	  f5(x1, x2) -> f4(x1, x2) 
0.95/1.15	  f4(I0, I1) -> f3(1, I1) 
0.95/1.15	  f3(I2, I3) -> f1(I2, I3) 
0.95/1.15	  f1(I4, I5) -> f3(1 + I4, 10 - I4) [I4 <= 10]
0.95/1.15	  f1(I6, I7) -> f2(I6, I7) [11 <= I6]
0.95/1.15	
0.95/1.15	The dependency graph for this problem is:
0.95/1.15	  2 -> 
0.95/1.15	Where:
0.95/1.15	  2) f3#(I2, I3) -> f1#(I2, I3) 
0.95/1.15	
0.95/1.15	We have the following SCCs.
0.95/1.15	  
0.95/4.13	EOF