1.22/1.22	YES
1.22/1.22	
1.22/1.22	DP problem for innermost termination.
1.22/1.22	P =
1.22/1.22	  f4#(x1, x2) -> f3#(x1, x2) 
1.22/1.22	  f3#(I0, I1) -> f1#(I0, I1) 
1.22/1.22	  f2#(I2, I3) -> f1#(I2, I3) 
1.22/1.22	  f1#(I4, I5) -> f2#(1 + I4, I5) [I5 <= I4 /\ I4 <= I5]
1.22/1.22	R = 
1.22/1.22	  f4(x1, x2) -> f3(x1, x2) 
1.22/1.22	  f3(I0, I1) -> f1(I0, I1) 
1.22/1.22	  f2(I2, I3) -> f1(I2, I3) 
1.22/1.22	  f1(I4, I5) -> f2(1 + I4, I5) [I5 <= I4 /\ I4 <= I5]
1.22/1.22	
1.22/1.22	The dependency graph for this problem is:
1.22/1.22	  0 -> 1
1.22/1.22	  1 -> 3
1.22/1.22	  2 -> 3
1.22/1.22	  3 -> 2
1.22/1.22	Where:
1.22/1.22	  0) f4#(x1, x2) -> f3#(x1, x2) 
1.22/1.22	  1) f3#(I0, I1) -> f1#(I0, I1) 
1.22/1.22	  2) f2#(I2, I3) -> f1#(I2, I3) 
1.22/1.22	  3) f1#(I4, I5) -> f2#(1 + I4, I5) [I5 <= I4 /\ I4 <= I5]
1.22/1.22	
1.22/1.22	We have the following SCCs.
1.22/1.22	  { 2, 3 }
1.22/1.22	
1.22/1.22	DP problem for innermost termination.
1.22/1.22	P =
1.22/1.22	  f2#(I2, I3) -> f1#(I2, I3) 
1.22/1.22	  f1#(I4, I5) -> f2#(1 + I4, I5) [I5 <= I4 /\ I4 <= I5]
1.22/1.22	R = 
1.22/1.22	  f4(x1, x2) -> f3(x1, x2) 
1.22/1.22	  f3(I0, I1) -> f1(I0, I1) 
1.22/1.22	  f2(I2, I3) -> f1(I2, I3) 
1.22/1.22	  f1(I4, I5) -> f2(1 + I4, I5) [I5 <= I4 /\ I4 <= I5]
1.22/1.22	
1.22/1.22	We use the reverse value criterion with the projection function NU:
1.22/1.22	  NU[f1#(z1,z2)] = z2 + -1 * z1
1.22/1.22	  NU[f2#(z1,z2)] = z2 + -1 * z1
1.22/1.22	
1.22/1.22	This gives the following inequalities:
1.22/1.22	    ==>  I3 + -1 * I2 >= I3 + -1 * I2
1.22/1.22	  I5 <= I4 /\ I4 <= I5  ==>  I5 + -1 * I4 > I5 + -1 * (1 + I4)  with  I5 + -1 * I4 >= 0
1.22/1.22	
1.22/1.22	We remove all the strictly oriented dependency pairs.
1.22/1.22	
1.22/1.22	DP problem for innermost termination.
1.22/1.22	P =
1.22/1.22	  f2#(I2, I3) -> f1#(I2, I3) 
1.22/1.22	R = 
1.22/1.22	  f4(x1, x2) -> f3(x1, x2) 
1.22/1.22	  f3(I0, I1) -> f1(I0, I1) 
1.22/1.22	  f2(I2, I3) -> f1(I2, I3) 
1.22/1.22	  f1(I4, I5) -> f2(1 + I4, I5) [I5 <= I4 /\ I4 <= I5]
1.22/1.22	
1.22/1.22	The dependency graph for this problem is:
1.22/1.22	  2 -> 
1.22/1.22	Where:
1.22/1.22	  2) f2#(I2, I3) -> f1#(I2, I3) 
1.22/1.22	
1.22/1.22	We have the following SCCs.
1.22/1.22	  
1.22/4.20	EOF