0.82/1.21	YES
0.82/1.21	
0.82/1.21	DP problem for innermost termination.
0.82/1.21	P =
0.82/1.21	  f4#(x1, x2) -> f3#(x1, x2) 
0.82/1.21	  f3#(I0, I1) -> f1#(I0, I1) 
0.82/1.21	  f2#(I2, I3) -> f1#(I2, I3) 
0.82/1.21	  f1#(I4, I5) -> f2#(I4, -1 + I5) [1 + I4 <= -1 + I5]
0.82/1.21	R = 
0.82/1.21	  f4(x1, x2) -> f3(x1, x2) 
0.82/1.21	  f3(I0, I1) -> f1(I0, I1) 
0.82/1.21	  f2(I2, I3) -> f1(I2, I3) 
0.82/1.21	  f1(I4, I5) -> f2(I4, -1 + I5) [1 + I4 <= -1 + I5]
0.82/1.21	
0.82/1.21	The dependency graph for this problem is:
0.82/1.21	  0 -> 1
0.82/1.21	  1 -> 3
0.82/1.21	  2 -> 3
0.82/1.21	  3 -> 2
0.82/1.21	Where:
0.82/1.21	  0) f4#(x1, x2) -> f3#(x1, x2) 
0.82/1.21	  1) f3#(I0, I1) -> f1#(I0, I1) 
0.82/1.21	  2) f2#(I2, I3) -> f1#(I2, I3) 
0.82/1.21	  3) f1#(I4, I5) -> f2#(I4, -1 + I5) [1 + I4 <= -1 + I5]
0.82/1.21	
0.82/1.21	We have the following SCCs.
0.82/1.21	  { 2, 3 }
0.82/1.21	
0.82/1.21	DP problem for innermost termination.
0.82/1.21	P =
0.82/1.21	  f2#(I2, I3) -> f1#(I2, I3) 
0.82/1.21	  f1#(I4, I5) -> f2#(I4, -1 + I5) [1 + I4 <= -1 + I5]
0.82/1.21	R = 
0.82/1.21	  f4(x1, x2) -> f3(x1, x2) 
0.82/1.21	  f3(I0, I1) -> f1(I0, I1) 
0.82/1.21	  f2(I2, I3) -> f1(I2, I3) 
0.82/1.21	  f1(I4, I5) -> f2(I4, -1 + I5) [1 + I4 <= -1 + I5]
0.82/1.21	
0.82/1.21	We use the reverse value criterion with the projection function NU:
0.82/1.21	  NU[f1#(z1,z2)] = -1 + z2 + -1 * (1 + z1)
0.82/1.21	  NU[f2#(z1,z2)] = -1 + z2 + -1 * (1 + z1)
0.82/1.21	
0.82/1.21	This gives the following inequalities:
0.82/1.21	    ==>  -1 + I3 + -1 * (1 + I2) >= -1 + I3 + -1 * (1 + I2)
0.82/1.21	  1 + I4 <= -1 + I5  ==>  -1 + I5 + -1 * (1 + I4) > -1 + (-1 + I5) + -1 * (1 + I4)  with  -1 + I5 + -1 * (1 + I4) >= 0
0.82/1.21	
0.82/1.21	We remove all the strictly oriented dependency pairs.
0.82/1.21	
0.82/1.21	DP problem for innermost termination.
0.82/1.21	P =
0.82/1.21	  f2#(I2, I3) -> f1#(I2, I3) 
0.82/1.21	R = 
0.82/1.21	  f4(x1, x2) -> f3(x1, x2) 
0.82/1.21	  f3(I0, I1) -> f1(I0, I1) 
0.82/1.21	  f2(I2, I3) -> f1(I2, I3) 
0.82/1.21	  f1(I4, I5) -> f2(I4, -1 + I5) [1 + I4 <= -1 + I5]
0.82/1.21	
0.82/1.21	The dependency graph for this problem is:
0.82/1.21	  2 -> 
0.82/1.21	Where:
0.82/1.21	  2) f2#(I2, I3) -> f1#(I2, I3) 
0.82/1.21	
0.82/1.21	We have the following SCCs.
0.82/1.21	  
0.82/4.19	EOF