2.92/2.93	YES
2.92/2.93	
2.92/2.93	DP problem for innermost termination.
2.92/2.93	P =
2.92/2.93	  f5#(x1, x2) -> f4#(x1, x2) 
2.92/2.93	  f2#(I0, I1) -> f1#(I0, I0) [1 <= I0]
2.92/2.93	  f4#(I2, I3) -> f2#(I2, I3) 
2.92/2.93	  f3#(I4, I5) -> f1#(I4, I5) 
2.92/2.93	  f1#(I6, I7) -> f3#(I6, -1 + I7) [1 <= I7]
2.92/2.93	  f1#(I8, I9) -> f2#(-1 + I8, I9) [I9 <= 0]
2.92/2.93	R = 
2.92/2.93	  f5(x1, x2) -> f4(x1, x2) 
2.92/2.93	  f2(I0, I1) -> f1(I0, I0) [1 <= I0]
2.92/2.93	  f4(I2, I3) -> f2(I2, I3) 
2.92/2.93	  f3(I4, I5) -> f1(I4, I5) 
2.92/2.93	  f1(I6, I7) -> f3(I6, -1 + I7) [1 <= I7]
2.92/2.93	  f1(I8, I9) -> f2(-1 + I8, I9) [I9 <= 0]
2.92/2.93	
2.92/2.93	The dependency graph for this problem is:
2.92/2.93	  0 -> 2
2.92/2.93	  1 -> 4
2.92/2.93	  2 -> 1
2.92/2.93	  3 -> 4, 5
2.92/2.93	  4 -> 3
2.92/2.93	  5 -> 1
2.92/2.93	Where:
2.92/2.93	  0) f5#(x1, x2) -> f4#(x1, x2) 
2.92/2.93	  1) f2#(I0, I1) -> f1#(I0, I0) [1 <= I0]
2.92/2.93	  2) f4#(I2, I3) -> f2#(I2, I3) 
2.92/2.93	  3) f3#(I4, I5) -> f1#(I4, I5) 
2.92/2.93	  4) f1#(I6, I7) -> f3#(I6, -1 + I7) [1 <= I7]
2.92/2.93	  5) f1#(I8, I9) -> f2#(-1 + I8, I9) [I9 <= 0]
2.92/2.93	
2.92/2.93	We have the following SCCs.
2.92/2.93	  { 1, 3, 4, 5 }
2.92/2.93	
2.92/2.93	DP problem for innermost termination.
2.92/2.93	P =
2.92/2.93	  f2#(I0, I1) -> f1#(I0, I0) [1 <= I0]
2.92/2.93	  f3#(I4, I5) -> f1#(I4, I5) 
2.92/2.93	  f1#(I6, I7) -> f3#(I6, -1 + I7) [1 <= I7]
2.92/2.93	  f1#(I8, I9) -> f2#(-1 + I8, I9) [I9 <= 0]
2.92/2.93	R = 
2.92/2.93	  f5(x1, x2) -> f4(x1, x2) 
2.92/2.93	  f2(I0, I1) -> f1(I0, I0) [1 <= I0]
2.92/2.93	  f4(I2, I3) -> f2(I2, I3) 
2.92/2.93	  f3(I4, I5) -> f1(I4, I5) 
2.92/2.93	  f1(I6, I7) -> f3(I6, -1 + I7) [1 <= I7]
2.92/2.93	  f1(I8, I9) -> f2(-1 + I8, I9) [I9 <= 0]
2.92/2.93	
2.92/2.93	We use the extended value criterion with the projection function NU:
2.92/2.93	  NU[f3#(x0,x1)] = x0 - 2
2.92/2.93	  NU[f1#(x0,x1)] = x0 - 2
2.92/2.93	  NU[f2#(x0,x1)] = x0 - 1
2.92/2.93	
2.92/2.93	This gives the following inequalities:
2.92/2.93	  1 <= I0  ==>  I0 - 1 > I0 - 2  with  I0 - 1 >= 0
2.92/2.93	    ==>  I4 - 2 >= I4 - 2
2.92/2.93	  1 <= I7  ==>  I6 - 2 >= I6 - 2
2.92/2.93	  I9 <= 0  ==>  I8 - 2 >= (-1 + I8) - 1
2.92/2.93	
2.92/2.93	We remove all the strictly oriented dependency pairs.
2.92/2.93	
2.92/2.93	DP problem for innermost termination.
2.92/2.93	P =
2.92/2.93	  f3#(I4, I5) -> f1#(I4, I5) 
2.92/2.93	  f1#(I6, I7) -> f3#(I6, -1 + I7) [1 <= I7]
2.92/2.93	  f1#(I8, I9) -> f2#(-1 + I8, I9) [I9 <= 0]
2.92/2.93	R = 
2.92/2.93	  f5(x1, x2) -> f4(x1, x2) 
2.92/2.93	  f2(I0, I1) -> f1(I0, I0) [1 <= I0]
2.92/2.93	  f4(I2, I3) -> f2(I2, I3) 
2.92/2.93	  f3(I4, I5) -> f1(I4, I5) 
2.92/2.93	  f1(I6, I7) -> f3(I6, -1 + I7) [1 <= I7]
2.92/2.93	  f1(I8, I9) -> f2(-1 + I8, I9) [I9 <= 0]
2.92/2.93	
2.92/2.93	The dependency graph for this problem is:
2.92/2.93	  3 -> 4, 5
2.92/2.93	  4 -> 3
2.92/2.93	  5 -> 
2.92/2.93	Where:
2.92/2.93	  3) f3#(I4, I5) -> f1#(I4, I5) 
2.92/2.93	  4) f1#(I6, I7) -> f3#(I6, -1 + I7) [1 <= I7]
2.92/2.93	  5) f1#(I8, I9) -> f2#(-1 + I8, I9) [I9 <= 0]
2.92/2.93	
2.92/2.93	We have the following SCCs.
2.92/2.93	  { 3, 4 }
2.92/2.93	
2.92/2.93	DP problem for innermost termination.
2.92/2.93	P =
2.92/2.93	  f3#(I4, I5) -> f1#(I4, I5) 
2.92/2.93	  f1#(I6, I7) -> f3#(I6, -1 + I7) [1 <= I7]
2.92/2.93	R = 
2.92/2.93	  f5(x1, x2) -> f4(x1, x2) 
2.92/2.93	  f2(I0, I1) -> f1(I0, I0) [1 <= I0]
2.92/2.93	  f4(I2, I3) -> f2(I2, I3) 
2.92/2.93	  f3(I4, I5) -> f1(I4, I5) 
2.92/2.93	  f1(I6, I7) -> f3(I6, -1 + I7) [1 <= I7]
2.92/2.93	  f1(I8, I9) -> f2(-1 + I8, I9) [I9 <= 0]
2.92/2.93	
2.92/2.93	We use the basic value criterion with the projection function NU:
2.92/2.93	  NU[f1#(z1,z2)] = z2
2.92/2.93	  NU[f3#(z1,z2)] = z2
2.92/2.93	
2.92/2.93	This gives the following inequalities:
2.92/2.93	    ==>  I5 (>! \union =) I5
2.92/2.93	  1 <= I7  ==>  I7 >! -1 + I7
2.92/2.93	
2.92/2.93	We remove all the strictly oriented dependency pairs.
2.92/2.93	
2.92/2.93	DP problem for innermost termination.
2.92/2.93	P =
2.92/2.93	  f3#(I4, I5) -> f1#(I4, I5) 
2.92/2.93	R = 
2.92/2.93	  f5(x1, x2) -> f4(x1, x2) 
2.92/2.93	  f2(I0, I1) -> f1(I0, I0) [1 <= I0]
2.92/2.93	  f4(I2, I3) -> f2(I2, I3) 
2.92/2.93	  f3(I4, I5) -> f1(I4, I5) 
2.92/2.93	  f1(I6, I7) -> f3(I6, -1 + I7) [1 <= I7]
2.92/2.93	  f1(I8, I9) -> f2(-1 + I8, I9) [I9 <= 0]
2.92/2.93	
2.92/2.93	The dependency graph for this problem is:
2.92/2.93	  3 -> 
2.92/2.93	Where:
2.92/2.93	  3) f3#(I4, I5) -> f1#(I4, I5) 
2.92/2.93	
2.92/2.93	We have the following SCCs.
2.92/2.93	  
2.92/5.91	EOF