2.76/2.78	YES
2.76/2.78	
2.76/2.78	DP problem for innermost termination.
2.76/2.78	P =
2.76/2.78	  f5#(x1, x2) -> f4#(x1, x2) 
2.76/2.78	  f4#(I0, I1) -> f1#(I0, I1) 
2.76/2.78	  f3#(I2, I3) -> f2#(I2, I3) 
2.76/2.78	  f2#(I4, I5) -> f3#(I4, -1 + I5) [1 <= I5]
2.76/2.78	  f2#(I6, I7) -> f1#(-1 + I6, 1 + I7) [I7 <= 0]
2.76/2.78	  f1#(I8, I9) -> f2#(I8, I8) [1 <= I8]
2.76/2.78	R = 
2.76/2.78	  f5(x1, x2) -> f4(x1, x2) 
2.76/2.78	  f4(I0, I1) -> f1(I0, I1) 
2.76/2.78	  f3(I2, I3) -> f2(I2, I3) 
2.76/2.78	  f2(I4, I5) -> f3(I4, -1 + I5) [1 <= I5]
2.76/2.78	  f2(I6, I7) -> f1(-1 + I6, 1 + I7) [I7 <= 0]
2.76/2.78	  f1(I8, I9) -> f2(I8, I8) [1 <= I8]
2.76/2.78	
2.76/2.78	The dependency graph for this problem is:
2.76/2.78	  0 -> 1
2.76/2.78	  1 -> 5
2.76/2.78	  2 -> 3, 4
2.76/2.78	  3 -> 2
2.76/2.78	  4 -> 5
2.76/2.78	  5 -> 3
2.76/2.78	Where:
2.76/2.78	  0) f5#(x1, x2) -> f4#(x1, x2) 
2.76/2.78	  1) f4#(I0, I1) -> f1#(I0, I1) 
2.76/2.78	  2) f3#(I2, I3) -> f2#(I2, I3) 
2.76/2.78	  3) f2#(I4, I5) -> f3#(I4, -1 + I5) [1 <= I5]
2.76/2.78	  4) f2#(I6, I7) -> f1#(-1 + I6, 1 + I7) [I7 <= 0]
2.76/2.78	  5) f1#(I8, I9) -> f2#(I8, I8) [1 <= I8]
2.76/2.78	
2.76/2.78	We have the following SCCs.
2.76/2.78	  { 2, 3, 4, 5 }
2.76/2.78	
2.76/2.78	DP problem for innermost termination.
2.76/2.78	P =
2.76/2.78	  f3#(I2, I3) -> f2#(I2, I3) 
2.76/2.78	  f2#(I4, I5) -> f3#(I4, -1 + I5) [1 <= I5]
2.76/2.78	  f2#(I6, I7) -> f1#(-1 + I6, 1 + I7) [I7 <= 0]
2.76/2.78	  f1#(I8, I9) -> f2#(I8, I8) [1 <= I8]
2.76/2.78	R = 
2.76/2.78	  f5(x1, x2) -> f4(x1, x2) 
2.76/2.78	  f4(I0, I1) -> f1(I0, I1) 
2.76/2.78	  f3(I2, I3) -> f2(I2, I3) 
2.76/2.78	  f2(I4, I5) -> f3(I4, -1 + I5) [1 <= I5]
2.76/2.78	  f2(I6, I7) -> f1(-1 + I6, 1 + I7) [I7 <= 0]
2.76/2.78	  f1(I8, I9) -> f2(I8, I8) [1 <= I8]
2.76/2.78	
2.76/2.78	We use the extended value criterion with the projection function NU:
2.76/2.78	  NU[f1#(x0,x1)] = x0 + 1
2.76/2.78	  NU[f2#(x0,x1)] = x0
2.76/2.78	  NU[f3#(x0,x1)] = x0
2.76/2.78	
2.76/2.78	This gives the following inequalities:
2.76/2.78	    ==>  I2 >= I2
2.76/2.78	  1 <= I5  ==>  I4 >= I4
2.76/2.78	  I7 <= 0  ==>  I6 >= (-1 + I6) + 1
2.76/2.78	  1 <= I8  ==>  I8 + 1 > I8  with  I8 + 1 >= 0
2.76/2.78	
2.76/2.78	We remove all the strictly oriented dependency pairs.
2.76/2.78	
2.76/2.78	DP problem for innermost termination.
2.76/2.78	P =
2.76/2.78	  f3#(I2, I3) -> f2#(I2, I3) 
2.76/2.78	  f2#(I4, I5) -> f3#(I4, -1 + I5) [1 <= I5]
2.76/2.78	  f2#(I6, I7) -> f1#(-1 + I6, 1 + I7) [I7 <= 0]
2.76/2.78	R = 
2.76/2.78	  f5(x1, x2) -> f4(x1, x2) 
2.76/2.78	  f4(I0, I1) -> f1(I0, I1) 
2.76/2.78	  f3(I2, I3) -> f2(I2, I3) 
2.76/2.78	  f2(I4, I5) -> f3(I4, -1 + I5) [1 <= I5]
2.76/2.78	  f2(I6, I7) -> f1(-1 + I6, 1 + I7) [I7 <= 0]
2.76/2.78	  f1(I8, I9) -> f2(I8, I8) [1 <= I8]
2.76/2.78	
2.76/2.78	The dependency graph for this problem is:
2.76/2.78	  2 -> 3, 4
2.76/2.78	  3 -> 2
2.76/2.78	  4 -> 
2.76/2.78	Where:
2.76/2.78	  2) f3#(I2, I3) -> f2#(I2, I3) 
2.76/2.78	  3) f2#(I4, I5) -> f3#(I4, -1 + I5) [1 <= I5]
2.76/2.78	  4) f2#(I6, I7) -> f1#(-1 + I6, 1 + I7) [I7 <= 0]
2.76/2.78	
2.76/2.78	We have the following SCCs.
2.76/2.78	  { 2, 3 }
2.76/2.78	
2.76/2.78	DP problem for innermost termination.
2.76/2.78	P =
2.76/2.78	  f3#(I2, I3) -> f2#(I2, I3) 
2.76/2.78	  f2#(I4, I5) -> f3#(I4, -1 + I5) [1 <= I5]
2.76/2.78	R = 
2.76/2.78	  f5(x1, x2) -> f4(x1, x2) 
2.76/2.78	  f4(I0, I1) -> f1(I0, I1) 
2.76/2.78	  f3(I2, I3) -> f2(I2, I3) 
2.76/2.78	  f2(I4, I5) -> f3(I4, -1 + I5) [1 <= I5]
2.76/2.78	  f2(I6, I7) -> f1(-1 + I6, 1 + I7) [I7 <= 0]
2.76/2.78	  f1(I8, I9) -> f2(I8, I8) [1 <= I8]
2.76/2.78	
2.76/2.78	We use the basic value criterion with the projection function NU:
2.76/2.78	  NU[f2#(z1,z2)] = z2
2.76/2.78	  NU[f3#(z1,z2)] = z2
2.76/2.78	
2.76/2.78	This gives the following inequalities:
2.76/2.78	    ==>  I3 (>! \union =) I3
2.76/2.78	  1 <= I5  ==>  I5 >! -1 + I5
2.76/2.78	
2.76/2.78	We remove all the strictly oriented dependency pairs.
2.76/2.78	
2.76/2.78	DP problem for innermost termination.
2.76/2.78	P =
2.76/2.78	  f3#(I2, I3) -> f2#(I2, I3) 
2.76/2.78	R = 
2.76/2.78	  f5(x1, x2) -> f4(x1, x2) 
2.76/2.78	  f4(I0, I1) -> f1(I0, I1) 
2.76/2.78	  f3(I2, I3) -> f2(I2, I3) 
2.76/2.78	  f2(I4, I5) -> f3(I4, -1 + I5) [1 <= I5]
2.76/2.78	  f2(I6, I7) -> f1(-1 + I6, 1 + I7) [I7 <= 0]
2.76/2.78	  f1(I8, I9) -> f2(I8, I8) [1 <= I8]
2.76/2.78	
2.76/2.78	The dependency graph for this problem is:
2.76/2.78	  2 -> 
2.76/2.78	Where:
2.76/2.78	  2) f3#(I2, I3) -> f2#(I2, I3) 
2.76/2.78	
2.76/2.78	We have the following SCCs.
2.76/2.78	  
2.76/5.76	EOF