3.00/3.05	YES
3.00/3.05	
3.00/3.05	DP problem for innermost termination.
3.00/3.05	P =
3.00/3.05	  f5#(x1, x2) -> f4#(x1, x2) 
3.00/3.05	  f4#(I0, I1) -> f2#(I0, I1) 
3.00/3.05	  f2#(I2, I3) -> f1#(I2, I2) [1 <= I2]
3.00/3.05	  f3#(I4, I5) -> f1#(I4, I5) 
3.00/3.05	  f1#(I6, I7) -> f3#(I6, -1 + I7) [1 <= I7]
3.00/3.05	  f1#(I8, I9) -> f2#(-1 + I8, I9) [I9 <= 0]
3.00/3.05	R = 
3.00/3.05	  f5(x1, x2) -> f4(x1, x2) 
3.00/3.05	  f4(I0, I1) -> f2(I0, I1) 
3.00/3.05	  f2(I2, I3) -> f1(I2, I2) [1 <= I2]
3.00/3.05	  f3(I4, I5) -> f1(I4, I5) 
3.00/3.05	  f1(I6, I7) -> f3(I6, -1 + I7) [1 <= I7]
3.00/3.05	  f1(I8, I9) -> f2(-1 + I8, I9) [I9 <= 0]
3.00/3.05	
3.00/3.05	The dependency graph for this problem is:
3.00/3.05	  0 -> 1
3.00/3.05	  1 -> 2
3.00/3.05	  2 -> 4
3.00/3.05	  3 -> 4, 5
3.00/3.05	  4 -> 3
3.00/3.05	  5 -> 2
3.00/3.05	Where:
3.00/3.05	  0) f5#(x1, x2) -> f4#(x1, x2) 
3.00/3.05	  1) f4#(I0, I1) -> f2#(I0, I1) 
3.00/3.05	  2) f2#(I2, I3) -> f1#(I2, I2) [1 <= I2]
3.00/3.05	  3) f3#(I4, I5) -> f1#(I4, I5) 
3.00/3.05	  4) f1#(I6, I7) -> f3#(I6, -1 + I7) [1 <= I7]
3.00/3.05	  5) f1#(I8, I9) -> f2#(-1 + I8, I9) [I9 <= 0]
3.00/3.05	
3.00/3.05	We have the following SCCs.
3.00/3.05	  { 2, 3, 4, 5 }
3.00/3.05	
3.00/3.05	DP problem for innermost termination.
3.00/3.05	P =
3.00/3.05	  f2#(I2, I3) -> f1#(I2, I2) [1 <= I2]
3.00/3.05	  f3#(I4, I5) -> f1#(I4, I5) 
3.00/3.05	  f1#(I6, I7) -> f3#(I6, -1 + I7) [1 <= I7]
3.00/3.05	  f1#(I8, I9) -> f2#(-1 + I8, I9) [I9 <= 0]
3.00/3.05	R = 
3.00/3.05	  f5(x1, x2) -> f4(x1, x2) 
3.00/3.05	  f4(I0, I1) -> f2(I0, I1) 
3.00/3.05	  f2(I2, I3) -> f1(I2, I2) [1 <= I2]
3.00/3.05	  f3(I4, I5) -> f1(I4, I5) 
3.00/3.05	  f1(I6, I7) -> f3(I6, -1 + I7) [1 <= I7]
3.00/3.05	  f1(I8, I9) -> f2(-1 + I8, I9) [I9 <= 0]
3.00/3.05	
3.00/3.05	We use the extended value criterion with the projection function NU:
3.00/3.05	  NU[f3#(x0,x1)] = x0 - 2
3.00/3.05	  NU[f1#(x0,x1)] = x0 - 2
3.00/3.05	  NU[f2#(x0,x1)] = x0 - 1
3.00/3.05	
3.00/3.05	This gives the following inequalities:
3.00/3.05	  1 <= I2  ==>  I2 - 1 > I2 - 2  with  I2 - 1 >= 0
3.00/3.05	    ==>  I4 - 2 >= I4 - 2
3.00/3.05	  1 <= I7  ==>  I6 - 2 >= I6 - 2
3.00/3.05	  I9 <= 0  ==>  I8 - 2 >= (-1 + I8) - 1
3.00/3.05	
3.00/3.05	We remove all the strictly oriented dependency pairs.
3.00/3.05	
3.00/3.05	DP problem for innermost termination.
3.00/3.05	P =
3.00/3.05	  f3#(I4, I5) -> f1#(I4, I5) 
3.00/3.05	  f1#(I6, I7) -> f3#(I6, -1 + I7) [1 <= I7]
3.00/3.05	  f1#(I8, I9) -> f2#(-1 + I8, I9) [I9 <= 0]
3.00/3.05	R = 
3.00/3.05	  f5(x1, x2) -> f4(x1, x2) 
3.00/3.05	  f4(I0, I1) -> f2(I0, I1) 
3.00/3.05	  f2(I2, I3) -> f1(I2, I2) [1 <= I2]
3.00/3.05	  f3(I4, I5) -> f1(I4, I5) 
3.00/3.05	  f1(I6, I7) -> f3(I6, -1 + I7) [1 <= I7]
3.00/3.05	  f1(I8, I9) -> f2(-1 + I8, I9) [I9 <= 0]
3.00/3.05	
3.00/3.05	The dependency graph for this problem is:
3.00/3.05	  3 -> 4, 5
3.00/3.05	  4 -> 3
3.00/3.05	  5 -> 
3.00/3.05	Where:
3.00/3.05	  3) f3#(I4, I5) -> f1#(I4, I5) 
3.00/3.05	  4) f1#(I6, I7) -> f3#(I6, -1 + I7) [1 <= I7]
3.00/3.05	  5) f1#(I8, I9) -> f2#(-1 + I8, I9) [I9 <= 0]
3.00/3.05	
3.00/3.05	We have the following SCCs.
3.00/3.05	  { 3, 4 }
3.00/3.05	
3.00/3.05	DP problem for innermost termination.
3.00/3.05	P =
3.00/3.05	  f3#(I4, I5) -> f1#(I4, I5) 
3.00/3.05	  f1#(I6, I7) -> f3#(I6, -1 + I7) [1 <= I7]
3.00/3.05	R = 
3.00/3.05	  f5(x1, x2) -> f4(x1, x2) 
3.00/3.05	  f4(I0, I1) -> f2(I0, I1) 
3.00/3.05	  f2(I2, I3) -> f1(I2, I2) [1 <= I2]
3.00/3.05	  f3(I4, I5) -> f1(I4, I5) 
3.00/3.05	  f1(I6, I7) -> f3(I6, -1 + I7) [1 <= I7]
3.00/3.05	  f1(I8, I9) -> f2(-1 + I8, I9) [I9 <= 0]
3.00/3.05	
3.00/3.05	We use the basic value criterion with the projection function NU:
3.00/3.05	  NU[f1#(z1,z2)] = z2
3.00/3.05	  NU[f3#(z1,z2)] = z2
3.00/3.05	
3.00/3.05	This gives the following inequalities:
3.00/3.05	    ==>  I5 (>! \union =) I5
3.00/3.05	  1 <= I7  ==>  I7 >! -1 + I7
3.00/3.05	
3.00/3.05	We remove all the strictly oriented dependency pairs.
3.00/3.05	
3.00/3.05	DP problem for innermost termination.
3.00/3.05	P =
3.00/3.05	  f3#(I4, I5) -> f1#(I4, I5) 
3.00/3.05	R = 
3.00/3.05	  f5(x1, x2) -> f4(x1, x2) 
3.00/3.05	  f4(I0, I1) -> f2(I0, I1) 
3.00/3.05	  f2(I2, I3) -> f1(I2, I2) [1 <= I2]
3.00/3.05	  f3(I4, I5) -> f1(I4, I5) 
3.00/3.05	  f1(I6, I7) -> f3(I6, -1 + I7) [1 <= I7]
3.00/3.05	  f1(I8, I9) -> f2(-1 + I8, I9) [I9 <= 0]
3.00/3.05	
3.00/3.05	The dependency graph for this problem is:
3.00/3.05	  3 -> 
3.00/3.05	Where:
3.00/3.05	  3) f3#(I4, I5) -> f1#(I4, I5) 
3.00/3.05	
3.00/3.05	We have the following SCCs.
3.00/3.05	  
3.00/6.03	EOF