0.00/0.48	YES
0.00/0.48	
0.00/0.48	DP problem for innermost termination.
0.00/0.48	P =
0.00/0.48	  f8#(x1) -> f7#(x1) 
0.00/0.48	  f7#(I0) -> f2#(0) 
0.00/0.48	  f2#(I1) -> f5#(I1) 
0.00/0.48	  f5#(I2) -> f4#(I2) [1 + I2 <= 42]
0.00/0.48	  f4#(I4) -> f3#(I4) 
0.00/0.48	  f4#(I5) -> f1#(I5) 
0.00/0.48	  f4#(I6) -> f3#(I6) 
0.00/0.48	  f3#(I7) -> f1#(I7) 
0.00/0.48	  f1#(I8) -> f2#(1 + I8) 
0.00/0.48	R = 
0.00/0.48	  f8(x1) -> f7(x1) 
0.00/0.48	  f7(I0) -> f2(0) 
0.00/0.48	  f2(I1) -> f5(I1) 
0.00/0.48	  f5(I2) -> f4(I2) [1 + I2 <= 42]
0.00/0.48	  f5(I3) -> f6(I3) [42 <= I3]
0.00/0.48	  f4(I4) -> f3(I4) 
0.00/0.48	  f4(I5) -> f1(I5) 
0.00/0.48	  f4(I6) -> f3(I6) 
0.00/0.48	  f3(I7) -> f1(I7) 
0.00/0.48	  f1(I8) -> f2(1 + I8) 
0.00/0.48	
0.00/0.48	The dependency graph for this problem is:
0.00/0.48	  0 -> 1
0.00/0.48	  1 -> 2
0.00/0.48	  2 -> 3
0.00/0.48	  3 -> 4, 5, 6
0.00/0.48	  4 -> 7
0.00/0.48	  5 -> 8
0.00/0.48	  6 -> 7
0.00/0.48	  7 -> 8
0.00/0.48	  8 -> 2
0.00/0.48	Where:
0.00/0.48	  0) f8#(x1) -> f7#(x1) 
0.00/0.48	  1) f7#(I0) -> f2#(0) 
0.00/0.48	  2) f2#(I1) -> f5#(I1) 
0.00/0.48	  3) f5#(I2) -> f4#(I2) [1 + I2 <= 42]
0.00/0.48	  4) f4#(I4) -> f3#(I4) 
0.00/0.48	  5) f4#(I5) -> f1#(I5) 
0.00/0.48	  6) f4#(I6) -> f3#(I6) 
0.00/0.48	  7) f3#(I7) -> f1#(I7) 
0.00/0.48	  8) f1#(I8) -> f2#(1 + I8) 
0.00/0.48	
0.00/0.48	We have the following SCCs.
0.00/0.48	  { 2, 3, 4, 5, 6, 7, 8 }
0.00/0.48	
0.00/0.48	DP problem for innermost termination.
0.00/0.48	P =
0.00/0.48	  f2#(I1) -> f5#(I1) 
0.00/0.48	  f5#(I2) -> f4#(I2) [1 + I2 <= 42]
0.00/0.48	  f4#(I4) -> f3#(I4) 
0.00/0.48	  f4#(I5) -> f1#(I5) 
0.00/0.48	  f4#(I6) -> f3#(I6) 
0.00/0.48	  f3#(I7) -> f1#(I7) 
0.00/0.48	  f1#(I8) -> f2#(1 + I8) 
0.00/0.48	R = 
0.00/0.48	  f8(x1) -> f7(x1) 
0.00/0.48	  f7(I0) -> f2(0) 
0.00/0.48	  f2(I1) -> f5(I1) 
0.00/0.48	  f5(I2) -> f4(I2) [1 + I2 <= 42]
0.00/0.48	  f5(I3) -> f6(I3) [42 <= I3]
0.00/0.48	  f4(I4) -> f3(I4) 
0.00/0.48	  f4(I5) -> f1(I5) 
0.00/0.48	  f4(I6) -> f3(I6) 
0.00/0.48	  f3(I7) -> f1(I7) 
0.00/0.48	  f1(I8) -> f2(1 + I8) 
0.00/0.48	
0.00/0.48	We use the extended value criterion with the projection function NU:
0.00/0.48	  NU[f1#(x0)] = -x0 + 40
0.00/0.48	  NU[f3#(x0)] = -x0 + 40
0.00/0.48	  NU[f4#(x0)] = -x0 + 40
0.00/0.48	  NU[f5#(x0)] = -x0 + 41
0.00/0.48	  NU[f2#(x0)] = -x0 + 41
0.00/0.48	
0.00/0.48	This gives the following inequalities:
0.00/0.48	    ==>  -I1 + 41 >= -I1 + 41
0.00/0.48	  1 + I2 <= 42  ==>  -I2 + 41 > -I2 + 40  with  -I2 + 41 >= 0
0.00/0.48	    ==>  -I4 + 40 >= -I4 + 40
0.00/0.48	    ==>  -I5 + 40 >= -I5 + 40
0.00/0.48	    ==>  -I6 + 40 >= -I6 + 40
0.00/0.48	    ==>  -I7 + 40 >= -I7 + 40
0.00/0.48	    ==>  -I8 + 40 >= -(1 + I8) + 41
0.00/0.48	
0.00/0.48	We remove all the strictly oriented dependency pairs.
0.00/0.48	
0.00/0.48	DP problem for innermost termination.
0.00/0.48	P =
0.00/0.48	  f2#(I1) -> f5#(I1) 
0.00/0.48	  f4#(I4) -> f3#(I4) 
0.00/0.48	  f4#(I5) -> f1#(I5) 
0.00/0.48	  f4#(I6) -> f3#(I6) 
0.00/0.48	  f3#(I7) -> f1#(I7) 
0.00/0.48	  f1#(I8) -> f2#(1 + I8) 
0.00/0.48	R = 
0.00/0.48	  f8(x1) -> f7(x1) 
0.00/0.48	  f7(I0) -> f2(0) 
0.00/0.48	  f2(I1) -> f5(I1) 
0.00/0.48	  f5(I2) -> f4(I2) [1 + I2 <= 42]
0.00/0.48	  f5(I3) -> f6(I3) [42 <= I3]
0.00/0.48	  f4(I4) -> f3(I4) 
0.00/0.48	  f4(I5) -> f1(I5) 
0.00/0.48	  f4(I6) -> f3(I6) 
0.00/0.48	  f3(I7) -> f1(I7) 
0.00/0.48	  f1(I8) -> f2(1 + I8) 
0.00/0.48	
0.00/0.48	The dependency graph for this problem is:
0.00/0.48	  2 -> 
0.00/0.48	  4 -> 7
0.00/0.48	  5 -> 8
0.00/0.48	  6 -> 7
0.00/0.48	  7 -> 8
0.00/0.48	  8 -> 2
0.00/0.48	Where:
0.00/0.48	  2) f2#(I1) -> f5#(I1) 
0.00/0.48	  4) f4#(I4) -> f3#(I4) 
0.00/0.48	  5) f4#(I5) -> f1#(I5) 
0.00/0.48	  6) f4#(I6) -> f3#(I6) 
0.00/0.48	  7) f3#(I7) -> f1#(I7) 
0.00/0.48	  8) f1#(I8) -> f2#(1 + I8) 
0.00/0.48	
0.00/0.48	We have the following SCCs.
0.00/0.48	  
0.00/3.46	EOF