0.62/0.63	YES
0.62/0.63	
0.62/0.63	DP problem for innermost termination.
0.62/0.63	P =
0.62/0.63	  f9#(x1) -> f8#(x1) 
0.62/0.63	  f8#(I0) -> f3#(0) 
0.62/0.63	  f7#(I1) -> f5#(I1) [2 <= I1]
0.62/0.63	  f7#(I2) -> f5#(I2) [1 + I2 <= 2]
0.62/0.63	  f2#(I3) -> f7#(I3) 
0.62/0.63	  f3#(I5) -> f4#(I5) 
0.62/0.63	  f4#(I6) -> f2#(I6) 
0.62/0.63	  f4#(I7) -> f1#(I7) 
0.62/0.63	  f4#(I8) -> f1#(I8) 
0.62/0.63	  f1#(I9) -> f3#(1 + I9) [1 + I9 <= 3]
0.62/0.63	  f1#(I10) -> f2#(I10) [3 <= I10]
0.62/0.63	R = 
0.62/0.63	  f9(x1) -> f8(x1) 
0.62/0.63	  f8(I0) -> f3(0) 
0.62/0.63	  f7(I1) -> f5(I1) [2 <= I1]
0.62/0.63	  f7(I2) -> f5(I2) [1 + I2 <= 2]
0.62/0.63	  f2(I3) -> f7(I3) 
0.62/0.63	  f5(I4) -> f6(I4) 
0.62/0.63	  f3(I5) -> f4(I5) 
0.62/0.63	  f4(I6) -> f2(I6) 
0.62/0.63	  f4(I7) -> f1(I7) 
0.62/0.63	  f4(I8) -> f1(I8) 
0.62/0.63	  f1(I9) -> f3(1 + I9) [1 + I9 <= 3]
0.62/0.63	  f1(I10) -> f2(I10) [3 <= I10]
0.62/0.63	
0.62/0.63	The dependency graph for this problem is:
0.62/0.63	  0 -> 1
0.62/0.63	  1 -> 5
0.62/0.63	  2 -> 
0.62/0.63	  3 -> 
0.62/0.63	  4 -> 2, 3
0.62/0.63	  5 -> 6, 7, 8
0.62/0.63	  6 -> 4
0.62/0.63	  7 -> 9, 10
0.62/0.63	  8 -> 9, 10
0.62/0.63	  9 -> 5
0.62/0.63	  10 -> 4
0.62/0.63	Where:
0.62/0.63	  0) f9#(x1) -> f8#(x1) 
0.62/0.63	  1) f8#(I0) -> f3#(0) 
0.62/0.63	  2) f7#(I1) -> f5#(I1) [2 <= I1]
0.62/0.63	  3) f7#(I2) -> f5#(I2) [1 + I2 <= 2]
0.62/0.63	  4) f2#(I3) -> f7#(I3) 
0.62/0.63	  5) f3#(I5) -> f4#(I5) 
0.62/0.63	  6) f4#(I6) -> f2#(I6) 
0.62/0.63	  7) f4#(I7) -> f1#(I7) 
0.62/0.63	  8) f4#(I8) -> f1#(I8) 
0.62/0.63	  9) f1#(I9) -> f3#(1 + I9) [1 + I9 <= 3]
0.62/0.63	  10) f1#(I10) -> f2#(I10) [3 <= I10]
0.62/0.63	
0.62/0.63	We have the following SCCs.
0.62/0.63	  { 5, 7, 8, 9 }
0.62/0.63	
0.62/0.63	DP problem for innermost termination.
0.62/0.63	P =
0.62/0.63	  f3#(I5) -> f4#(I5) 
0.62/0.63	  f4#(I7) -> f1#(I7) 
0.62/0.63	  f4#(I8) -> f1#(I8) 
0.62/0.63	  f1#(I9) -> f3#(1 + I9) [1 + I9 <= 3]
0.62/0.63	R = 
0.62/0.63	  f9(x1) -> f8(x1) 
0.62/0.63	  f8(I0) -> f3(0) 
0.62/0.63	  f7(I1) -> f5(I1) [2 <= I1]
0.62/0.63	  f7(I2) -> f5(I2) [1 + I2 <= 2]
0.62/0.63	  f2(I3) -> f7(I3) 
0.62/0.63	  f5(I4) -> f6(I4) 
0.62/0.63	  f3(I5) -> f4(I5) 
0.62/0.63	  f4(I6) -> f2(I6) 
0.62/0.63	  f4(I7) -> f1(I7) 
0.62/0.63	  f4(I8) -> f1(I8) 
0.62/0.63	  f1(I9) -> f3(1 + I9) [1 + I9 <= 3]
0.62/0.63	  f1(I10) -> f2(I10) [3 <= I10]
0.62/0.63	
0.62/0.63	We use the extended value criterion with the projection function NU:
0.62/0.63	  NU[f1#(x0)] = -x0 + 2
0.62/0.63	  NU[f4#(x0)] = -x0 + 2
0.62/0.63	  NU[f3#(x0)] = -x0 + 2
0.62/0.63	
0.62/0.63	This gives the following inequalities:
0.62/0.63	    ==>  -I5 + 2 >= -I5 + 2
0.62/0.63	    ==>  -I7 + 2 >= -I7 + 2
0.62/0.63	    ==>  -I8 + 2 >= -I8 + 2
0.62/0.63	  1 + I9 <= 3  ==>  -I9 + 2 > -(1 + I9) + 2  with  -I9 + 2 >= 0
0.62/0.63	
0.62/0.63	We remove all the strictly oriented dependency pairs.
0.62/0.63	
0.62/0.63	DP problem for innermost termination.
0.62/0.63	P =
0.62/0.63	  f3#(I5) -> f4#(I5) 
0.62/0.63	  f4#(I7) -> f1#(I7) 
0.62/0.63	  f4#(I8) -> f1#(I8) 
0.62/0.63	R = 
0.62/0.63	  f9(x1) -> f8(x1) 
0.62/0.63	  f8(I0) -> f3(0) 
0.62/0.63	  f7(I1) -> f5(I1) [2 <= I1]
0.62/0.63	  f7(I2) -> f5(I2) [1 + I2 <= 2]
0.62/0.63	  f2(I3) -> f7(I3) 
0.62/0.63	  f5(I4) -> f6(I4) 
0.62/0.63	  f3(I5) -> f4(I5) 
0.62/0.63	  f4(I6) -> f2(I6) 
0.62/0.63	  f4(I7) -> f1(I7) 
0.62/0.63	  f4(I8) -> f1(I8) 
0.62/0.63	  f1(I9) -> f3(1 + I9) [1 + I9 <= 3]
0.62/0.63	  f1(I10) -> f2(I10) [3 <= I10]
0.62/0.63	
0.62/0.63	The dependency graph for this problem is:
0.62/0.63	  5 -> 7, 8
0.62/0.63	  7 -> 
0.62/0.63	  8 -> 
0.62/0.63	Where:
0.62/0.63	  5) f3#(I5) -> f4#(I5) 
0.62/0.63	  7) f4#(I7) -> f1#(I7) 
0.62/0.63	  8) f4#(I8) -> f1#(I8) 
0.62/0.63	
0.62/0.63	We have the following SCCs.
0.62/0.63	  
0.62/3.61	EOF