0.00/0.41	YES
0.00/0.41	
0.00/0.41	DP problem for innermost termination.
0.00/0.41	P =
0.00/0.41	  f7#(x1) -> f6#(x1) 
0.00/0.41	  f6#(I0) -> f4#(0) 
0.00/0.41	  f4#(I2) -> f3#(I2) 
0.00/0.41	  f3#(I3) -> f4#(1 + I3) [1 + I3 <= 10]
0.00/0.41	  f3#(I4) -> f1#(I4) [10 <= I4]
0.00/0.41	  f1#(I5) -> f2#(I5) 
0.00/0.41	  f1#(I6) -> f2#(I6) 
0.00/0.41	R = 
0.00/0.41	  f7(x1) -> f6(x1) 
0.00/0.41	  f6(I0) -> f4(0) 
0.00/0.41	  f2(I1) -> f5(I1) 
0.00/0.41	  f4(I2) -> f3(I2) 
0.00/0.41	  f3(I3) -> f4(1 + I3) [1 + I3 <= 10]
0.00/0.41	  f3(I4) -> f1(I4) [10 <= I4]
0.00/0.41	  f1(I5) -> f2(I5) 
0.00/0.41	  f1(I6) -> f2(I6) 
0.00/0.41	
0.00/0.41	The dependency graph for this problem is:
0.00/0.41	  0 -> 1
0.00/0.41	  1 -> 2
0.00/0.41	  2 -> 3, 4
0.00/0.41	  3 -> 2
0.00/0.41	  4 -> 5, 6
0.00/0.41	  5 -> 
0.00/0.41	  6 -> 
0.00/0.41	Where:
0.00/0.41	  0) f7#(x1) -> f6#(x1) 
0.00/0.41	  1) f6#(I0) -> f4#(0) 
0.00/0.41	  2) f4#(I2) -> f3#(I2) 
0.00/0.41	  3) f3#(I3) -> f4#(1 + I3) [1 + I3 <= 10]
0.00/0.41	  4) f3#(I4) -> f1#(I4) [10 <= I4]
0.00/0.41	  5) f1#(I5) -> f2#(I5) 
0.00/0.41	  6) f1#(I6) -> f2#(I6) 
0.00/0.41	
0.00/0.41	We have the following SCCs.
0.00/0.41	  { 2, 3 }
0.00/0.41	
0.00/0.41	DP problem for innermost termination.
0.00/0.41	P =
0.00/0.41	  f4#(I2) -> f3#(I2) 
0.00/0.41	  f3#(I3) -> f4#(1 + I3) [1 + I3 <= 10]
0.00/0.41	R = 
0.00/0.41	  f7(x1) -> f6(x1) 
0.00/0.41	  f6(I0) -> f4(0) 
0.00/0.41	  f2(I1) -> f5(I1) 
0.00/0.41	  f4(I2) -> f3(I2) 
0.00/0.41	  f3(I3) -> f4(1 + I3) [1 + I3 <= 10]
0.00/0.41	  f3(I4) -> f1(I4) [10 <= I4]
0.00/0.41	  f1(I5) -> f2(I5) 
0.00/0.41	  f1(I6) -> f2(I6) 
0.00/0.41	
0.00/0.41	We use the reverse value criterion with the projection function NU:
0.00/0.41	  NU[f3#(z1)] = 10 + -1 * (1 + z1)
0.00/0.41	  NU[f4#(z1)] = 10 + -1 * (1 + z1)
0.00/0.41	
0.00/0.41	This gives the following inequalities:
0.00/0.41	    ==>  10 + -1 * (1 + I2) >= 10 + -1 * (1 + I2)
0.00/0.41	  1 + I3 <= 10  ==>  10 + -1 * (1 + I3) > 10 + -1 * (1 + (1 + I3))  with  10 + -1 * (1 + I3) >= 0
0.00/0.41	
0.00/0.41	We remove all the strictly oriented dependency pairs.
0.00/0.41	
0.00/0.41	DP problem for innermost termination.
0.00/0.41	P =
0.00/0.41	  f4#(I2) -> f3#(I2) 
0.00/0.41	R = 
0.00/0.41	  f7(x1) -> f6(x1) 
0.00/0.41	  f6(I0) -> f4(0) 
0.00/0.41	  f2(I1) -> f5(I1) 
0.00/0.41	  f4(I2) -> f3(I2) 
0.00/0.41	  f3(I3) -> f4(1 + I3) [1 + I3 <= 10]
0.00/0.41	  f3(I4) -> f1(I4) [10 <= I4]
0.00/0.41	  f1(I5) -> f2(I5) 
0.00/0.41	  f1(I6) -> f2(I6) 
0.00/0.41	
0.00/0.41	The dependency graph for this problem is:
0.00/0.41	  2 -> 
0.00/0.41	Where:
0.00/0.41	  2) f4#(I2) -> f3#(I2) 
0.00/0.41	
0.00/0.41	We have the following SCCs.
0.00/0.41	  
0.00/3.40	EOF