0.83/1.20	YES
0.83/1.20	
0.83/1.20	DP problem for innermost termination.
0.83/1.20	P =
0.83/1.20	  f7#(x1, x2) -> f6#(x1, x2) 
0.83/1.20	  f6#(I0, I1) -> f4#(I0, 0) 
0.83/1.20	  f4#(I4, I5) -> f3#(I4, I5) 
0.83/1.20	  f3#(I6, I7) -> f4#(I6, 1 + I7) [1 + I7 <= I6]
0.83/1.20	  f3#(I8, I9) -> f1#(I8, I9) [I8 <= I9]
0.83/1.20	  f1#(I10, I11) -> f2#(I10, I11) 
0.83/1.20	  f1#(I12, I13) -> f2#(I12, I13) 
0.83/1.20	R = 
0.83/1.20	  f7(x1, x2) -> f6(x1, x2) 
0.83/1.20	  f6(I0, I1) -> f4(I0, 0) 
0.83/1.20	  f2(I2, I3) -> f5(I2, I3) 
0.83/1.20	  f4(I4, I5) -> f3(I4, I5) 
0.83/1.20	  f3(I6, I7) -> f4(I6, 1 + I7) [1 + I7 <= I6]
0.83/1.20	  f3(I8, I9) -> f1(I8, I9) [I8 <= I9]
0.83/1.20	  f1(I10, I11) -> f2(I10, I11) 
0.83/1.20	  f1(I12, I13) -> f2(I12, I13) 
0.83/1.20	
0.83/1.20	The dependency graph for this problem is:
0.83/1.20	  0 -> 1
0.83/1.20	  1 -> 2
0.83/1.20	  2 -> 3, 4
0.83/1.20	  3 -> 2
0.83/1.20	  4 -> 5, 6
0.83/1.20	  5 -> 
0.83/1.20	  6 -> 
0.83/1.20	Where:
0.83/1.20	  0) f7#(x1, x2) -> f6#(x1, x2) 
0.83/1.20	  1) f6#(I0, I1) -> f4#(I0, 0) 
0.83/1.20	  2) f4#(I4, I5) -> f3#(I4, I5) 
0.83/1.20	  3) f3#(I6, I7) -> f4#(I6, 1 + I7) [1 + I7 <= I6]
0.83/1.20	  4) f3#(I8, I9) -> f1#(I8, I9) [I8 <= I9]
0.83/1.20	  5) f1#(I10, I11) -> f2#(I10, I11) 
0.83/1.20	  6) f1#(I12, I13) -> f2#(I12, I13) 
0.83/1.20	
0.83/1.20	We have the following SCCs.
0.83/1.20	  { 2, 3 }
0.83/1.20	
0.83/1.20	DP problem for innermost termination.
0.83/1.20	P =
0.83/1.20	  f4#(I4, I5) -> f3#(I4, I5) 
0.83/1.20	  f3#(I6, I7) -> f4#(I6, 1 + I7) [1 + I7 <= I6]
0.83/1.20	R = 
0.83/1.20	  f7(x1, x2) -> f6(x1, x2) 
0.83/1.20	  f6(I0, I1) -> f4(I0, 0) 
0.83/1.20	  f2(I2, I3) -> f5(I2, I3) 
0.83/1.20	  f4(I4, I5) -> f3(I4, I5) 
0.83/1.20	  f3(I6, I7) -> f4(I6, 1 + I7) [1 + I7 <= I6]
0.83/1.20	  f3(I8, I9) -> f1(I8, I9) [I8 <= I9]
0.83/1.20	  f1(I10, I11) -> f2(I10, I11) 
0.83/1.20	  f1(I12, I13) -> f2(I12, I13) 
0.83/1.20	
0.83/1.20	We use the reverse value criterion with the projection function NU:
0.83/1.20	  NU[f3#(z1,z2)] = z1 + -1 * (1 + z2)
0.83/1.20	  NU[f4#(z1,z2)] = z1 + -1 * (1 + z2)
0.83/1.20	
0.83/1.20	This gives the following inequalities:
0.83/1.20	    ==>  I4 + -1 * (1 + I5) >= I4 + -1 * (1 + I5)
0.83/1.20	  1 + I7 <= I6  ==>  I6 + -1 * (1 + I7) > I6 + -1 * (1 + (1 + I7))  with  I6 + -1 * (1 + I7) >= 0
0.83/1.20	
0.83/1.20	We remove all the strictly oriented dependency pairs.
0.83/1.20	
0.83/1.20	DP problem for innermost termination.
0.83/1.20	P =
0.83/1.20	  f4#(I4, I5) -> f3#(I4, I5) 
0.83/1.20	R = 
0.83/1.20	  f7(x1, x2) -> f6(x1, x2) 
0.83/1.20	  f6(I0, I1) -> f4(I0, 0) 
0.83/1.20	  f2(I2, I3) -> f5(I2, I3) 
0.83/1.20	  f4(I4, I5) -> f3(I4, I5) 
0.83/1.20	  f3(I6, I7) -> f4(I6, 1 + I7) [1 + I7 <= I6]
0.83/1.20	  f3(I8, I9) -> f1(I8, I9) [I8 <= I9]
0.83/1.20	  f1(I10, I11) -> f2(I10, I11) 
0.83/1.20	  f1(I12, I13) -> f2(I12, I13) 
0.83/1.20	
0.83/1.20	The dependency graph for this problem is:
0.83/1.20	  2 -> 
0.83/1.20	Where:
0.83/1.20	  2) f4#(I4, I5) -> f3#(I4, I5) 
0.83/1.20	
0.83/1.20	We have the following SCCs.
0.83/1.20	  
0.83/4.18	EOF