1.82/1.81	YES
1.82/1.81	
1.82/1.81	DP problem for innermost termination.
1.82/1.81	P =
1.82/1.81	  f5#(x1, x2, x3) -> f4#(x1, x2, x3) 
1.82/1.81	  f4#(I0, I1, I2) -> f3#(I0, I1, 0) 
1.82/1.81	  f3#(I3, I4, I5) -> f1#(I3, rnd2, 2 + I5) [rnd2 = 2 + 2 + I5]
1.82/1.81	  f1#(I6, I7, I8) -> f3#(I6, I7, I8) [1 + I8 <= I6]
1.82/1.81	R = 
1.82/1.81	  f5(x1, x2, x3) -> f4(x1, x2, x3) 
1.82/1.81	  f4(I0, I1, I2) -> f3(I0, I1, 0) 
1.82/1.81	  f3(I3, I4, I5) -> f1(I3, rnd2, 2 + I5) [rnd2 = 2 + 2 + I5]
1.82/1.81	  f1(I6, I7, I8) -> f3(I6, I7, I8) [1 + I8 <= I6]
1.82/1.81	  f1(I9, I10, I11) -> f2(I9, I10, I11) [I9 <= I11]
1.82/1.81	
1.82/1.81	The dependency graph for this problem is:
1.82/1.81	  0 -> 1
1.82/1.81	  1 -> 2
1.82/1.81	  2 -> 3
1.82/1.81	  3 -> 2
1.82/1.81	Where:
1.82/1.81	  0) f5#(x1, x2, x3) -> f4#(x1, x2, x3) 
1.82/1.81	  1) f4#(I0, I1, I2) -> f3#(I0, I1, 0) 
1.82/1.81	  2) f3#(I3, I4, I5) -> f1#(I3, rnd2, 2 + I5) [rnd2 = 2 + 2 + I5]
1.82/1.81	  3) f1#(I6, I7, I8) -> f3#(I6, I7, I8) [1 + I8 <= I6]
1.82/1.81	
1.82/1.81	We have the following SCCs.
1.82/1.81	  { 2, 3 }
1.82/1.81	
1.82/1.81	DP problem for innermost termination.
1.82/1.81	P =
1.82/1.81	  f3#(I3, I4, I5) -> f1#(I3, rnd2, 2 + I5) [rnd2 = 2 + 2 + I5]
1.82/1.81	  f1#(I6, I7, I8) -> f3#(I6, I7, I8) [1 + I8 <= I6]
1.82/1.81	R = 
1.82/1.81	  f5(x1, x2, x3) -> f4(x1, x2, x3) 
1.82/1.81	  f4(I0, I1, I2) -> f3(I0, I1, 0) 
1.82/1.81	  f3(I3, I4, I5) -> f1(I3, rnd2, 2 + I5) [rnd2 = 2 + 2 + I5]
1.82/1.81	  f1(I6, I7, I8) -> f3(I6, I7, I8) [1 + I8 <= I6]
1.82/1.81	  f1(I9, I10, I11) -> f2(I9, I10, I11) [I9 <= I11]
1.82/1.81	
1.82/1.81	We use the reverse value criterion with the projection function NU:
1.82/1.81	  NU[f1#(z1,z2,z3)] = z1 + -1 * (1 + z3)
1.82/1.81	  NU[f3#(z1,z2,z3)] = z1 + -1 * (1 + (2 + z3))
1.82/1.81	
1.82/1.81	This gives the following inequalities:
1.82/1.81	  rnd2 = 2 + 2 + I5  ==>  I3 + -1 * (1 + (2 + I5)) >= I3 + -1 * (1 + (2 + I5))
1.82/1.81	  1 + I8 <= I6  ==>  I6 + -1 * (1 + I8) > I6 + -1 * (1 + (2 + I8))  with  I6 + -1 * (1 + I8) >= 0
1.82/1.81	
1.82/1.81	We remove all the strictly oriented dependency pairs.
1.82/1.81	
1.82/1.81	DP problem for innermost termination.
1.82/1.81	P =
1.82/1.81	  f3#(I3, I4, I5) -> f1#(I3, rnd2, 2 + I5) [rnd2 = 2 + 2 + I5]
1.82/1.81	R = 
1.82/1.81	  f5(x1, x2, x3) -> f4(x1, x2, x3) 
1.82/1.81	  f4(I0, I1, I2) -> f3(I0, I1, 0) 
1.82/1.81	  f3(I3, I4, I5) -> f1(I3, rnd2, 2 + I5) [rnd2 = 2 + 2 + I5]
1.82/1.81	  f1(I6, I7, I8) -> f3(I6, I7, I8) [1 + I8 <= I6]
1.82/1.81	  f1(I9, I10, I11) -> f2(I9, I10, I11) [I9 <= I11]
1.82/1.81	
1.82/1.81	The dependency graph for this problem is:
1.82/1.81	  2 -> 
1.82/1.81	Where:
1.82/1.81	  2) f3#(I3, I4, I5) -> f1#(I3, rnd2, 2 + I5) [rnd2 = 2 + 2 + I5]
1.82/1.81	
1.82/1.81	We have the following SCCs.
1.82/1.81	  
1.82/4.79	EOF