2.21/2.27	YES
2.21/2.27	
2.21/2.27	DP problem for innermost termination.
2.21/2.27	P =
2.21/2.27	  f6#(x1, x2) -> f5#(x1, x2) 
2.21/2.27	  f5#(I0, I1) -> f2#(I0, rnd2) [y1 = I0 /\ rnd2 = rnd2]
2.21/2.27	  f2#(I2, I3) -> f3#(I2, I3) 
2.21/2.27	  f3#(I4, I5) -> f1#(I4, I5) [1 + I5 <= 4]
2.21/2.27	  f1#(I8, I9) -> f2#(I8, 1 + I9) [1 <= I9]
2.21/2.27	  f1#(I10, I11) -> f2#(I10, 1) [I11 <= 0]
2.21/2.27	R = 
2.21/2.27	  f6(x1, x2) -> f5(x1, x2) 
2.21/2.27	  f5(I0, I1) -> f2(I0, rnd2) [y1 = I0 /\ rnd2 = rnd2]
2.21/2.27	  f2(I2, I3) -> f3(I2, I3) 
2.21/2.27	  f3(I4, I5) -> f1(I4, I5) [1 + I5 <= 4]
2.21/2.27	  f3(I6, I7) -> f4(I6, I7) [4 <= I7]
2.21/2.27	  f1(I8, I9) -> f2(I8, 1 + I9) [1 <= I9]
2.21/2.27	  f1(I10, I11) -> f2(I10, 1) [I11 <= 0]
2.21/2.27	
2.21/2.27	The dependency graph for this problem is:
2.21/2.27	  0 -> 1
2.21/2.27	  1 -> 2
2.21/2.27	  2 -> 3
2.21/2.27	  3 -> 4, 5
2.21/2.27	  4 -> 2
2.21/2.27	  5 -> 2
2.21/2.27	Where:
2.21/2.27	  0) f6#(x1, x2) -> f5#(x1, x2) 
2.21/2.27	  1) f5#(I0, I1) -> f2#(I0, rnd2) [y1 = I0 /\ rnd2 = rnd2]
2.21/2.27	  2) f2#(I2, I3) -> f3#(I2, I3) 
2.21/2.27	  3) f3#(I4, I5) -> f1#(I4, I5) [1 + I5 <= 4]
2.21/2.27	  4) f1#(I8, I9) -> f2#(I8, 1 + I9) [1 <= I9]
2.21/2.27	  5) f1#(I10, I11) -> f2#(I10, 1) [I11 <= 0]
2.21/2.27	
2.21/2.27	We have the following SCCs.
2.21/2.27	  { 2, 3, 4, 5 }
2.21/2.27	
2.21/2.27	DP problem for innermost termination.
2.21/2.27	P =
2.21/2.27	  f2#(I2, I3) -> f3#(I2, I3) 
2.21/2.27	  f3#(I4, I5) -> f1#(I4, I5) [1 + I5 <= 4]
2.21/2.27	  f1#(I8, I9) -> f2#(I8, 1 + I9) [1 <= I9]
2.21/2.27	  f1#(I10, I11) -> f2#(I10, 1) [I11 <= 0]
2.21/2.27	R = 
2.21/2.27	  f6(x1, x2) -> f5(x1, x2) 
2.21/2.27	  f5(I0, I1) -> f2(I0, rnd2) [y1 = I0 /\ rnd2 = rnd2]
2.21/2.27	  f2(I2, I3) -> f3(I2, I3) 
2.21/2.27	  f3(I4, I5) -> f1(I4, I5) [1 + I5 <= 4]
2.21/2.27	  f3(I6, I7) -> f4(I6, I7) [4 <= I7]
2.21/2.27	  f1(I8, I9) -> f2(I8, 1 + I9) [1 <= I9]
2.21/2.27	  f1(I10, I11) -> f2(I10, 1) [I11 <= 0]
2.21/2.27	
2.21/2.27	We use the extended value criterion with the projection function NU:
2.21/2.27	  NU[f1#(x0,x1)] = -x1 + 2
2.21/2.27	  NU[f3#(x0,x1)] = -x1 + 3
2.21/2.27	  NU[f2#(x0,x1)] = -x1 + 3
2.21/2.27	
2.21/2.27	This gives the following inequalities:
2.21/2.27	    ==>  -I3 + 3 >= -I3 + 3
2.21/2.27	  1 + I5 <= 4  ==>  -I5 + 3 > -I5 + 2  with  -I5 + 3 >= 0
2.21/2.27	  1 <= I9  ==>  -I9 + 2 >= -(1 + I9) + 3
2.21/2.27	  I11 <= 0  ==>  -I11 + 2 >= -1 + 3
2.21/2.27	
2.21/2.27	We remove all the strictly oriented dependency pairs.
2.21/2.27	
2.21/2.27	DP problem for innermost termination.
2.21/2.27	P =
2.21/2.27	  f2#(I2, I3) -> f3#(I2, I3) 
2.21/2.27	  f1#(I8, I9) -> f2#(I8, 1 + I9) [1 <= I9]
2.21/2.27	  f1#(I10, I11) -> f2#(I10, 1) [I11 <= 0]
2.21/2.27	R = 
2.21/2.27	  f6(x1, x2) -> f5(x1, x2) 
2.21/2.27	  f5(I0, I1) -> f2(I0, rnd2) [y1 = I0 /\ rnd2 = rnd2]
2.21/2.27	  f2(I2, I3) -> f3(I2, I3) 
2.21/2.27	  f3(I4, I5) -> f1(I4, I5) [1 + I5 <= 4]
2.21/2.27	  f3(I6, I7) -> f4(I6, I7) [4 <= I7]
2.21/2.27	  f1(I8, I9) -> f2(I8, 1 + I9) [1 <= I9]
2.21/2.27	  f1(I10, I11) -> f2(I10, 1) [I11 <= 0]
2.21/2.27	
2.21/2.27	The dependency graph for this problem is:
2.21/2.27	  2 -> 
2.21/2.27	  4 -> 2
2.21/2.27	  5 -> 2
2.21/2.27	Where:
2.21/2.27	  2) f2#(I2, I3) -> f3#(I2, I3) 
2.21/2.27	  4) f1#(I8, I9) -> f2#(I8, 1 + I9) [1 <= I9]
2.21/2.27	  5) f1#(I10, I11) -> f2#(I10, 1) [I11 <= 0]
2.21/2.27	
2.21/2.27	We have the following SCCs.
2.21/2.27	  
2.21/2.27	EOF