0.98/1.46	YES
0.98/1.46	
0.98/1.46	DP problem for innermost termination.
0.98/1.46	P =
0.98/1.46	  f5#(x1, x2) -> f4#(x1, x2) 
0.98/1.46	  f4#(I0, I1) -> f3#(I0, I1) 
0.98/1.46	  f3#(I2, I3) -> f1#(-1 + I2, rnd2) [rnd2 = rnd2 /\ 1 <= I2]
0.98/1.46	  f1#(I4, I5) -> f3#(I4, I5) [I5 <= 0]
0.98/1.46	  f2#(I6, I7) -> f1#(I6, I7) 
0.98/1.46	  f1#(I8, I9) -> f2#(I8, -1 + I9) [1 <= I9]
0.98/1.46	R = 
0.98/1.46	  f5(x1, x2) -> f4(x1, x2) 
0.98/1.46	  f4(I0, I1) -> f3(I0, I1) 
0.98/1.46	  f3(I2, I3) -> f1(-1 + I2, rnd2) [rnd2 = rnd2 /\ 1 <= I2]
0.98/1.46	  f1(I4, I5) -> f3(I4, I5) [I5 <= 0]
0.98/1.46	  f2(I6, I7) -> f1(I6, I7) 
0.98/1.46	  f1(I8, I9) -> f2(I8, -1 + I9) [1 <= I9]
0.98/1.46	
0.98/1.46	The dependency graph for this problem is:
0.98/1.46	  0 -> 1
0.98/1.46	  1 -> 2
0.98/1.46	  2 -> 3, 5
0.98/1.46	  3 -> 2
0.98/1.46	  4 -> 3, 5
0.98/1.46	  5 -> 4
0.98/1.46	Where:
0.98/1.46	  0) f5#(x1, x2) -> f4#(x1, x2) 
0.98/1.46	  1) f4#(I0, I1) -> f3#(I0, I1) 
0.98/1.46	  2) f3#(I2, I3) -> f1#(-1 + I2, rnd2) [rnd2 = rnd2 /\ 1 <= I2]
0.98/1.46	  3) f1#(I4, I5) -> f3#(I4, I5) [I5 <= 0]
0.98/1.46	  4) f2#(I6, I7) -> f1#(I6, I7) 
0.98/1.46	  5) f1#(I8, I9) -> f2#(I8, -1 + I9) [1 <= I9]
0.98/1.46	
0.98/1.46	We have the following SCCs.
0.98/1.46	  { 2, 3, 4, 5 }
0.98/1.46	
0.98/1.46	DP problem for innermost termination.
0.98/1.46	P =
0.98/1.46	  f3#(I2, I3) -> f1#(-1 + I2, rnd2) [rnd2 = rnd2 /\ 1 <= I2]
0.98/1.46	  f1#(I4, I5) -> f3#(I4, I5) [I5 <= 0]
0.98/1.46	  f2#(I6, I7) -> f1#(I6, I7) 
0.98/1.46	  f1#(I8, I9) -> f2#(I8, -1 + I9) [1 <= I9]
0.98/1.46	R = 
0.98/1.46	  f5(x1, x2) -> f4(x1, x2) 
0.98/1.46	  f4(I0, I1) -> f3(I0, I1) 
0.98/1.46	  f3(I2, I3) -> f1(-1 + I2, rnd2) [rnd2 = rnd2 /\ 1 <= I2]
0.98/1.46	  f1(I4, I5) -> f3(I4, I5) [I5 <= 0]
0.98/1.46	  f2(I6, I7) -> f1(I6, I7) 
0.98/1.46	  f1(I8, I9) -> f2(I8, -1 + I9) [1 <= I9]
0.98/1.46	
0.98/1.46	We use the basic value criterion with the projection function NU:
0.98/1.46	  NU[f2#(z1,z2)] = z1
0.98/1.46	  NU[f1#(z1,z2)] = z1
0.98/1.46	  NU[f3#(z1,z2)] = z1
0.98/1.46	
0.98/1.46	This gives the following inequalities:
0.98/1.46	  rnd2 = rnd2 /\ 1 <= I2  ==>  I2 >! -1 + I2
0.98/1.46	  I5 <= 0  ==>  I4 (>! \union =) I4
0.98/1.46	    ==>  I6 (>! \union =) I6
0.98/1.46	  1 <= I9  ==>  I8 (>! \union =) I8
0.98/1.46	
0.98/1.46	We remove all the strictly oriented dependency pairs.
0.98/1.46	
0.98/1.46	DP problem for innermost termination.
0.98/1.46	P =
0.98/1.46	  f1#(I4, I5) -> f3#(I4, I5) [I5 <= 0]
0.98/1.46	  f2#(I6, I7) -> f1#(I6, I7) 
0.98/1.46	  f1#(I8, I9) -> f2#(I8, -1 + I9) [1 <= I9]
0.98/1.46	R = 
0.98/1.46	  f5(x1, x2) -> f4(x1, x2) 
0.98/1.46	  f4(I0, I1) -> f3(I0, I1) 
0.98/1.46	  f3(I2, I3) -> f1(-1 + I2, rnd2) [rnd2 = rnd2 /\ 1 <= I2]
0.98/1.46	  f1(I4, I5) -> f3(I4, I5) [I5 <= 0]
0.98/1.46	  f2(I6, I7) -> f1(I6, I7) 
0.98/1.46	  f1(I8, I9) -> f2(I8, -1 + I9) [1 <= I9]
0.98/1.46	
0.98/1.46	The dependency graph for this problem is:
0.98/1.46	  3 -> 
0.98/1.46	  4 -> 3, 5
0.98/1.46	  5 -> 4
0.98/1.46	Where:
0.98/1.46	  3) f1#(I4, I5) -> f3#(I4, I5) [I5 <= 0]
0.98/1.46	  4) f2#(I6, I7) -> f1#(I6, I7) 
0.98/1.46	  5) f1#(I8, I9) -> f2#(I8, -1 + I9) [1 <= I9]
0.98/1.46	
0.98/1.46	We have the following SCCs.
0.98/1.46	  { 4, 5 }
0.98/1.46	
0.98/1.46	DP problem for innermost termination.
0.98/1.46	P =
0.98/1.46	  f2#(I6, I7) -> f1#(I6, I7) 
0.98/1.46	  f1#(I8, I9) -> f2#(I8, -1 + I9) [1 <= I9]
0.98/1.46	R = 
0.98/1.46	  f5(x1, x2) -> f4(x1, x2) 
0.98/1.46	  f4(I0, I1) -> f3(I0, I1) 
0.98/1.46	  f3(I2, I3) -> f1(-1 + I2, rnd2) [rnd2 = rnd2 /\ 1 <= I2]
0.98/1.46	  f1(I4, I5) -> f3(I4, I5) [I5 <= 0]
0.98/1.46	  f2(I6, I7) -> f1(I6, I7) 
0.98/1.46	  f1(I8, I9) -> f2(I8, -1 + I9) [1 <= I9]
0.98/1.46	
0.98/1.46	We use the basic value criterion with the projection function NU:
0.98/1.46	  NU[f1#(z1,z2)] = z2
0.98/1.46	  NU[f2#(z1,z2)] = z2
0.98/1.46	
0.98/1.46	This gives the following inequalities:
0.98/1.46	    ==>  I7 (>! \union =) I7
0.98/1.46	  1 <= I9  ==>  I9 >! -1 + I9
0.98/1.46	
0.98/1.46	We remove all the strictly oriented dependency pairs.
0.98/1.46	
0.98/1.46	DP problem for innermost termination.
0.98/1.46	P =
0.98/1.46	  f2#(I6, I7) -> f1#(I6, I7) 
0.98/1.46	R = 
0.98/1.46	  f5(x1, x2) -> f4(x1, x2) 
0.98/1.46	  f4(I0, I1) -> f3(I0, I1) 
0.98/1.46	  f3(I2, I3) -> f1(-1 + I2, rnd2) [rnd2 = rnd2 /\ 1 <= I2]
0.98/1.46	  f1(I4, I5) -> f3(I4, I5) [I5 <= 0]
0.98/1.46	  f2(I6, I7) -> f1(I6, I7) 
0.98/1.46	  f1(I8, I9) -> f2(I8, -1 + I9) [1 <= I9]
0.98/1.46	
0.98/1.46	The dependency graph for this problem is:
0.98/1.46	  4 -> 
0.98/1.46	Where:
0.98/1.46	  4) f2#(I6, I7) -> f1#(I6, I7) 
0.98/1.46	
0.98/1.46	We have the following SCCs.
0.98/1.46	  
0.98/4.44	EOF