0.00/0.35	YES
0.00/0.35	
0.00/0.35	DP problem for innermost termination.
0.00/0.35	P =
0.00/0.35	  f6#(x1, x2) -> f5#(x1, x2) 
0.00/0.35	  f5#(I0, I1) -> f4#(I0, I1) 
0.00/0.35	  f4#(I4, I5) -> f1#(I4, -1 + I5) [0 <= -1 + I5]
0.00/0.35	  f2#(I9, I10) -> f1#(I9, I10) 
0.00/0.35	  f1#(I11, I12) -> f2#(I11, -1 + I12) [0 <= -1 + I12]
0.00/0.35	R = 
0.00/0.35	  f6(x1, x2) -> f5(x1, x2) 
0.00/0.35	  f5(I0, I1) -> f4(I0, I1) 
0.00/0.35	  f4(I2, I3) -> f3(rnd1, I3) [I3 <= 0 /\ y1 = y1 /\ rnd1 = rnd1]
0.00/0.35	  f4(I4, I5) -> f1(I4, -1 + I5) [0 <= -1 + I5]
0.00/0.35	  f1(I6, I7) -> f3(I8, I7) [I7 <= 0 /\ B0 = B0 /\ I8 = I8]
0.00/0.35	  f2(I9, I10) -> f1(I9, I10) 
0.00/0.35	  f1(I11, I12) -> f2(I11, -1 + I12) [0 <= -1 + I12]
0.00/0.35	
0.00/0.35	The dependency graph for this problem is:
0.00/0.35	  0 -> 1
0.00/0.35	  1 -> 2
0.00/0.35	  2 -> 4
0.00/0.35	  3 -> 4
0.00/0.35	  4 -> 3
0.00/0.35	Where:
0.00/0.35	  0) f6#(x1, x2) -> f5#(x1, x2) 
0.00/0.35	  1) f5#(I0, I1) -> f4#(I0, I1) 
0.00/0.35	  2) f4#(I4, I5) -> f1#(I4, -1 + I5) [0 <= -1 + I5]
0.00/0.35	  3) f2#(I9, I10) -> f1#(I9, I10) 
0.00/0.35	  4) f1#(I11, I12) -> f2#(I11, -1 + I12) [0 <= -1 + I12]
0.00/0.35	
0.00/0.35	We have the following SCCs.
0.00/0.35	  { 3, 4 }
0.00/0.35	
0.00/0.35	DP problem for innermost termination.
0.00/0.35	P =
0.00/0.35	  f2#(I9, I10) -> f1#(I9, I10) 
0.00/0.35	  f1#(I11, I12) -> f2#(I11, -1 + I12) [0 <= -1 + I12]
0.00/0.35	R = 
0.00/0.35	  f6(x1, x2) -> f5(x1, x2) 
0.00/0.35	  f5(I0, I1) -> f4(I0, I1) 
0.00/0.35	  f4(I2, I3) -> f3(rnd1, I3) [I3 <= 0 /\ y1 = y1 /\ rnd1 = rnd1]
0.00/0.35	  f4(I4, I5) -> f1(I4, -1 + I5) [0 <= -1 + I5]
0.00/0.35	  f1(I6, I7) -> f3(I8, I7) [I7 <= 0 /\ B0 = B0 /\ I8 = I8]
0.00/0.35	  f2(I9, I10) -> f1(I9, I10) 
0.00/0.35	  f1(I11, I12) -> f2(I11, -1 + I12) [0 <= -1 + I12]
0.00/0.35	
0.00/0.35	We use the basic value criterion with the projection function NU:
0.00/0.35	  NU[f1#(z1,z2)] = z2
0.00/0.35	  NU[f2#(z1,z2)] = z2
0.00/0.35	
0.00/0.35	This gives the following inequalities:
0.00/0.35	    ==>  I10 (>! \union =) I10
0.00/0.35	  0 <= -1 + I12  ==>  I12 >! -1 + I12
0.00/0.35	
0.00/0.35	We remove all the strictly oriented dependency pairs.
0.00/0.35	
0.00/0.35	DP problem for innermost termination.
0.00/0.35	P =
0.00/0.35	  f2#(I9, I10) -> f1#(I9, I10) 
0.00/0.35	R = 
0.00/0.35	  f6(x1, x2) -> f5(x1, x2) 
0.00/0.35	  f5(I0, I1) -> f4(I0, I1) 
0.00/0.35	  f4(I2, I3) -> f3(rnd1, I3) [I3 <= 0 /\ y1 = y1 /\ rnd1 = rnd1]
0.00/0.35	  f4(I4, I5) -> f1(I4, -1 + I5) [0 <= -1 + I5]
0.00/0.35	  f1(I6, I7) -> f3(I8, I7) [I7 <= 0 /\ B0 = B0 /\ I8 = I8]
0.00/0.35	  f2(I9, I10) -> f1(I9, I10) 
0.00/0.35	  f1(I11, I12) -> f2(I11, -1 + I12) [0 <= -1 + I12]
0.00/0.35	
0.00/0.35	The dependency graph for this problem is:
0.00/0.35	  3 -> 
0.00/0.35	Where:
0.00/0.35	  3) f2#(I9, I10) -> f1#(I9, I10) 
0.00/0.35	
0.00/0.35	We have the following SCCs.
0.00/0.35	  
0.00/3.33	EOF