0.86/0.95	YES
0.86/0.95	
0.86/0.95	DP problem for innermost termination.
0.86/0.95	P =
0.86/0.95	  f7#(x1, x2) -> f6#(x1, x2) 
0.86/0.95	  f6#(I0, I1) -> f5#(I0, rnd2) [y1 = 0 /\ rnd2 = rnd2]
0.86/0.95	  f4#(I2, I3) -> f3#(I2, I3) 
0.86/0.95	  f5#(I4, I5) -> f4#(I4, I5) [1 <= I5]
0.86/0.95	  f5#(I6, I7) -> f1#(I6, I7) [I7 <= 0]
0.86/0.95	  f3#(I8, I9) -> f4#(I8, 1 + I9) [1 + I9 <= I8]
0.86/0.95	  f3#(I10, I11) -> f1#(I10, I11) [I10 <= I11]
0.86/0.95	R = 
0.86/0.95	  f7(x1, x2) -> f6(x1, x2) 
0.86/0.95	  f6(I0, I1) -> f5(I0, rnd2) [y1 = 0 /\ rnd2 = rnd2]
0.86/0.95	  f4(I2, I3) -> f3(I2, I3) 
0.86/0.95	  f5(I4, I5) -> f4(I4, I5) [1 <= I5]
0.86/0.95	  f5(I6, I7) -> f1(I6, I7) [I7 <= 0]
0.86/0.95	  f3(I8, I9) -> f4(I8, 1 + I9) [1 + I9 <= I8]
0.86/0.95	  f3(I10, I11) -> f1(I10, I11) [I10 <= I11]
0.86/0.95	  f1(I12, I13) -> f2(I12, I13) 
0.86/0.95	
0.86/0.95	The dependency graph for this problem is:
0.86/0.95	  0 -> 1
0.86/0.95	  1 -> 3, 4
0.86/0.95	  2 -> 5, 6
0.86/0.95	  3 -> 2
0.86/0.95	  4 -> 
0.86/0.95	  5 -> 2
0.86/0.95	  6 -> 
0.86/0.95	Where:
0.86/0.95	  0) f7#(x1, x2) -> f6#(x1, x2) 
0.86/0.95	  1) f6#(I0, I1) -> f5#(I0, rnd2) [y1 = 0 /\ rnd2 = rnd2]
0.86/0.95	  2) f4#(I2, I3) -> f3#(I2, I3) 
0.86/0.95	  3) f5#(I4, I5) -> f4#(I4, I5) [1 <= I5]
0.86/0.95	  4) f5#(I6, I7) -> f1#(I6, I7) [I7 <= 0]
0.86/0.95	  5) f3#(I8, I9) -> f4#(I8, 1 + I9) [1 + I9 <= I8]
0.86/0.95	  6) f3#(I10, I11) -> f1#(I10, I11) [I10 <= I11]
0.86/0.95	
0.86/0.95	We have the following SCCs.
0.86/0.95	  { 2, 5 }
0.86/0.95	
0.86/0.95	DP problem for innermost termination.
0.86/0.95	P =
0.86/0.95	  f4#(I2, I3) -> f3#(I2, I3) 
0.86/0.95	  f3#(I8, I9) -> f4#(I8, 1 + I9) [1 + I9 <= I8]
0.86/0.95	R = 
0.86/0.95	  f7(x1, x2) -> f6(x1, x2) 
0.86/0.95	  f6(I0, I1) -> f5(I0, rnd2) [y1 = 0 /\ rnd2 = rnd2]
0.86/0.95	  f4(I2, I3) -> f3(I2, I3) 
0.86/0.95	  f5(I4, I5) -> f4(I4, I5) [1 <= I5]
0.86/0.95	  f5(I6, I7) -> f1(I6, I7) [I7 <= 0]
0.86/0.95	  f3(I8, I9) -> f4(I8, 1 + I9) [1 + I9 <= I8]
0.86/0.95	  f3(I10, I11) -> f1(I10, I11) [I10 <= I11]
0.86/0.95	  f1(I12, I13) -> f2(I12, I13) 
0.86/0.95	
0.86/0.95	We use the reverse value criterion with the projection function NU:
0.86/0.95	  NU[f3#(z1,z2)] = z1 + -1 * (1 + z2)
0.86/0.95	  NU[f4#(z1,z2)] = z1 + -1 * (1 + z2)
0.86/0.95	
0.86/0.95	This gives the following inequalities:
0.86/0.95	    ==>  I2 + -1 * (1 + I3) >= I2 + -1 * (1 + I3)
0.86/0.95	  1 + I9 <= I8  ==>  I8 + -1 * (1 + I9) > I8 + -1 * (1 + (1 + I9))  with  I8 + -1 * (1 + I9) >= 0
0.86/0.95	
0.86/0.95	We remove all the strictly oriented dependency pairs.
0.86/0.95	
0.86/0.95	DP problem for innermost termination.
0.86/0.95	P =
0.86/0.95	  f4#(I2, I3) -> f3#(I2, I3) 
0.86/0.95	R = 
0.86/0.95	  f7(x1, x2) -> f6(x1, x2) 
0.86/0.95	  f6(I0, I1) -> f5(I0, rnd2) [y1 = 0 /\ rnd2 = rnd2]
0.86/0.95	  f4(I2, I3) -> f3(I2, I3) 
0.86/0.95	  f5(I4, I5) -> f4(I4, I5) [1 <= I5]
0.86/0.95	  f5(I6, I7) -> f1(I6, I7) [I7 <= 0]
0.86/0.95	  f3(I8, I9) -> f4(I8, 1 + I9) [1 + I9 <= I8]
0.86/0.95	  f3(I10, I11) -> f1(I10, I11) [I10 <= I11]
0.86/0.95	  f1(I12, I13) -> f2(I12, I13) 
0.86/0.95	
0.86/0.95	The dependency graph for this problem is:
0.86/0.95	  2 -> 
0.86/0.95	Where:
0.86/0.95	  2) f4#(I2, I3) -> f3#(I2, I3) 
0.86/0.95	
0.86/0.95	We have the following SCCs.
0.86/0.95	  
0.86/3.93	EOF