0.77/0.82	YES
0.77/0.82	
0.77/0.82	DP problem for innermost termination.
0.77/0.82	P =
0.77/0.82	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.77/0.82	  f3#(I0, I1) -> f2#(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1]
0.77/0.82	  f4#(I3, I4) -> f3#(I3, I3 - 2) [0 <= I3 - 1]
0.77/0.82	  f2#(I5, I6) -> f3#(I5, I5 - 2) [I5 - 1 <= I5 - 1 /\ 1 <= I5 - 1]
0.77/0.82	  f2#(I7, I8) -> f3#(2, 0) [2 = I7]
0.77/0.82	  f2#(I9, I10) -> f2#(I9 - 1, I11) [I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1]
0.77/0.82	  f1#(I12, I13) -> f2#(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1]
0.77/0.82	R = 
0.77/0.82	  init(x1, x2) -> f1(rnd1, rnd2) 
0.77/0.82	  f3(I0, I1) -> f2(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1]
0.77/0.82	  f4(I3, I4) -> f3(I3, I3 - 2) [0 <= I3 - 1]
0.77/0.82	  f2(I5, I6) -> f3(I5, I5 - 2) [I5 - 1 <= I5 - 1 /\ 1 <= I5 - 1]
0.77/0.82	  f2(I7, I8) -> f3(2, 0) [2 = I7]
0.77/0.82	  f2(I9, I10) -> f2(I9 - 1, I11) [I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1]
0.77/0.82	  f1(I12, I13) -> f2(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1]
0.77/0.82	
0.77/0.82	The dependency graph for this problem is:
0.77/0.82	  0 -> 6
0.77/0.82	  1 -> 3, 4, 5
0.77/0.82	  2 -> 1
0.77/0.82	  3 -> 1
0.77/0.82	  4 -> 1
0.77/0.82	  5 -> 3, 4, 5
0.77/0.82	  6 -> 3, 4, 5
0.77/0.82	Where:
0.77/0.82	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.77/0.82	  1) f3#(I0, I1) -> f2#(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1]
0.77/0.82	  2) f4#(I3, I4) -> f3#(I3, I3 - 2) [0 <= I3 - 1]
0.77/0.82	  3) f2#(I5, I6) -> f3#(I5, I5 - 2) [I5 - 1 <= I5 - 1 /\ 1 <= I5 - 1]
0.77/0.82	  4) f2#(I7, I8) -> f3#(2, 0) [2 = I7]
0.77/0.82	  5) f2#(I9, I10) -> f2#(I9 - 1, I11) [I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1]
0.77/0.82	  6) f1#(I12, I13) -> f2#(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1]
0.77/0.82	
0.77/0.82	We have the following SCCs.
0.77/0.82	  { 1, 3, 4, 5 }
0.77/0.82	
0.77/0.82	DP problem for innermost termination.
0.77/0.82	P =
0.77/0.82	  f3#(I0, I1) -> f2#(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1]
0.77/0.82	  f2#(I5, I6) -> f3#(I5, I5 - 2) [I5 - 1 <= I5 - 1 /\ 1 <= I5 - 1]
0.77/0.82	  f2#(I7, I8) -> f3#(2, 0) [2 = I7]
0.77/0.82	  f2#(I9, I10) -> f2#(I9 - 1, I11) [I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1]
0.77/0.82	R = 
0.77/0.82	  init(x1, x2) -> f1(rnd1, rnd2) 
0.77/0.82	  f3(I0, I1) -> f2(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1]
0.77/0.82	  f4(I3, I4) -> f3(I3, I3 - 2) [0 <= I3 - 1]
0.77/0.82	  f2(I5, I6) -> f3(I5, I5 - 2) [I5 - 1 <= I5 - 1 /\ 1 <= I5 - 1]
0.77/0.82	  f2(I7, I8) -> f3(2, 0) [2 = I7]
0.77/0.82	  f2(I9, I10) -> f2(I9 - 1, I11) [I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1]
0.77/0.82	  f1(I12, I13) -> f2(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1]
0.77/0.82	
0.77/0.82	We use the basic value criterion with the projection function NU:
0.77/0.82	  NU[f2#(z1,z2)] = z1
0.77/0.82	  NU[f3#(z1,z2)] = z2
0.77/0.82	
0.77/0.82	This gives the following inequalities:
0.77/0.82	  I1 <= I0 - 1 /\ 1 <= I0 - 1  ==>  I1 (>! \union =) I1
0.77/0.82	  I5 - 1 <= I5 - 1 /\ 1 <= I5 - 1  ==>  I5 >! I5 - 2
0.77/0.82	  2 = I7  ==>  I7 >! 0
0.77/0.82	  I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1  ==>  I9 >! I9 - 1
0.77/0.82	
0.77/0.82	We remove all the strictly oriented dependency pairs.
0.77/0.82	
0.77/0.82	DP problem for innermost termination.
0.77/0.82	P =
0.77/0.82	  f3#(I0, I1) -> f2#(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1]
0.77/0.82	R = 
0.77/0.82	  init(x1, x2) -> f1(rnd1, rnd2) 
0.77/0.82	  f3(I0, I1) -> f2(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1]
0.77/0.82	  f4(I3, I4) -> f3(I3, I3 - 2) [0 <= I3 - 1]
0.77/0.82	  f2(I5, I6) -> f3(I5, I5 - 2) [I5 - 1 <= I5 - 1 /\ 1 <= I5 - 1]
0.77/0.82	  f2(I7, I8) -> f3(2, 0) [2 = I7]
0.77/0.82	  f2(I9, I10) -> f2(I9 - 1, I11) [I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1]
0.77/0.82	  f1(I12, I13) -> f2(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1]
0.77/0.82	
0.77/0.82	The dependency graph for this problem is:
0.77/0.82	  1 -> 
0.77/0.82	Where:
0.77/0.82	  1) f3#(I0, I1) -> f2#(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1]
0.77/0.82	
0.77/0.82	We have the following SCCs.
0.77/0.82	  
0.77/3.80	EOF