0.75/0.78	YES
0.75/0.78	
0.75/0.78	DP problem for innermost termination.
0.75/0.78	P =
0.75/0.78	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.75/0.78	  f3#(I0, I1) -> f2#(I1, I0) [I0 <= I1 - 1]
0.75/0.78	  f3#(I2, I3) -> f3#(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2 - 1]
0.75/0.78	  f2#(I4, I5) -> f2#(I4, 0) [I4 = I5 /\ 0 <= I4 - 1]
0.75/0.78	  f2#(I6, I7) -> f2#(I7, 0) [0 = I6 /\ 0 <= I7 - 1]
0.75/0.78	  f2#(I8, I9) -> f3#(I8, I9) [0 <= I8 - 1 /\ I9 <= I8 - 1 /\ 0 <= I9 - 1]
0.75/0.78	  f2#(I10, I11) -> f3#(I10, I11) [0 <= I10 - 1 /\ I10 <= I11 - 1 /\ 0 <= I11 - 1]
0.75/0.78	  f1#(I12, I13) -> f2#(I14, I15) [0 <= I12 - 1 /\ -1 <= I14 - 1 /\ -1 <= I13 - 1 /\ -1 <= I15 - 1]
0.75/0.78	R = 
0.75/0.78	  init(x1, x2) -> f1(rnd1, rnd2) 
0.75/0.78	  f3(I0, I1) -> f2(I1, I0) [I0 <= I1 - 1]
0.75/0.78	  f3(I2, I3) -> f3(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2 - 1]
0.75/0.78	  f2(I4, I5) -> f2(I4, 0) [I4 = I5 /\ 0 <= I4 - 1]
0.75/0.78	  f2(I6, I7) -> f2(I7, 0) [0 = I6 /\ 0 <= I7 - 1]
0.75/0.78	  f2(I8, I9) -> f3(I8, I9) [0 <= I8 - 1 /\ I9 <= I8 - 1 /\ 0 <= I9 - 1]
0.75/0.78	  f2(I10, I11) -> f3(I10, I11) [0 <= I10 - 1 /\ I10 <= I11 - 1 /\ 0 <= I11 - 1]
0.75/0.78	  f1(I12, I13) -> f2(I14, I15) [0 <= I12 - 1 /\ -1 <= I14 - 1 /\ -1 <= I13 - 1 /\ -1 <= I15 - 1]
0.75/0.78	
0.75/0.78	The dependency graph for this problem is:
0.75/0.78	  0 -> 7
0.75/0.78	  1 -> 5
0.75/0.78	  2 -> 1, 2
0.75/0.78	  3 -> 
0.75/0.78	  4 -> 
0.75/0.78	  5 -> 2
0.75/0.78	  6 -> 1
0.75/0.78	  7 -> 3, 4, 5, 6
0.75/0.78	Where:
0.75/0.78	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.75/0.78	  1) f3#(I0, I1) -> f2#(I1, I0) [I0 <= I1 - 1]
0.75/0.78	  2) f3#(I2, I3) -> f3#(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2 - 1]
0.75/0.78	  3) f2#(I4, I5) -> f2#(I4, 0) [I4 = I5 /\ 0 <= I4 - 1]
0.75/0.78	  4) f2#(I6, I7) -> f2#(I7, 0) [0 = I6 /\ 0 <= I7 - 1]
0.75/0.78	  5) f2#(I8, I9) -> f3#(I8, I9) [0 <= I8 - 1 /\ I9 <= I8 - 1 /\ 0 <= I9 - 1]
0.75/0.78	  6) f2#(I10, I11) -> f3#(I10, I11) [0 <= I10 - 1 /\ I10 <= I11 - 1 /\ 0 <= I11 - 1]
0.75/0.78	  7) f1#(I12, I13) -> f2#(I14, I15) [0 <= I12 - 1 /\ -1 <= I14 - 1 /\ -1 <= I13 - 1 /\ -1 <= I15 - 1]
0.75/0.78	
0.75/0.78	We have the following SCCs.
0.75/0.78	  { 1, 2, 5 }
0.75/0.78	
0.75/0.78	DP problem for innermost termination.
0.75/0.78	P =
0.75/0.78	  f3#(I0, I1) -> f2#(I1, I0) [I0 <= I1 - 1]
0.75/0.78	  f3#(I2, I3) -> f3#(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2 - 1]
0.75/0.78	  f2#(I8, I9) -> f3#(I8, I9) [0 <= I8 - 1 /\ I9 <= I8 - 1 /\ 0 <= I9 - 1]
0.75/0.78	R = 
0.75/0.78	  init(x1, x2) -> f1(rnd1, rnd2) 
0.75/0.78	  f3(I0, I1) -> f2(I1, I0) [I0 <= I1 - 1]
0.75/0.78	  f3(I2, I3) -> f3(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2 - 1]
0.75/0.78	  f2(I4, I5) -> f2(I4, 0) [I4 = I5 /\ 0 <= I4 - 1]
0.75/0.78	  f2(I6, I7) -> f2(I7, 0) [0 = I6 /\ 0 <= I7 - 1]
0.75/0.78	  f2(I8, I9) -> f3(I8, I9) [0 <= I8 - 1 /\ I9 <= I8 - 1 /\ 0 <= I9 - 1]
0.75/0.78	  f2(I10, I11) -> f3(I10, I11) [0 <= I10 - 1 /\ I10 <= I11 - 1 /\ 0 <= I11 - 1]
0.75/0.78	  f1(I12, I13) -> f2(I14, I15) [0 <= I12 - 1 /\ -1 <= I14 - 1 /\ -1 <= I13 - 1 /\ -1 <= I15 - 1]
0.75/0.78	
0.75/0.78	We use the basic value criterion with the projection function NU:
0.75/0.78	  NU[f2#(z1,z2)] = z1
0.75/0.78	  NU[f3#(z1,z2)] = z2
0.75/0.78	
0.75/0.78	This gives the following inequalities:
0.75/0.78	  I0 <= I1 - 1  ==>  I1 (>! \union =) I1
0.75/0.78	  0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2 - 1  ==>  I3 (>! \union =) I3
0.75/0.78	  0 <= I8 - 1 /\ I9 <= I8 - 1 /\ 0 <= I9 - 1  ==>  I8 >! I9
0.75/0.78	
0.75/0.78	We remove all the strictly oriented dependency pairs.
0.75/0.78	
0.75/0.78	DP problem for innermost termination.
0.75/0.78	P =
0.75/0.78	  f3#(I0, I1) -> f2#(I1, I0) [I0 <= I1 - 1]
0.75/0.78	  f3#(I2, I3) -> f3#(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2 - 1]
0.75/0.78	R = 
0.75/0.78	  init(x1, x2) -> f1(rnd1, rnd2) 
0.75/0.78	  f3(I0, I1) -> f2(I1, I0) [I0 <= I1 - 1]
0.75/0.78	  f3(I2, I3) -> f3(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2 - 1]
0.75/0.78	  f2(I4, I5) -> f2(I4, 0) [I4 = I5 /\ 0 <= I4 - 1]
0.75/0.78	  f2(I6, I7) -> f2(I7, 0) [0 = I6 /\ 0 <= I7 - 1]
0.75/0.78	  f2(I8, I9) -> f3(I8, I9) [0 <= I8 - 1 /\ I9 <= I8 - 1 /\ 0 <= I9 - 1]
0.75/0.78	  f2(I10, I11) -> f3(I10, I11) [0 <= I10 - 1 /\ I10 <= I11 - 1 /\ 0 <= I11 - 1]
0.75/0.78	  f1(I12, I13) -> f2(I14, I15) [0 <= I12 - 1 /\ -1 <= I14 - 1 /\ -1 <= I13 - 1 /\ -1 <= I15 - 1]
0.75/0.78	
0.75/0.78	The dependency graph for this problem is:
0.75/0.78	  1 -> 
0.75/0.78	  2 -> 1, 2
0.75/0.78	Where:
0.75/0.78	  1) f3#(I0, I1) -> f2#(I1, I0) [I0 <= I1 - 1]
0.75/0.78	  2) f3#(I2, I3) -> f3#(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2 - 1]
0.75/0.78	
0.75/0.78	We have the following SCCs.
0.75/0.78	  { 2 }
0.75/0.78	
0.75/0.78	DP problem for innermost termination.
0.75/0.78	P =
0.75/0.78	  f3#(I2, I3) -> f3#(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2 - 1]
0.75/0.78	R = 
0.75/0.78	  init(x1, x2) -> f1(rnd1, rnd2) 
0.75/0.78	  f3(I0, I1) -> f2(I1, I0) [I0 <= I1 - 1]
0.75/0.78	  f3(I2, I3) -> f3(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2 - 1]
0.75/0.78	  f2(I4, I5) -> f2(I4, 0) [I4 = I5 /\ 0 <= I4 - 1]
0.75/0.78	  f2(I6, I7) -> f2(I7, 0) [0 = I6 /\ 0 <= I7 - 1]
0.75/0.78	  f2(I8, I9) -> f3(I8, I9) [0 <= I8 - 1 /\ I9 <= I8 - 1 /\ 0 <= I9 - 1]
0.75/0.78	  f2(I10, I11) -> f3(I10, I11) [0 <= I10 - 1 /\ I10 <= I11 - 1 /\ 0 <= I11 - 1]
0.75/0.78	  f1(I12, I13) -> f2(I14, I15) [0 <= I12 - 1 /\ -1 <= I14 - 1 /\ -1 <= I13 - 1 /\ -1 <= I15 - 1]
0.75/0.78	
0.75/0.78	We use the basic value criterion with the projection function NU:
0.75/0.78	  NU[f3#(z1,z2)] = z1
0.75/0.78	
0.75/0.78	This gives the following inequalities:
0.75/0.78	  0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2 - 1  ==>  I2 >! I2 - I3
0.75/0.78	
0.75/0.78	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.75/3.76	EOF