1.53/1.59	YES
1.53/1.59	
1.53/1.59	DP problem for innermost termination.
1.53/1.59	P =
1.53/1.59	  init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
1.53/1.59	  f3#(I0, I1, I2) -> f3#(I0, I1 - 1, I3) [0 <= I0 - 1 /\ 0 <= I1 - 1]
1.53/1.59	  f3#(I4, I5, I6) -> f3#(I4, I5 - 1, I6 - 1) [0 <= I4 - 1 /\ 0 <= I5 - 1]
1.53/1.59	  f3#(I7, I8, I9) -> f2#(I7 - I8, I9 - 1, I10) [0 <= I7 - 1 /\ 0 <= I8 - 1]
1.53/1.59	  f2#(I11, I12, I13) -> f3#(I11, 2, I12) [0 <= I11 - 1]
1.53/1.59	  f1#(I14, I15, I16) -> f2#(I15, I15, I17) [-1 <= I15 - 1 /\ 0 <= I14 - 1]
1.53/1.59	R = 
1.53/1.59	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
1.53/1.59	  f3(I0, I1, I2) -> f3(I0, I1 - 1, I3) [0 <= I0 - 1 /\ 0 <= I1 - 1]
1.53/1.59	  f3(I4, I5, I6) -> f3(I4, I5 - 1, I6 - 1) [0 <= I4 - 1 /\ 0 <= I5 - 1]
1.53/1.59	  f3(I7, I8, I9) -> f2(I7 - I8, I9 - 1, I10) [0 <= I7 - 1 /\ 0 <= I8 - 1]
1.53/1.59	  f2(I11, I12, I13) -> f3(I11, 2, I12) [0 <= I11 - 1]
1.53/1.59	  f1(I14, I15, I16) -> f2(I15, I15, I17) [-1 <= I15 - 1 /\ 0 <= I14 - 1]
1.53/1.59	
1.53/1.59	The dependency graph for this problem is:
1.53/1.59	  0 -> 5
1.53/1.59	  1 -> 1, 2, 3
1.53/1.59	  2 -> 1, 2, 3
1.53/1.59	  3 -> 4
1.53/1.59	  4 -> 1, 2, 3
1.53/1.59	  5 -> 4
1.53/1.59	Where:
1.53/1.59	  0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
1.53/1.59	  1) f3#(I0, I1, I2) -> f3#(I0, I1 - 1, I3) [0 <= I0 - 1 /\ 0 <= I1 - 1]
1.53/1.59	  2) f3#(I4, I5, I6) -> f3#(I4, I5 - 1, I6 - 1) [0 <= I4 - 1 /\ 0 <= I5 - 1]
1.53/1.59	  3) f3#(I7, I8, I9) -> f2#(I7 - I8, I9 - 1, I10) [0 <= I7 - 1 /\ 0 <= I8 - 1]
1.53/1.59	  4) f2#(I11, I12, I13) -> f3#(I11, 2, I12) [0 <= I11 - 1]
1.53/1.59	  5) f1#(I14, I15, I16) -> f2#(I15, I15, I17) [-1 <= I15 - 1 /\ 0 <= I14 - 1]
1.53/1.59	
1.53/1.59	We have the following SCCs.
1.53/1.59	  { 1, 2, 3, 4 }
1.53/1.59	
1.53/1.59	DP problem for innermost termination.
1.53/1.59	P =
1.53/1.59	  f3#(I0, I1, I2) -> f3#(I0, I1 - 1, I3) [0 <= I0 - 1 /\ 0 <= I1 - 1]
1.53/1.59	  f3#(I4, I5, I6) -> f3#(I4, I5 - 1, I6 - 1) [0 <= I4 - 1 /\ 0 <= I5 - 1]
1.53/1.59	  f3#(I7, I8, I9) -> f2#(I7 - I8, I9 - 1, I10) [0 <= I7 - 1 /\ 0 <= I8 - 1]
1.53/1.59	  f2#(I11, I12, I13) -> f3#(I11, 2, I12) [0 <= I11 - 1]
1.53/1.59	R = 
1.53/1.59	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
1.53/1.59	  f3(I0, I1, I2) -> f3(I0, I1 - 1, I3) [0 <= I0 - 1 /\ 0 <= I1 - 1]
1.53/1.59	  f3(I4, I5, I6) -> f3(I4, I5 - 1, I6 - 1) [0 <= I4 - 1 /\ 0 <= I5 - 1]
1.53/1.59	  f3(I7, I8, I9) -> f2(I7 - I8, I9 - 1, I10) [0 <= I7 - 1 /\ 0 <= I8 - 1]
1.53/1.59	  f2(I11, I12, I13) -> f3(I11, 2, I12) [0 <= I11 - 1]
1.53/1.59	  f1(I14, I15, I16) -> f2(I15, I15, I17) [-1 <= I15 - 1 /\ 0 <= I14 - 1]
1.53/1.59	
1.53/1.59	We use the basic value criterion with the projection function NU:
1.53/1.59	  NU[f2#(z1,z2,z3)] = z1
1.53/1.59	  NU[f3#(z1,z2,z3)] = z1
1.53/1.59	
1.53/1.59	This gives the following inequalities:
1.53/1.59	  0 <= I0 - 1 /\ 0 <= I1 - 1  ==>  I0 (>! \union =) I0
1.53/1.59	  0 <= I4 - 1 /\ 0 <= I5 - 1  ==>  I4 (>! \union =) I4
1.53/1.59	  0 <= I7 - 1 /\ 0 <= I8 - 1  ==>  I7 >! I7 - I8
1.53/1.59	  0 <= I11 - 1  ==>  I11 (>! \union =) I11
1.53/1.59	
1.53/1.59	We remove all the strictly oriented dependency pairs.
1.53/1.59	
1.53/1.59	DP problem for innermost termination.
1.53/1.59	P =
1.53/1.59	  f3#(I0, I1, I2) -> f3#(I0, I1 - 1, I3) [0 <= I0 - 1 /\ 0 <= I1 - 1]
1.53/1.59	  f3#(I4, I5, I6) -> f3#(I4, I5 - 1, I6 - 1) [0 <= I4 - 1 /\ 0 <= I5 - 1]
1.53/1.59	  f2#(I11, I12, I13) -> f3#(I11, 2, I12) [0 <= I11 - 1]
1.53/1.59	R = 
1.53/1.59	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
1.53/1.59	  f3(I0, I1, I2) -> f3(I0, I1 - 1, I3) [0 <= I0 - 1 /\ 0 <= I1 - 1]
1.53/1.59	  f3(I4, I5, I6) -> f3(I4, I5 - 1, I6 - 1) [0 <= I4 - 1 /\ 0 <= I5 - 1]
1.53/1.59	  f3(I7, I8, I9) -> f2(I7 - I8, I9 - 1, I10) [0 <= I7 - 1 /\ 0 <= I8 - 1]
1.53/1.59	  f2(I11, I12, I13) -> f3(I11, 2, I12) [0 <= I11 - 1]
1.53/1.59	  f1(I14, I15, I16) -> f2(I15, I15, I17) [-1 <= I15 - 1 /\ 0 <= I14 - 1]
1.53/1.59	
1.53/1.59	The dependency graph for this problem is:
1.53/1.59	  1 -> 1, 2
1.53/1.59	  2 -> 1, 2
1.53/1.59	  4 -> 1, 2
1.53/1.59	Where:
1.53/1.59	  1) f3#(I0, I1, I2) -> f3#(I0, I1 - 1, I3) [0 <= I0 - 1 /\ 0 <= I1 - 1]
1.53/1.59	  2) f3#(I4, I5, I6) -> f3#(I4, I5 - 1, I6 - 1) [0 <= I4 - 1 /\ 0 <= I5 - 1]
1.53/1.59	  4) f2#(I11, I12, I13) -> f3#(I11, 2, I12) [0 <= I11 - 1]
1.53/1.59	
1.53/1.59	We have the following SCCs.
1.53/1.59	  { 1, 2 }
1.53/1.59	
1.53/1.59	DP problem for innermost termination.
1.53/1.59	P =
1.53/1.59	  f3#(I0, I1, I2) -> f3#(I0, I1 - 1, I3) [0 <= I0 - 1 /\ 0 <= I1 - 1]
1.53/1.59	  f3#(I4, I5, I6) -> f3#(I4, I5 - 1, I6 - 1) [0 <= I4 - 1 /\ 0 <= I5 - 1]
1.53/1.59	R = 
1.53/1.59	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
1.53/1.59	  f3(I0, I1, I2) -> f3(I0, I1 - 1, I3) [0 <= I0 - 1 /\ 0 <= I1 - 1]
1.53/1.59	  f3(I4, I5, I6) -> f3(I4, I5 - 1, I6 - 1) [0 <= I4 - 1 /\ 0 <= I5 - 1]
1.53/1.59	  f3(I7, I8, I9) -> f2(I7 - I8, I9 - 1, I10) [0 <= I7 - 1 /\ 0 <= I8 - 1]
1.53/1.59	  f2(I11, I12, I13) -> f3(I11, 2, I12) [0 <= I11 - 1]
1.53/1.59	  f1(I14, I15, I16) -> f2(I15, I15, I17) [-1 <= I15 - 1 /\ 0 <= I14 - 1]
1.53/1.59	
1.53/1.59	We use the basic value criterion with the projection function NU:
1.53/1.59	  NU[f3#(z1,z2,z3)] = z2
1.53/1.59	
1.53/1.59	This gives the following inequalities:
1.53/1.59	  0 <= I0 - 1 /\ 0 <= I1 - 1  ==>  I1 >! I1 - 1
1.53/1.59	  0 <= I4 - 1 /\ 0 <= I5 - 1  ==>  I5 >! I5 - 1
1.53/1.59	
1.53/1.59	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
1.53/4.57	EOF