0.65/0.74	MAYBE
0.65/0.74	
0.65/0.74	DP problem for innermost termination.
0.65/0.74	P =
0.65/0.74	  init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
0.65/0.74	  f3#(I0, I1, I2) -> f3#(I3, I4, I5) [-1 <= I3 - 1 /\ 0 <= I0 - 1 /\ I3 + 1 <= I0]
0.65/0.74	  f2#(I6, I7, I8) -> f3#(I9, I10, I11) [-1 <= I9 - 1 /\ 0 <= I6 - 1 /\ I8 <= 0 /\ I9 + 1 <= I6]
0.65/0.74	  f2#(I12, I13, I14) -> f2#(I15, I13 - 1, I13) [3 <= I15 - 1 /\ 0 <= I12 - 1 /\ 0 <= I14 - 1 /\ I15 - 3 <= I12]
0.65/0.74	  f1#(I16, I17, I18) -> f2#(I19, 27, 28) [0 <= I19 - 1]
0.65/0.74	R = 
0.65/0.74	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.65/0.74	  f3(I0, I1, I2) -> f3(I3, I4, I5) [-1 <= I3 - 1 /\ 0 <= I0 - 1 /\ I3 + 1 <= I0]
0.65/0.74	  f2(I6, I7, I8) -> f3(I9, I10, I11) [-1 <= I9 - 1 /\ 0 <= I6 - 1 /\ I8 <= 0 /\ I9 + 1 <= I6]
0.65/0.74	  f2(I12, I13, I14) -> f2(I15, I13 - 1, I13) [3 <= I15 - 1 /\ 0 <= I12 - 1 /\ 0 <= I14 - 1 /\ I15 - 3 <= I12]
0.65/0.74	  f1(I16, I17, I18) -> f2(I19, 27, 28) [0 <= I19 - 1]
0.65/0.74	
0.65/0.74	The dependency graph for this problem is:
0.65/0.74	  0 -> 4
0.65/0.74	  1 -> 1
0.65/0.74	  2 -> 1
0.65/0.74	  3 -> 2, 3
0.65/0.74	  4 -> 3
0.65/0.74	Where:
0.65/0.74	  0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
0.65/0.74	  1) f3#(I0, I1, I2) -> f3#(I3, I4, I5) [-1 <= I3 - 1 /\ 0 <= I0 - 1 /\ I3 + 1 <= I0]
0.65/0.74	  2) f2#(I6, I7, I8) -> f3#(I9, I10, I11) [-1 <= I9 - 1 /\ 0 <= I6 - 1 /\ I8 <= 0 /\ I9 + 1 <= I6]
0.65/0.74	  3) f2#(I12, I13, I14) -> f2#(I15, I13 - 1, I13) [3 <= I15 - 1 /\ 0 <= I12 - 1 /\ 0 <= I14 - 1 /\ I15 - 3 <= I12]
0.65/0.74	  4) f1#(I16, I17, I18) -> f2#(I19, 27, 28) [0 <= I19 - 1]
0.65/0.74	
0.65/0.74	We have the following SCCs.
0.65/0.74	  { 3 }
0.65/0.74	  { 1 }
0.65/0.74	
0.65/0.74	DP problem for innermost termination.
0.65/0.74	P =
0.65/0.74	  f3#(I0, I1, I2) -> f3#(I3, I4, I5) [-1 <= I3 - 1 /\ 0 <= I0 - 1 /\ I3 + 1 <= I0]
0.65/0.74	R = 
0.65/0.74	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.65/0.74	  f3(I0, I1, I2) -> f3(I3, I4, I5) [-1 <= I3 - 1 /\ 0 <= I0 - 1 /\ I3 + 1 <= I0]
0.65/0.74	  f2(I6, I7, I8) -> f3(I9, I10, I11) [-1 <= I9 - 1 /\ 0 <= I6 - 1 /\ I8 <= 0 /\ I9 + 1 <= I6]
0.65/0.74	  f2(I12, I13, I14) -> f2(I15, I13 - 1, I13) [3 <= I15 - 1 /\ 0 <= I12 - 1 /\ 0 <= I14 - 1 /\ I15 - 3 <= I12]
0.65/0.74	  f1(I16, I17, I18) -> f2(I19, 27, 28) [0 <= I19 - 1]
0.65/0.74	
0.65/0.74	We use the basic value criterion with the projection function NU:
0.65/0.74	  NU[f3#(z1,z2,z3)] = z1
0.65/0.74	
0.65/0.74	This gives the following inequalities:
0.65/0.74	  -1 <= I3 - 1 /\ 0 <= I0 - 1 /\ I3 + 1 <= I0  ==>  I0 >! I3
0.65/0.74	
0.65/0.74	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.65/0.74	
0.65/0.74	DP problem for innermost termination.
0.65/0.74	P =
0.65/0.74	  f2#(I12, I13, I14) -> f2#(I15, I13 - 1, I13) [3 <= I15 - 1 /\ 0 <= I12 - 1 /\ 0 <= I14 - 1 /\ I15 - 3 <= I12]
0.65/0.74	R = 
0.65/0.74	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.65/0.74	  f3(I0, I1, I2) -> f3(I3, I4, I5) [-1 <= I3 - 1 /\ 0 <= I0 - 1 /\ I3 + 1 <= I0]
0.65/0.74	  f2(I6, I7, I8) -> f3(I9, I10, I11) [-1 <= I9 - 1 /\ 0 <= I6 - 1 /\ I8 <= 0 /\ I9 + 1 <= I6]
0.65/0.74	  f2(I12, I13, I14) -> f2(I15, I13 - 1, I13) [3 <= I15 - 1 /\ 0 <= I12 - 1 /\ 0 <= I14 - 1 /\ I15 - 3 <= I12]
0.65/0.74	  f1(I16, I17, I18) -> f2(I19, 27, 28) [0 <= I19 - 1]
0.65/0.74	
0.65/3.72	EOF