0.91/1.20	MAYBE
0.91/1.20	
0.91/1.20	DP problem for innermost termination.
0.91/1.20	P =
0.91/1.20	  init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
0.91/1.20	  f2#(I0, I1, I2) -> f2#(-1, 1, 0) [1 = I2 /\ I1 <= 1 /\ 5 <= I0 - 1]
0.91/1.20	  f2#(I3, I4, I5) -> f2#(I3 + 1, 0 - (I3 + 1), I5 - 1) [1 <= I5 - 1 /\ 5 <= I3 + I5 /\ 5 <= I3 - I5 /\ I4 <= I5 /\ I5 <= I3]
0.91/1.20	  f2#(I6, I7, I8) -> f2#(I6 + 1, 0 - (I6 + 1), I8 - 1) [I8 <= 0 /\ 5 <= I6 + I8 /\ 5 <= I6 - I8 /\ I7 <= I8 /\ I8 <= I6]
0.91/1.20	  f2#(I9, I10, I11) -> f2#(I9, 0 - I9, -1 * I11) [I9 + I11 <= 4 /\ 5 <= I9 - I11 /\ I10 <= I11 /\ I11 <= I9]
0.91/1.20	  f2#(I12, I13, I14) -> f2#(I12, 0 - I12, -1 * I14) [I13 <= I14 /\ I14 <= I12 /\ I12 - I14 <= 4]
0.91/1.20	  f1#(I15, I16, I17) -> f2#(20, -20, I16) [-1 <= I16 - 1 /\ 0 <= I15 - 1]
0.91/1.20	R = 
0.91/1.20	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.91/1.20	  f2(I0, I1, I2) -> f2(-1, 1, 0) [1 = I2 /\ I1 <= 1 /\ 5 <= I0 - 1]
0.91/1.20	  f2(I3, I4, I5) -> f2(I3 + 1, 0 - (I3 + 1), I5 - 1) [1 <= I5 - 1 /\ 5 <= I3 + I5 /\ 5 <= I3 - I5 /\ I4 <= I5 /\ I5 <= I3]
0.91/1.20	  f2(I6, I7, I8) -> f2(I6 + 1, 0 - (I6 + 1), I8 - 1) [I8 <= 0 /\ 5 <= I6 + I8 /\ 5 <= I6 - I8 /\ I7 <= I8 /\ I8 <= I6]
0.91/1.20	  f2(I9, I10, I11) -> f2(I9, 0 - I9, -1 * I11) [I9 + I11 <= 4 /\ 5 <= I9 - I11 /\ I10 <= I11 /\ I11 <= I9]
0.91/1.20	  f2(I12, I13, I14) -> f2(I12, 0 - I12, -1 * I14) [I13 <= I14 /\ I14 <= I12 /\ I12 - I14 <= 4]
0.91/1.20	  f1(I15, I16, I17) -> f2(20, -20, I16) [-1 <= I16 - 1 /\ 0 <= I15 - 1]
0.91/1.20	
0.91/1.20	The dependency graph for this problem is:
0.91/1.20	  0 -> 6
0.91/1.20	  1 -> 
0.91/1.20	  2 -> 1, 2
0.91/1.20	  3 -> 3
0.91/1.20	  4 -> 5
0.91/1.20	  5 -> 4, 5
0.91/1.20	  6 -> 1, 2, 3, 5
0.91/1.20	Where:
0.91/1.20	  0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
0.91/1.20	  1) f2#(I0, I1, I2) -> f2#(-1, 1, 0) [1 = I2 /\ I1 <= 1 /\ 5 <= I0 - 1]
0.91/1.20	  2) f2#(I3, I4, I5) -> f2#(I3 + 1, 0 - (I3 + 1), I5 - 1) [1 <= I5 - 1 /\ 5 <= I3 + I5 /\ 5 <= I3 - I5 /\ I4 <= I5 /\ I5 <= I3]
0.91/1.20	  3) f2#(I6, I7, I8) -> f2#(I6 + 1, 0 - (I6 + 1), I8 - 1) [I8 <= 0 /\ 5 <= I6 + I8 /\ 5 <= I6 - I8 /\ I7 <= I8 /\ I8 <= I6]
0.91/1.20	  4) f2#(I9, I10, I11) -> f2#(I9, 0 - I9, -1 * I11) [I9 + I11 <= 4 /\ 5 <= I9 - I11 /\ I10 <= I11 /\ I11 <= I9]
0.91/1.20	  5) f2#(I12, I13, I14) -> f2#(I12, 0 - I12, -1 * I14) [I13 <= I14 /\ I14 <= I12 /\ I12 - I14 <= 4]
0.91/1.20	  6) f1#(I15, I16, I17) -> f2#(20, -20, I16) [-1 <= I16 - 1 /\ 0 <= I15 - 1]
0.91/1.20	
0.91/1.20	We have the following SCCs.
0.91/1.20	  { 4, 5 }
0.91/1.20	  { 3 }
0.91/1.20	  { 2 }
0.91/1.20	
0.91/1.20	DP problem for innermost termination.
0.91/1.20	P =
0.91/1.20	  f2#(I3, I4, I5) -> f2#(I3 + 1, 0 - (I3 + 1), I5 - 1) [1 <= I5 - 1 /\ 5 <= I3 + I5 /\ 5 <= I3 - I5 /\ I4 <= I5 /\ I5 <= I3]
0.91/1.20	R = 
0.91/1.20	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.91/1.20	  f2(I0, I1, I2) -> f2(-1, 1, 0) [1 = I2 /\ I1 <= 1 /\ 5 <= I0 - 1]
0.91/1.20	  f2(I3, I4, I5) -> f2(I3 + 1, 0 - (I3 + 1), I5 - 1) [1 <= I5 - 1 /\ 5 <= I3 + I5 /\ 5 <= I3 - I5 /\ I4 <= I5 /\ I5 <= I3]
0.91/1.20	  f2(I6, I7, I8) -> f2(I6 + 1, 0 - (I6 + 1), I8 - 1) [I8 <= 0 /\ 5 <= I6 + I8 /\ 5 <= I6 - I8 /\ I7 <= I8 /\ I8 <= I6]
0.91/1.20	  f2(I9, I10, I11) -> f2(I9, 0 - I9, -1 * I11) [I9 + I11 <= 4 /\ 5 <= I9 - I11 /\ I10 <= I11 /\ I11 <= I9]
0.91/1.20	  f2(I12, I13, I14) -> f2(I12, 0 - I12, -1 * I14) [I13 <= I14 /\ I14 <= I12 /\ I12 - I14 <= 4]
0.91/1.20	  f1(I15, I16, I17) -> f2(20, -20, I16) [-1 <= I16 - 1 /\ 0 <= I15 - 1]
0.91/1.20	
0.91/1.20	We use the basic value criterion with the projection function NU:
0.91/1.20	  NU[f2#(z1,z2,z3)] = z3
0.91/1.20	
0.91/1.20	This gives the following inequalities:
0.91/1.20	  1 <= I5 - 1 /\ 5 <= I3 + I5 /\ 5 <= I3 - I5 /\ I4 <= I5 /\ I5 <= I3  ==>  I5 >! I5 - 1
0.91/1.20	
0.91/1.20	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.91/1.20	
0.91/1.20	DP problem for innermost termination.
0.91/1.20	P =
0.91/1.20	  f2#(I6, I7, I8) -> f2#(I6 + 1, 0 - (I6 + 1), I8 - 1) [I8 <= 0 /\ 5 <= I6 + I8 /\ 5 <= I6 - I8 /\ I7 <= I8 /\ I8 <= I6]
0.91/1.20	R = 
0.91/1.20	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.91/1.20	  f2(I0, I1, I2) -> f2(-1, 1, 0) [1 = I2 /\ I1 <= 1 /\ 5 <= I0 - 1]
0.91/1.20	  f2(I3, I4, I5) -> f2(I3 + 1, 0 - (I3 + 1), I5 - 1) [1 <= I5 - 1 /\ 5 <= I3 + I5 /\ 5 <= I3 - I5 /\ I4 <= I5 /\ I5 <= I3]
0.91/1.20	  f2(I6, I7, I8) -> f2(I6 + 1, 0 - (I6 + 1), I8 - 1) [I8 <= 0 /\ 5 <= I6 + I8 /\ 5 <= I6 - I8 /\ I7 <= I8 /\ I8 <= I6]
0.91/1.20	  f2(I9, I10, I11) -> f2(I9, 0 - I9, -1 * I11) [I9 + I11 <= 4 /\ 5 <= I9 - I11 /\ I10 <= I11 /\ I11 <= I9]
0.91/1.20	  f2(I12, I13, I14) -> f2(I12, 0 - I12, -1 * I14) [I13 <= I14 /\ I14 <= I12 /\ I12 - I14 <= 4]
0.91/1.20	  f1(I15, I16, I17) -> f2(20, -20, I16) [-1 <= I16 - 1 /\ 0 <= I15 - 1]
0.91/1.20	
0.91/4.18	EOF