0.00/0.39	YES
0.00/0.39	
0.00/0.39	DP problem for innermost termination.
0.00/0.39	P =
0.00/0.39	  init#(x1, x2) -> f3#(rnd1, rnd2) 
0.00/0.39	  f5#(I0, I1) -> f5#(I0 - 1, I2) [0 <= I0 - 1]
0.00/0.39	  f3#(I3, I4) -> f5#(I5, I6) [0 <= I3 - 1 /\ -1 <= I5 - 1 /\ -1 <= I4 - 1]
0.00/0.39	  f4#(I7, I8) -> f4#(I9, I10) [-1 <= I10 - 1 /\ -1 <= I9 - 1 /\ 2 <= I8 - 1 /\ 0 <= I7 - 1 /\ I10 + 3 <= I8 /\ I9 + 1 <= I7]
0.00/0.39	  f2#(I11, I12) -> f4#(I13, I14) [-1 <= I14 - 1 /\ 0 <= I13 - 1 /\ 0 <= I11 - 1 /\ I14 + 1 <= I11 /\ I13 <= I11]
0.00/0.39	  f3#(I15, I16) -> f2#(I17, I18) [-1 <= I17 - 1 /\ 0 <= I15 - 1]
0.00/0.39	  f1#(I19, I20) -> f2#(I21, I22) [-1 <= I21 - 1 /\ -1 <= I19 - 1 /\ I21 <= I19]
0.00/0.39	R = 
0.00/0.39	  init(x1, x2) -> f3(rnd1, rnd2) 
0.00/0.39	  f5(I0, I1) -> f5(I0 - 1, I2) [0 <= I0 - 1]
0.00/0.39	  f3(I3, I4) -> f5(I5, I6) [0 <= I3 - 1 /\ -1 <= I5 - 1 /\ -1 <= I4 - 1]
0.00/0.39	  f4(I7, I8) -> f4(I9, I10) [-1 <= I10 - 1 /\ -1 <= I9 - 1 /\ 2 <= I8 - 1 /\ 0 <= I7 - 1 /\ I10 + 3 <= I8 /\ I9 + 1 <= I7]
0.00/0.39	  f2(I11, I12) -> f4(I13, I14) [-1 <= I14 - 1 /\ 0 <= I13 - 1 /\ 0 <= I11 - 1 /\ I14 + 1 <= I11 /\ I13 <= I11]
0.00/0.39	  f3(I15, I16) -> f2(I17, I18) [-1 <= I17 - 1 /\ 0 <= I15 - 1]
0.00/0.39	  f1(I19, I20) -> f2(I21, I22) [-1 <= I21 - 1 /\ -1 <= I19 - 1 /\ I21 <= I19]
0.00/0.39	
0.00/0.39	The dependency graph for this problem is:
0.00/0.39	  0 -> 2, 5
0.00/0.39	  1 -> 1
0.00/0.39	  2 -> 1
0.00/0.39	  3 -> 3
0.00/0.39	  4 -> 3
0.00/0.39	  5 -> 4
0.00/0.39	  6 -> 4
0.00/0.39	Where:
0.00/0.39	  0) init#(x1, x2) -> f3#(rnd1, rnd2) 
0.00/0.39	  1) f5#(I0, I1) -> f5#(I0 - 1, I2) [0 <= I0 - 1]
0.00/0.39	  2) f3#(I3, I4) -> f5#(I5, I6) [0 <= I3 - 1 /\ -1 <= I5 - 1 /\ -1 <= I4 - 1]
0.00/0.39	  3) f4#(I7, I8) -> f4#(I9, I10) [-1 <= I10 - 1 /\ -1 <= I9 - 1 /\ 2 <= I8 - 1 /\ 0 <= I7 - 1 /\ I10 + 3 <= I8 /\ I9 + 1 <= I7]
0.00/0.39	  4) f2#(I11, I12) -> f4#(I13, I14) [-1 <= I14 - 1 /\ 0 <= I13 - 1 /\ 0 <= I11 - 1 /\ I14 + 1 <= I11 /\ I13 <= I11]
0.00/0.39	  5) f3#(I15, I16) -> f2#(I17, I18) [-1 <= I17 - 1 /\ 0 <= I15 - 1]
0.00/0.39	  6) f1#(I19, I20) -> f2#(I21, I22) [-1 <= I21 - 1 /\ -1 <= I19 - 1 /\ I21 <= I19]
0.00/0.39	
0.00/0.39	We have the following SCCs.
0.00/0.39	  { 1 }
0.00/0.39	  { 3 }
0.00/0.39	
0.00/0.39	DP problem for innermost termination.
0.00/0.39	P =
0.00/0.39	  f4#(I7, I8) -> f4#(I9, I10) [-1 <= I10 - 1 /\ -1 <= I9 - 1 /\ 2 <= I8 - 1 /\ 0 <= I7 - 1 /\ I10 + 3 <= I8 /\ I9 + 1 <= I7]
0.00/0.39	R = 
0.00/0.39	  init(x1, x2) -> f3(rnd1, rnd2) 
0.00/0.39	  f5(I0, I1) -> f5(I0 - 1, I2) [0 <= I0 - 1]
0.00/0.39	  f3(I3, I4) -> f5(I5, I6) [0 <= I3 - 1 /\ -1 <= I5 - 1 /\ -1 <= I4 - 1]
0.00/0.39	  f4(I7, I8) -> f4(I9, I10) [-1 <= I10 - 1 /\ -1 <= I9 - 1 /\ 2 <= I8 - 1 /\ 0 <= I7 - 1 /\ I10 + 3 <= I8 /\ I9 + 1 <= I7]
0.00/0.39	  f2(I11, I12) -> f4(I13, I14) [-1 <= I14 - 1 /\ 0 <= I13 - 1 /\ 0 <= I11 - 1 /\ I14 + 1 <= I11 /\ I13 <= I11]
0.00/0.39	  f3(I15, I16) -> f2(I17, I18) [-1 <= I17 - 1 /\ 0 <= I15 - 1]
0.00/0.39	  f1(I19, I20) -> f2(I21, I22) [-1 <= I21 - 1 /\ -1 <= I19 - 1 /\ I21 <= I19]
0.00/0.39	
0.00/0.39	We use the basic value criterion with the projection function NU:
0.00/0.39	  NU[f4#(z1,z2)] = z2
0.00/0.39	
0.00/0.39	This gives the following inequalities:
0.00/0.39	  -1 <= I10 - 1 /\ -1 <= I9 - 1 /\ 2 <= I8 - 1 /\ 0 <= I7 - 1 /\ I10 + 3 <= I8 /\ I9 + 1 <= I7  ==>  I8 >! I10
0.00/0.39	
0.00/0.39	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/0.39	
0.00/0.39	DP problem for innermost termination.
0.00/0.39	P =
0.00/0.39	  f5#(I0, I1) -> f5#(I0 - 1, I2) [0 <= I0 - 1]
0.00/0.39	R = 
0.00/0.39	  init(x1, x2) -> f3(rnd1, rnd2) 
0.00/0.39	  f5(I0, I1) -> f5(I0 - 1, I2) [0 <= I0 - 1]
0.00/0.39	  f3(I3, I4) -> f5(I5, I6) [0 <= I3 - 1 /\ -1 <= I5 - 1 /\ -1 <= I4 - 1]
0.00/0.39	  f4(I7, I8) -> f4(I9, I10) [-1 <= I10 - 1 /\ -1 <= I9 - 1 /\ 2 <= I8 - 1 /\ 0 <= I7 - 1 /\ I10 + 3 <= I8 /\ I9 + 1 <= I7]
0.00/0.39	  f2(I11, I12) -> f4(I13, I14) [-1 <= I14 - 1 /\ 0 <= I13 - 1 /\ 0 <= I11 - 1 /\ I14 + 1 <= I11 /\ I13 <= I11]
0.00/0.39	  f3(I15, I16) -> f2(I17, I18) [-1 <= I17 - 1 /\ 0 <= I15 - 1]
0.00/0.39	  f1(I19, I20) -> f2(I21, I22) [-1 <= I21 - 1 /\ -1 <= I19 - 1 /\ I21 <= I19]
0.00/0.39	
0.00/0.39	We use the basic value criterion with the projection function NU:
0.00/0.39	  NU[f5#(z1,z2)] = z1
0.00/0.39	
0.00/0.39	This gives the following inequalities:
0.00/0.39	  0 <= I0 - 1  ==>  I0 >! I0 - 1
0.00/0.39	
0.00/0.39	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.37	EOF