0.00/0.35	YES
0.00/0.35	
0.00/0.35	DP problem for innermost termination.
0.00/0.35	P =
0.00/0.35	  init#(x1, x2) -> f3#(rnd1, rnd2) 
0.00/0.35	  f5#(I0, I1) -> f5#(I0 - 1, I2) [0 <= I0 - 1]
0.00/0.35	  f2#(I3, I4) -> f5#(I5, I6) [1 <= I3 - 1 /\ 0 <= I5 - 1]
0.00/0.35	  f4#(I7, I8) -> f4#(I7 - 1, I9) [0 <= I7 - 1]
0.00/0.35	  f3#(I10, I11) -> f4#(I12, I13) [0 <= I10 - 1 /\ -1 <= I12 - 1 /\ -1 <= I11 - 1]
0.00/0.35	  f3#(I14, I15) -> f2#(I15, I16) [0 <= I14 - 1]
0.00/0.35	  f1#(I17, I18) -> f2#(I19, I20) [0 <= I17 - 1]
0.00/0.35	R = 
0.00/0.35	  init(x1, x2) -> f3(rnd1, rnd2) 
0.00/0.35	  f5(I0, I1) -> f5(I0 - 1, I2) [0 <= I0 - 1]
0.00/0.35	  f2(I3, I4) -> f5(I5, I6) [1 <= I3 - 1 /\ 0 <= I5 - 1]
0.00/0.35	  f4(I7, I8) -> f4(I7 - 1, I9) [0 <= I7 - 1]
0.00/0.35	  f3(I10, I11) -> f4(I12, I13) [0 <= I10 - 1 /\ -1 <= I12 - 1 /\ -1 <= I11 - 1]
0.00/0.35	  f3(I14, I15) -> f2(I15, I16) [0 <= I14 - 1]
0.00/0.35	  f1(I17, I18) -> f2(I19, I20) [0 <= I17 - 1]
0.00/0.35	
0.00/0.35	The dependency graph for this problem is:
0.00/0.35	  0 -> 4, 5
0.00/0.35	  1 -> 1
0.00/0.35	  2 -> 1
0.00/0.35	  3 -> 3
0.00/0.35	  4 -> 3
0.00/0.35	  5 -> 2
0.00/0.35	  6 -> 2
0.00/0.35	Where:
0.00/0.35	  0) init#(x1, x2) -> f3#(rnd1, rnd2) 
0.00/0.35	  1) f5#(I0, I1) -> f5#(I0 - 1, I2) [0 <= I0 - 1]
0.00/0.35	  2) f2#(I3, I4) -> f5#(I5, I6) [1 <= I3 - 1 /\ 0 <= I5 - 1]
0.00/0.35	  3) f4#(I7, I8) -> f4#(I7 - 1, I9) [0 <= I7 - 1]
0.00/0.35	  4) f3#(I10, I11) -> f4#(I12, I13) [0 <= I10 - 1 /\ -1 <= I12 - 1 /\ -1 <= I11 - 1]
0.00/0.35	  5) f3#(I14, I15) -> f2#(I15, I16) [0 <= I14 - 1]
0.00/0.35	  6) f1#(I17, I18) -> f2#(I19, I20) [0 <= I17 - 1]
0.00/0.35	
0.00/0.35	We have the following SCCs.
0.00/0.35	  { 3 }
0.00/0.35	  { 1 }
0.00/0.35	
0.00/0.35	DP problem for innermost termination.
0.00/0.35	P =
0.00/0.35	  f5#(I0, I1) -> f5#(I0 - 1, I2) [0 <= I0 - 1]
0.00/0.35	R = 
0.00/0.35	  init(x1, x2) -> f3(rnd1, rnd2) 
0.00/0.35	  f5(I0, I1) -> f5(I0 - 1, I2) [0 <= I0 - 1]
0.00/0.35	  f2(I3, I4) -> f5(I5, I6) [1 <= I3 - 1 /\ 0 <= I5 - 1]
0.00/0.35	  f4(I7, I8) -> f4(I7 - 1, I9) [0 <= I7 - 1]
0.00/0.35	  f3(I10, I11) -> f4(I12, I13) [0 <= I10 - 1 /\ -1 <= I12 - 1 /\ -1 <= I11 - 1]
0.00/0.35	  f3(I14, I15) -> f2(I15, I16) [0 <= I14 - 1]
0.00/0.35	  f1(I17, I18) -> f2(I19, I20) [0 <= I17 - 1]
0.00/0.35	
0.00/0.35	We use the basic value criterion with the projection function NU:
0.00/0.35	  NU[f5#(z1,z2)] = z1
0.00/0.35	
0.00/0.35	This gives the following inequalities:
0.00/0.35	  0 <= I0 - 1  ==>  I0 >! I0 - 1
0.00/0.35	
0.00/0.35	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/0.35	
0.00/0.35	DP problem for innermost termination.
0.00/0.35	P =
0.00/0.35	  f4#(I7, I8) -> f4#(I7 - 1, I9) [0 <= I7 - 1]
0.00/0.35	R = 
0.00/0.35	  init(x1, x2) -> f3(rnd1, rnd2) 
0.00/0.35	  f5(I0, I1) -> f5(I0 - 1, I2) [0 <= I0 - 1]
0.00/0.35	  f2(I3, I4) -> f5(I5, I6) [1 <= I3 - 1 /\ 0 <= I5 - 1]
0.00/0.35	  f4(I7, I8) -> f4(I7 - 1, I9) [0 <= I7 - 1]
0.00/0.35	  f3(I10, I11) -> f4(I12, I13) [0 <= I10 - 1 /\ -1 <= I12 - 1 /\ -1 <= I11 - 1]
0.00/0.35	  f3(I14, I15) -> f2(I15, I16) [0 <= I14 - 1]
0.00/0.35	  f1(I17, I18) -> f2(I19, I20) [0 <= I17 - 1]
0.00/0.35	
0.00/0.35	We use the basic value criterion with the projection function NU:
0.00/0.35	  NU[f4#(z1,z2)] = z1
0.00/0.35	
0.00/0.35	This gives the following inequalities:
0.00/0.35	  0 <= I7 - 1  ==>  I7 >! I7 - 1
0.00/0.35	
0.00/0.35	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.34	EOF