0.00/0.36	YES
0.00/0.36	
0.00/0.36	DP problem for innermost termination.
0.00/0.36	P =
0.00/0.36	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.36	  f4#(I0, I1) -> f4#(I0, I1 - 1) [0 <= I1 - 1]
0.00/0.36	  f4#(I2, I3) -> f2#(I2, 0) [0 = I3]
0.00/0.36	  f3#(I4, I5) -> f3#(I4, I5 - 1) [0 <= I5 - 1]
0.00/0.36	  f3#(I6, I7) -> f2#(0, I6) [0 = I7]
0.00/0.36	  f2#(I8, I9) -> f4#(I8, I9) [0 <= I8 - 1 /\ 0 <= I9 - 1 /\ I8 <= I9 - 1]
0.00/0.36	  f2#(I10, I11) -> f3#(I11, I10) [0 <= I10 - 1 /\ 0 <= I11 - 1 /\ I11 <= I10]
0.00/0.36	  f1#(I12, I13) -> f2#(I14, I15) [0 <= I12 - 1 /\ -1 <= I15 - 1 /\ -1 <= I13 - 1 /\ -1 <= I14 - 1]
0.00/0.36	R = 
0.00/0.36	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.36	  f4(I0, I1) -> f4(I0, I1 - 1) [0 <= I1 - 1]
0.00/0.36	  f4(I2, I3) -> f2(I2, 0) [0 = I3]
0.00/0.36	  f3(I4, I5) -> f3(I4, I5 - 1) [0 <= I5 - 1]
0.00/0.36	  f3(I6, I7) -> f2(0, I6) [0 = I7]
0.00/0.36	  f2(I8, I9) -> f4(I8, I9) [0 <= I8 - 1 /\ 0 <= I9 - 1 /\ I8 <= I9 - 1]
0.00/0.36	  f2(I10, I11) -> f3(I11, I10) [0 <= I10 - 1 /\ 0 <= I11 - 1 /\ I11 <= I10]
0.00/0.36	  f1(I12, I13) -> f2(I14, I15) [0 <= I12 - 1 /\ -1 <= I15 - 1 /\ -1 <= I13 - 1 /\ -1 <= I14 - 1]
0.00/0.36	
0.00/0.36	The dependency graph for this problem is:
0.00/0.36	  0 -> 7
0.00/0.36	  1 -> 1, 2
0.00/0.36	  2 -> 
0.00/0.36	  3 -> 3, 4
0.00/0.36	  4 -> 
0.00/0.36	  5 -> 1
0.00/0.36	  6 -> 3
0.00/0.36	  7 -> 5, 6
0.00/0.36	Where:
0.00/0.36	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.36	  1) f4#(I0, I1) -> f4#(I0, I1 - 1) [0 <= I1 - 1]
0.00/0.36	  2) f4#(I2, I3) -> f2#(I2, 0) [0 = I3]
0.00/0.36	  3) f3#(I4, I5) -> f3#(I4, I5 - 1) [0 <= I5 - 1]
0.00/0.36	  4) f3#(I6, I7) -> f2#(0, I6) [0 = I7]
0.00/0.36	  5) f2#(I8, I9) -> f4#(I8, I9) [0 <= I8 - 1 /\ 0 <= I9 - 1 /\ I8 <= I9 - 1]
0.00/0.36	  6) f2#(I10, I11) -> f3#(I11, I10) [0 <= I10 - 1 /\ 0 <= I11 - 1 /\ I11 <= I10]
0.00/0.36	  7) f1#(I12, I13) -> f2#(I14, I15) [0 <= I12 - 1 /\ -1 <= I15 - 1 /\ -1 <= I13 - 1 /\ -1 <= I14 - 1]
0.00/0.36	
0.00/0.36	We have the following SCCs.
0.00/0.36	  { 3 }
0.00/0.36	  { 1 }
0.00/0.36	
0.00/0.36	DP problem for innermost termination.
0.00/0.36	P =
0.00/0.36	  f4#(I0, I1) -> f4#(I0, I1 - 1) [0 <= I1 - 1]
0.00/0.36	R = 
0.00/0.36	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.36	  f4(I0, I1) -> f4(I0, I1 - 1) [0 <= I1 - 1]
0.00/0.36	  f4(I2, I3) -> f2(I2, 0) [0 = I3]
0.00/0.36	  f3(I4, I5) -> f3(I4, I5 - 1) [0 <= I5 - 1]
0.00/0.36	  f3(I6, I7) -> f2(0, I6) [0 = I7]
0.00/0.36	  f2(I8, I9) -> f4(I8, I9) [0 <= I8 - 1 /\ 0 <= I9 - 1 /\ I8 <= I9 - 1]
0.00/0.36	  f2(I10, I11) -> f3(I11, I10) [0 <= I10 - 1 /\ 0 <= I11 - 1 /\ I11 <= I10]
0.00/0.36	  f1(I12, I13) -> f2(I14, I15) [0 <= I12 - 1 /\ -1 <= I15 - 1 /\ -1 <= I13 - 1 /\ -1 <= I14 - 1]
0.00/0.36	
0.00/0.36	We use the basic value criterion with the projection function NU:
0.00/0.36	  NU[f4#(z1,z2)] = z2
0.00/0.36	
0.00/0.36	This gives the following inequalities:
0.00/0.36	  0 <= I1 - 1  ==>  I1 >! I1 - 1
0.00/0.36	
0.00/0.36	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/0.36	
0.00/0.36	DP problem for innermost termination.
0.00/0.36	P =
0.00/0.36	  f3#(I4, I5) -> f3#(I4, I5 - 1) [0 <= I5 - 1]
0.00/0.36	R = 
0.00/0.36	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.36	  f4(I0, I1) -> f4(I0, I1 - 1) [0 <= I1 - 1]
0.00/0.36	  f4(I2, I3) -> f2(I2, 0) [0 = I3]
0.00/0.36	  f3(I4, I5) -> f3(I4, I5 - 1) [0 <= I5 - 1]
0.00/0.36	  f3(I6, I7) -> f2(0, I6) [0 = I7]
0.00/0.36	  f2(I8, I9) -> f4(I8, I9) [0 <= I8 - 1 /\ 0 <= I9 - 1 /\ I8 <= I9 - 1]
0.00/0.36	  f2(I10, I11) -> f3(I11, I10) [0 <= I10 - 1 /\ 0 <= I11 - 1 /\ I11 <= I10]
0.00/0.36	  f1(I12, I13) -> f2(I14, I15) [0 <= I12 - 1 /\ -1 <= I15 - 1 /\ -1 <= I13 - 1 /\ -1 <= I14 - 1]
0.00/0.36	
0.00/0.36	We use the basic value criterion with the projection function NU:
0.00/0.36	  NU[f3#(z1,z2)] = z2
0.00/0.36	
0.00/0.36	This gives the following inequalities:
0.00/0.36	  0 <= I5 - 1  ==>  I5 >! I5 - 1
0.00/0.36	
0.00/0.36	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.34	EOF