7.71/7.66	YES
7.71/7.66	
7.71/7.66	DP problem for innermost termination.
7.71/7.66	P =
7.71/7.66	  init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
7.71/7.66	  f3#(I0, I1, I2) -> f3#(I0, I1 + 1, 100 - I0) [I1 <= 99 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I1 <= 98 /\ 0 <= I0 - 1 /\ y1 <= y2 - 1]
7.71/7.66	  f3#(I3, I4, I5) -> f3#(I3, I4 + 1, 100 - I3) [I4 <= 99 /\ I4 <= I5 - 1 /\ -1 <= I4 - 1 /\ I4 <= 98 /\ I6 <= I7 /\ 0 <= I3 - 1]
7.71/7.66	  f3#(I8, I9, I10) -> f2#(I8 + 1, I11, I12) [I10 <= I9]
7.71/7.66	  f2#(I13, I14, I15) -> f3#(I13, 0, 100 - I13) [I13 <= 99 /\ 0 <= I13 - 1]
7.71/7.66	  f1#(I16, I17, I18) -> f2#(1, I19, I20) 
7.71/7.66	R = 
7.71/7.66	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
7.71/7.66	  f3(I0, I1, I2) -> f3(I0, I1 + 1, 100 - I0) [I1 <= 99 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I1 <= 98 /\ 0 <= I0 - 1 /\ y1 <= y2 - 1]
7.71/7.66	  f3(I3, I4, I5) -> f3(I3, I4 + 1, 100 - I3) [I4 <= 99 /\ I4 <= I5 - 1 /\ -1 <= I4 - 1 /\ I4 <= 98 /\ I6 <= I7 /\ 0 <= I3 - 1]
7.71/7.66	  f3(I8, I9, I10) -> f2(I8 + 1, I11, I12) [I10 <= I9]
7.71/7.66	  f2(I13, I14, I15) -> f3(I13, 0, 100 - I13) [I13 <= 99 /\ 0 <= I13 - 1]
7.71/7.66	  f1(I16, I17, I18) -> f2(1, I19, I20) 
7.71/7.66	
7.71/7.66	The dependency graph for this problem is:
7.71/7.66	  0 -> 5
7.71/7.66	  1 -> 1, 2, 3
7.71/7.66	  2 -> 1, 2, 3
7.71/7.66	  3 -> 4
7.71/7.66	  4 -> 1, 2
7.71/7.66	  5 -> 4
7.71/7.66	Where:
7.71/7.66	  0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
7.71/7.66	  1) f3#(I0, I1, I2) -> f3#(I0, I1 + 1, 100 - I0) [I1 <= 99 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I1 <= 98 /\ 0 <= I0 - 1 /\ y1 <= y2 - 1]
7.71/7.66	  2) f3#(I3, I4, I5) -> f3#(I3, I4 + 1, 100 - I3) [I4 <= 99 /\ I4 <= I5 - 1 /\ -1 <= I4 - 1 /\ I4 <= 98 /\ I6 <= I7 /\ 0 <= I3 - 1]
7.71/7.66	  3) f3#(I8, I9, I10) -> f2#(I8 + 1, I11, I12) [I10 <= I9]
7.71/7.66	  4) f2#(I13, I14, I15) -> f3#(I13, 0, 100 - I13) [I13 <= 99 /\ 0 <= I13 - 1]
7.71/7.66	  5) f1#(I16, I17, I18) -> f2#(1, I19, I20) 
7.71/7.66	
7.71/7.66	We have the following SCCs.
7.71/7.66	  { 1, 2, 3, 4 }
7.71/7.66	
7.71/7.66	DP problem for innermost termination.
7.71/7.66	P =
7.71/7.66	  f3#(I0, I1, I2) -> f3#(I0, I1 + 1, 100 - I0) [I1 <= 99 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I1 <= 98 /\ 0 <= I0 - 1 /\ y1 <= y2 - 1]
7.71/7.66	  f3#(I3, I4, I5) -> f3#(I3, I4 + 1, 100 - I3) [I4 <= 99 /\ I4 <= I5 - 1 /\ -1 <= I4 - 1 /\ I4 <= 98 /\ I6 <= I7 /\ 0 <= I3 - 1]
7.71/7.66	  f3#(I8, I9, I10) -> f2#(I8 + 1, I11, I12) [I10 <= I9]
7.71/7.66	  f2#(I13, I14, I15) -> f3#(I13, 0, 100 - I13) [I13 <= 99 /\ 0 <= I13 - 1]
7.71/7.66	R = 
7.71/7.66	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
7.71/7.66	  f3(I0, I1, I2) -> f3(I0, I1 + 1, 100 - I0) [I1 <= 99 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I1 <= 98 /\ 0 <= I0 - 1 /\ y1 <= y2 - 1]
7.71/7.66	  f3(I3, I4, I5) -> f3(I3, I4 + 1, 100 - I3) [I4 <= 99 /\ I4 <= I5 - 1 /\ -1 <= I4 - 1 /\ I4 <= 98 /\ I6 <= I7 /\ 0 <= I3 - 1]
7.71/7.66	  f3(I8, I9, I10) -> f2(I8 + 1, I11, I12) [I10 <= I9]
7.71/7.66	  f2(I13, I14, I15) -> f3(I13, 0, 100 - I13) [I13 <= 99 /\ 0 <= I13 - 1]
7.71/7.66	  f1(I16, I17, I18) -> f2(1, I19, I20) 
7.71/7.66	
7.71/7.66	We use the reverse value criterion with the projection function NU:
7.71/7.66	  NU[f2#(z1,z2,z3)] = 99 + -1 * z1
7.71/7.66	  NU[f3#(z1,z2,z3)] = 99 + -1 * (z1 + 1)
7.71/7.66	
7.71/7.66	This gives the following inequalities:
7.71/7.66	  I1 <= 99 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I1 <= 98 /\ 0 <= I0 - 1 /\ y1 <= y2 - 1  ==>  99 + -1 * (I0 + 1) >= 99 + -1 * (I0 + 1)
7.71/7.66	  I4 <= 99 /\ I4 <= I5 - 1 /\ -1 <= I4 - 1 /\ I4 <= 98 /\ I6 <= I7 /\ 0 <= I3 - 1  ==>  99 + -1 * (I3 + 1) >= 99 + -1 * (I3 + 1)
7.71/7.66	  I10 <= I9  ==>  99 + -1 * (I8 + 1) >= 99 + -1 * (I8 + 1)
7.71/7.66	  I13 <= 99 /\ 0 <= I13 - 1  ==>  99 + -1 * I13 > 99 + -1 * (I13 + 1)  with  99 + -1 * I13 >= 0
7.71/7.66	
7.71/7.66	We remove all the strictly oriented dependency pairs.
7.71/7.66	
7.71/7.66	DP problem for innermost termination.
7.71/7.66	P =
7.71/7.66	  f3#(I0, I1, I2) -> f3#(I0, I1 + 1, 100 - I0) [I1 <= 99 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I1 <= 98 /\ 0 <= I0 - 1 /\ y1 <= y2 - 1]
7.71/7.66	  f3#(I3, I4, I5) -> f3#(I3, I4 + 1, 100 - I3) [I4 <= 99 /\ I4 <= I5 - 1 /\ -1 <= I4 - 1 /\ I4 <= 98 /\ I6 <= I7 /\ 0 <= I3 - 1]
7.71/7.66	  f3#(I8, I9, I10) -> f2#(I8 + 1, I11, I12) [I10 <= I9]
7.71/7.66	R = 
7.71/7.66	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
7.71/7.66	  f3(I0, I1, I2) -> f3(I0, I1 + 1, 100 - I0) [I1 <= 99 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I1 <= 98 /\ 0 <= I0 - 1 /\ y1 <= y2 - 1]
7.71/7.66	  f3(I3, I4, I5) -> f3(I3, I4 + 1, 100 - I3) [I4 <= 99 /\ I4 <= I5 - 1 /\ -1 <= I4 - 1 /\ I4 <= 98 /\ I6 <= I7 /\ 0 <= I3 - 1]
7.71/7.66	  f3(I8, I9, I10) -> f2(I8 + 1, I11, I12) [I10 <= I9]
7.71/7.66	  f2(I13, I14, I15) -> f3(I13, 0, 100 - I13) [I13 <= 99 /\ 0 <= I13 - 1]
7.71/7.66	  f1(I16, I17, I18) -> f2(1, I19, I20) 
7.71/7.66	
7.71/7.66	The dependency graph for this problem is:
7.71/7.66	  1 -> 1, 2, 3
7.71/7.66	  2 -> 1, 2, 3
7.71/7.66	  3 -> 
7.71/7.66	Where:
7.71/7.66	  1) f3#(I0, I1, I2) -> f3#(I0, I1 + 1, 100 - I0) [I1 <= 99 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I1 <= 98 /\ 0 <= I0 - 1 /\ y1 <= y2 - 1]
7.71/7.66	  2) f3#(I3, I4, I5) -> f3#(I3, I4 + 1, 100 - I3) [I4 <= 99 /\ I4 <= I5 - 1 /\ -1 <= I4 - 1 /\ I4 <= 98 /\ I6 <= I7 /\ 0 <= I3 - 1]
7.71/7.66	  3) f3#(I8, I9, I10) -> f2#(I8 + 1, I11, I12) [I10 <= I9]
7.71/7.66	
7.71/7.66	We have the following SCCs.
7.71/7.66	  { 1, 2 }
7.71/7.66	
7.71/7.66	DP problem for innermost termination.
7.71/7.66	P =
7.71/7.66	  f3#(I0, I1, I2) -> f3#(I0, I1 + 1, 100 - I0) [I1 <= 99 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I1 <= 98 /\ 0 <= I0 - 1 /\ y1 <= y2 - 1]
7.71/7.66	  f3#(I3, I4, I5) -> f3#(I3, I4 + 1, 100 - I3) [I4 <= 99 /\ I4 <= I5 - 1 /\ -1 <= I4 - 1 /\ I4 <= 98 /\ I6 <= I7 /\ 0 <= I3 - 1]
7.71/7.66	R = 
7.71/7.66	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
7.71/7.66	  f3(I0, I1, I2) -> f3(I0, I1 + 1, 100 - I0) [I1 <= 99 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I1 <= 98 /\ 0 <= I0 - 1 /\ y1 <= y2 - 1]
7.71/7.66	  f3(I3, I4, I5) -> f3(I3, I4 + 1, 100 - I3) [I4 <= 99 /\ I4 <= I5 - 1 /\ -1 <= I4 - 1 /\ I4 <= 98 /\ I6 <= I7 /\ 0 <= I3 - 1]
7.71/7.66	  f3(I8, I9, I10) -> f2(I8 + 1, I11, I12) [I10 <= I9]
7.71/7.66	  f2(I13, I14, I15) -> f3(I13, 0, 100 - I13) [I13 <= 99 /\ 0 <= I13 - 1]
7.71/7.66	  f1(I16, I17, I18) -> f2(1, I19, I20) 
7.71/7.66	
7.71/7.66	We use the reverse value criterion with the projection function NU:
7.71/7.66	  NU[f3#(z1,z2,z3)] = 99 + -1 * z2
7.71/7.66	
7.71/7.66	This gives the following inequalities:
7.71/7.66	  I1 <= 99 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I1 <= 98 /\ 0 <= I0 - 1 /\ y1 <= y2 - 1  ==>  99 + -1 * I1 > 99 + -1 * (I1 + 1)  with  99 + -1 * I1 >= 0
7.71/7.66	  I4 <= 99 /\ I4 <= I5 - 1 /\ -1 <= I4 - 1 /\ I4 <= 98 /\ I6 <= I7 /\ 0 <= I3 - 1  ==>  99 + -1 * I4 > 99 + -1 * (I4 + 1)  with  99 + -1 * I4 >= 0
7.71/7.66	
7.71/7.66	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
7.71/10.64	EOF