1.68/1.69	YES
1.68/1.69	
1.68/1.69	DP problem for innermost termination.
1.68/1.69	P =
1.68/1.69	  init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
1.68/1.69	  f4#(I0, I1, I2) -> f4#(I0 - 1, I3, I4) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1]
1.68/1.69	  f3#(I5, I6, I7) -> f3#(I5, I6 + 1, I6 + 1) [I6 = I7 /\ 0 <= I6 - 1 /\ I6 <= 99]
1.68/1.69	  f3#(I8, I9, I10) -> f4#(I9, I11, I12) [I9 = I10 /\ 0 <= I9 - 1 /\ I9 <= 99]
1.68/1.69	  f3#(I13, I14, I15) -> f4#(I14, I16, I17) [I14 = I15 /\ I14 <= 99 /\ 0 <= I14 - 1]
1.68/1.69	  f3#(I18, I19, I20) -> f2#(I18 - 1, I21, I22) [I19 = I20 /\ 99 <= I19 - 1 /\ 0 <= I18 - 1 /\ I18 - 1 <= I18 - 1]
1.68/1.69	  f2#(I23, I24, I25) -> f3#(I23, I23, I23) 
1.68/1.69	  f1#(I26, I27, I28) -> f2#(I27, I29, I30) [-1 <= I27 - 1 /\ 0 <= I26 - 1]
1.68/1.69	R = 
1.68/1.69	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
1.68/1.69	  f4(I0, I1, I2) -> f4(I0 - 1, I3, I4) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1]
1.68/1.69	  f3(I5, I6, I7) -> f3(I5, I6 + 1, I6 + 1) [I6 = I7 /\ 0 <= I6 - 1 /\ I6 <= 99]
1.68/1.69	  f3(I8, I9, I10) -> f4(I9, I11, I12) [I9 = I10 /\ 0 <= I9 - 1 /\ I9 <= 99]
1.68/1.69	  f3(I13, I14, I15) -> f4(I14, I16, I17) [I14 = I15 /\ I14 <= 99 /\ 0 <= I14 - 1]
1.68/1.69	  f3(I18, I19, I20) -> f2(I18 - 1, I21, I22) [I19 = I20 /\ 99 <= I19 - 1 /\ 0 <= I18 - 1 /\ I18 - 1 <= I18 - 1]
1.68/1.69	  f2(I23, I24, I25) -> f3(I23, I23, I23) 
1.68/1.69	  f1(I26, I27, I28) -> f2(I27, I29, I30) [-1 <= I27 - 1 /\ 0 <= I26 - 1]
1.68/1.69	
1.68/1.69	The dependency graph for this problem is:
1.68/1.69	  0 -> 7
1.68/1.69	  1 -> 1
1.68/1.69	  2 -> 2, 3, 4, 5
1.68/1.69	  3 -> 1
1.68/1.69	  4 -> 1
1.68/1.69	  5 -> 6
1.68/1.69	  6 -> 2, 3, 4, 5
1.68/1.69	  7 -> 6
1.68/1.69	Where:
1.68/1.69	  0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
1.68/1.69	  1) f4#(I0, I1, I2) -> f4#(I0 - 1, I3, I4) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1]
1.68/1.69	  2) f3#(I5, I6, I7) -> f3#(I5, I6 + 1, I6 + 1) [I6 = I7 /\ 0 <= I6 - 1 /\ I6 <= 99]
1.68/1.69	  3) f3#(I8, I9, I10) -> f4#(I9, I11, I12) [I9 = I10 /\ 0 <= I9 - 1 /\ I9 <= 99]
1.68/1.69	  4) f3#(I13, I14, I15) -> f4#(I14, I16, I17) [I14 = I15 /\ I14 <= 99 /\ 0 <= I14 - 1]
1.68/1.69	  5) f3#(I18, I19, I20) -> f2#(I18 - 1, I21, I22) [I19 = I20 /\ 99 <= I19 - 1 /\ 0 <= I18 - 1 /\ I18 - 1 <= I18 - 1]
1.68/1.69	  6) f2#(I23, I24, I25) -> f3#(I23, I23, I23) 
1.68/1.69	  7) f1#(I26, I27, I28) -> f2#(I27, I29, I30) [-1 <= I27 - 1 /\ 0 <= I26 - 1]
1.68/1.69	
1.68/1.69	We have the following SCCs.
1.68/1.69	  { 2, 5, 6 }
1.68/1.69	  { 1 }
1.68/1.69	
1.68/1.69	DP problem for innermost termination.
1.68/1.69	P =
1.68/1.69	  f4#(I0, I1, I2) -> f4#(I0 - 1, I3, I4) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1]
1.68/1.69	R = 
1.68/1.69	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
1.68/1.69	  f4(I0, I1, I2) -> f4(I0 - 1, I3, I4) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1]
1.68/1.69	  f3(I5, I6, I7) -> f3(I5, I6 + 1, I6 + 1) [I6 = I7 /\ 0 <= I6 - 1 /\ I6 <= 99]
1.68/1.69	  f3(I8, I9, I10) -> f4(I9, I11, I12) [I9 = I10 /\ 0 <= I9 - 1 /\ I9 <= 99]
1.68/1.69	  f3(I13, I14, I15) -> f4(I14, I16, I17) [I14 = I15 /\ I14 <= 99 /\ 0 <= I14 - 1]
1.68/1.69	  f3(I18, I19, I20) -> f2(I18 - 1, I21, I22) [I19 = I20 /\ 99 <= I19 - 1 /\ 0 <= I18 - 1 /\ I18 - 1 <= I18 - 1]
1.68/1.69	  f2(I23, I24, I25) -> f3(I23, I23, I23) 
1.68/1.69	  f1(I26, I27, I28) -> f2(I27, I29, I30) [-1 <= I27 - 1 /\ 0 <= I26 - 1]
1.68/1.69	
1.68/1.69	We use the basic value criterion with the projection function NU:
1.68/1.69	  NU[f4#(z1,z2,z3)] = z1
1.68/1.69	
1.68/1.69	This gives the following inequalities:
1.68/1.69	  I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1  ==>  I0 >! I0 - 1
1.68/1.69	
1.68/1.69	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
1.68/1.69	
1.68/1.69	DP problem for innermost termination.
1.68/1.69	P =
1.68/1.69	  f3#(I5, I6, I7) -> f3#(I5, I6 + 1, I6 + 1) [I6 = I7 /\ 0 <= I6 - 1 /\ I6 <= 99]
1.68/1.69	  f3#(I18, I19, I20) -> f2#(I18 - 1, I21, I22) [I19 = I20 /\ 99 <= I19 - 1 /\ 0 <= I18 - 1 /\ I18 - 1 <= I18 - 1]
1.68/1.69	  f2#(I23, I24, I25) -> f3#(I23, I23, I23) 
1.68/1.69	R = 
1.68/1.69	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
1.68/1.69	  f4(I0, I1, I2) -> f4(I0 - 1, I3, I4) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1]
1.68/1.69	  f3(I5, I6, I7) -> f3(I5, I6 + 1, I6 + 1) [I6 = I7 /\ 0 <= I6 - 1 /\ I6 <= 99]
1.68/1.69	  f3(I8, I9, I10) -> f4(I9, I11, I12) [I9 = I10 /\ 0 <= I9 - 1 /\ I9 <= 99]
1.68/1.69	  f3(I13, I14, I15) -> f4(I14, I16, I17) [I14 = I15 /\ I14 <= 99 /\ 0 <= I14 - 1]
1.68/1.69	  f3(I18, I19, I20) -> f2(I18 - 1, I21, I22) [I19 = I20 /\ 99 <= I19 - 1 /\ 0 <= I18 - 1 /\ I18 - 1 <= I18 - 1]
1.68/1.69	  f2(I23, I24, I25) -> f3(I23, I23, I23) 
1.68/1.69	  f1(I26, I27, I28) -> f2(I27, I29, I30) [-1 <= I27 - 1 /\ 0 <= I26 - 1]
1.68/1.69	
1.68/1.69	We use the basic value criterion with the projection function NU:
1.68/1.69	  NU[f2#(z1,z2,z3)] = z1
1.68/1.69	  NU[f3#(z1,z2,z3)] = z1
1.68/1.69	
1.68/1.69	This gives the following inequalities:
1.68/1.69	  I6 = I7 /\ 0 <= I6 - 1 /\ I6 <= 99  ==>  I5 (>! \union =) I5
1.68/1.69	  I19 = I20 /\ 99 <= I19 - 1 /\ 0 <= I18 - 1 /\ I18 - 1 <= I18 - 1  ==>  I18 >! I18 - 1
1.68/1.69	    ==>  I23 (>! \union =) I23
1.68/1.69	
1.68/1.69	We remove all the strictly oriented dependency pairs.
1.68/1.69	
1.68/1.69	DP problem for innermost termination.
1.68/1.69	P =
1.68/1.69	  f3#(I5, I6, I7) -> f3#(I5, I6 + 1, I6 + 1) [I6 = I7 /\ 0 <= I6 - 1 /\ I6 <= 99]
1.68/1.69	  f2#(I23, I24, I25) -> f3#(I23, I23, I23) 
1.68/1.69	R = 
1.68/1.69	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
1.68/1.69	  f4(I0, I1, I2) -> f4(I0 - 1, I3, I4) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1]
1.68/1.69	  f3(I5, I6, I7) -> f3(I5, I6 + 1, I6 + 1) [I6 = I7 /\ 0 <= I6 - 1 /\ I6 <= 99]
1.68/1.69	  f3(I8, I9, I10) -> f4(I9, I11, I12) [I9 = I10 /\ 0 <= I9 - 1 /\ I9 <= 99]
1.68/1.69	  f3(I13, I14, I15) -> f4(I14, I16, I17) [I14 = I15 /\ I14 <= 99 /\ 0 <= I14 - 1]
1.68/1.69	  f3(I18, I19, I20) -> f2(I18 - 1, I21, I22) [I19 = I20 /\ 99 <= I19 - 1 /\ 0 <= I18 - 1 /\ I18 - 1 <= I18 - 1]
1.68/1.69	  f2(I23, I24, I25) -> f3(I23, I23, I23) 
1.68/1.69	  f1(I26, I27, I28) -> f2(I27, I29, I30) [-1 <= I27 - 1 /\ 0 <= I26 - 1]
1.68/1.69	
1.68/1.69	The dependency graph for this problem is:
1.68/1.69	  2 -> 2
1.68/1.69	  6 -> 2
1.68/1.69	Where:
1.68/1.69	  2) f3#(I5, I6, I7) -> f3#(I5, I6 + 1, I6 + 1) [I6 = I7 /\ 0 <= I6 - 1 /\ I6 <= 99]
1.68/1.69	  6) f2#(I23, I24, I25) -> f3#(I23, I23, I23) 
1.68/1.69	
1.68/1.69	We have the following SCCs.
1.68/1.69	  { 2 }
1.68/1.69	
1.68/1.69	DP problem for innermost termination.
1.68/1.69	P =
1.68/1.69	  f3#(I5, I6, I7) -> f3#(I5, I6 + 1, I6 + 1) [I6 = I7 /\ 0 <= I6 - 1 /\ I6 <= 99]
1.68/1.69	R = 
1.68/1.69	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
1.68/1.69	  f4(I0, I1, I2) -> f4(I0 - 1, I3, I4) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1]
1.68/1.69	  f3(I5, I6, I7) -> f3(I5, I6 + 1, I6 + 1) [I6 = I7 /\ 0 <= I6 - 1 /\ I6 <= 99]
1.68/1.69	  f3(I8, I9, I10) -> f4(I9, I11, I12) [I9 = I10 /\ 0 <= I9 - 1 /\ I9 <= 99]
1.68/1.69	  f3(I13, I14, I15) -> f4(I14, I16, I17) [I14 = I15 /\ I14 <= 99 /\ 0 <= I14 - 1]
1.68/1.69	  f3(I18, I19, I20) -> f2(I18 - 1, I21, I22) [I19 = I20 /\ 99 <= I19 - 1 /\ 0 <= I18 - 1 /\ I18 - 1 <= I18 - 1]
1.68/1.69	  f2(I23, I24, I25) -> f3(I23, I23, I23) 
1.68/1.69	  f1(I26, I27, I28) -> f2(I27, I29, I30) [-1 <= I27 - 1 /\ 0 <= I26 - 1]
1.68/1.69	
1.68/1.69	We use the reverse value criterion with the projection function NU:
1.68/1.69	  NU[f3#(z1,z2,z3)] = 99 + -1 * z2
1.68/1.69	
1.68/1.69	This gives the following inequalities:
1.68/1.69	  I6 = I7 /\ 0 <= I6 - 1 /\ I6 <= 99  ==>  99 + -1 * I6 > 99 + -1 * (I6 + 1)  with  99 + -1 * I6 >= 0
1.68/1.69	
1.68/1.69	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
1.68/4.67	EOF