4.86/4.83	YES
4.86/4.83	
4.86/4.83	DP problem for innermost termination.
4.86/4.83	P =
4.86/4.83	  init#(x1, x2, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) 
4.86/4.83	  f3#(I0, I1, I2, I3) -> f3#(I0, I1 - 1, I2 - 1, I2 - 1) [I2 = I3 /\ 0 <= I2 - 1]
4.86/4.83	  f3#(I4, I5, I6, I7) -> f3#(I4, I5 + 1, I6 + 1, I6 + 1) [I6 = I7 /\ I6 <= 0 /\ 0 <= I6 - 1]
4.86/4.83	  f3#(I8, I9, I10, I11) -> f3#(I8, I9 + 1, I10 + 1, I10 + 1) [I10 = I11 /\ I10 <= 0 /\ I10 <= -1]
4.86/4.83	  f3#(I12, I13, I14, I15) -> f2#(I13, I12, I16, I17) [0 = I15 /\ 0 = I14]
4.86/4.83	  f2#(I18, I19, I20, I21) -> f3#(I19, I18, I19, I19) [0 <= I19 - 1 /\ I19 <= I18]
4.86/4.83	  f1#(I22, I23, I24, I25) -> f2#(I26, I27, I28, I29) [0 <= I22 - 1 /\ -1 <= I26 - 1 /\ 1 <= I23 - 1 /\ -1 <= I27 - 1]
4.86/4.83	R = 
4.86/4.83	  init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 
4.86/4.83	  f3(I0, I1, I2, I3) -> f3(I0, I1 - 1, I2 - 1, I2 - 1) [I2 = I3 /\ 0 <= I2 - 1]
4.86/4.83	  f3(I4, I5, I6, I7) -> f3(I4, I5 + 1, I6 + 1, I6 + 1) [I6 = I7 /\ I6 <= 0 /\ 0 <= I6 - 1]
4.86/4.83	  f3(I8, I9, I10, I11) -> f3(I8, I9 + 1, I10 + 1, I10 + 1) [I10 = I11 /\ I10 <= 0 /\ I10 <= -1]
4.86/4.83	  f3(I12, I13, I14, I15) -> f2(I13, I12, I16, I17) [0 = I15 /\ 0 = I14]
4.86/4.83	  f2(I18, I19, I20, I21) -> f3(I19, I18, I19, I19) [0 <= I19 - 1 /\ I19 <= I18]
4.86/4.83	  f1(I22, I23, I24, I25) -> f2(I26, I27, I28, I29) [0 <= I22 - 1 /\ -1 <= I26 - 1 /\ 1 <= I23 - 1 /\ -1 <= I27 - 1]
4.86/4.83	
4.86/4.83	The dependency graph for this problem is:
4.86/4.83	  0 -> 6
4.86/4.83	  1 -> 1, 4
4.86/4.83	  2 -> 
4.86/4.83	  3 -> 3, 4
4.86/4.83	  4 -> 5
4.86/4.83	  5 -> 1
4.86/4.83	  6 -> 5
4.86/4.83	Where:
4.86/4.83	  0) init#(x1, x2, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) 
4.86/4.83	  1) f3#(I0, I1, I2, I3) -> f3#(I0, I1 - 1, I2 - 1, I2 - 1) [I2 = I3 /\ 0 <= I2 - 1]
4.86/4.83	  2) f3#(I4, I5, I6, I7) -> f3#(I4, I5 + 1, I6 + 1, I6 + 1) [I6 = I7 /\ I6 <= 0 /\ 0 <= I6 - 1]
4.86/4.83	  3) f3#(I8, I9, I10, I11) -> f3#(I8, I9 + 1, I10 + 1, I10 + 1) [I10 = I11 /\ I10 <= 0 /\ I10 <= -1]
4.86/4.83	  4) f3#(I12, I13, I14, I15) -> f2#(I13, I12, I16, I17) [0 = I15 /\ 0 = I14]
4.86/4.83	  5) f2#(I18, I19, I20, I21) -> f3#(I19, I18, I19, I19) [0 <= I19 - 1 /\ I19 <= I18]
4.86/4.83	  6) f1#(I22, I23, I24, I25) -> f2#(I26, I27, I28, I29) [0 <= I22 - 1 /\ -1 <= I26 - 1 /\ 1 <= I23 - 1 /\ -1 <= I27 - 1]
4.86/4.83	
4.86/4.83	We have the following SCCs.
4.86/4.83	  { 3 }
4.86/4.83	  { 1, 4, 5 }
4.86/4.83	
4.86/4.83	DP problem for innermost termination.
4.86/4.83	P =
4.86/4.83	  f3#(I0, I1, I2, I3) -> f3#(I0, I1 - 1, I2 - 1, I2 - 1) [I2 = I3 /\ 0 <= I2 - 1]
4.86/4.83	  f3#(I12, I13, I14, I15) -> f2#(I13, I12, I16, I17) [0 = I15 /\ 0 = I14]
4.86/4.83	  f2#(I18, I19, I20, I21) -> f3#(I19, I18, I19, I19) [0 <= I19 - 1 /\ I19 <= I18]
4.86/4.83	R = 
4.86/4.83	  init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 
4.86/4.83	  f3(I0, I1, I2, I3) -> f3(I0, I1 - 1, I2 - 1, I2 - 1) [I2 = I3 /\ 0 <= I2 - 1]
4.86/4.83	  f3(I4, I5, I6, I7) -> f3(I4, I5 + 1, I6 + 1, I6 + 1) [I6 = I7 /\ I6 <= 0 /\ 0 <= I6 - 1]
4.86/4.83	  f3(I8, I9, I10, I11) -> f3(I8, I9 + 1, I10 + 1, I10 + 1) [I10 = I11 /\ I10 <= 0 /\ I10 <= -1]
4.86/4.83	  f3(I12, I13, I14, I15) -> f2(I13, I12, I16, I17) [0 = I15 /\ 0 = I14]
4.86/4.83	  f2(I18, I19, I20, I21) -> f3(I19, I18, I19, I19) [0 <= I19 - 1 /\ I19 <= I18]
4.86/4.83	  f1(I22, I23, I24, I25) -> f2(I26, I27, I28, I29) [0 <= I22 - 1 /\ -1 <= I26 - 1 /\ 1 <= I23 - 1 /\ -1 <= I27 - 1]
4.86/4.83	
4.86/4.83	We use the extended value criterion with the projection function NU:
4.86/4.83	  NU[f2#(x0,x1,x2,x3)] = x0 - x1
4.86/4.83	  NU[f3#(x0,x1,x2,x3)] = -x0 + x1 - x2
4.86/4.83	
4.86/4.83	This gives the following inequalities:
4.86/4.83	  I2 = I3 /\ 0 <= I2 - 1  ==>  -I0 + I1 - I2 >= -I0 + (I1 - 1) - (I2 - 1)
4.86/4.83	  0 = I15 /\ 0 = I14  ==>  -I12 + I13 - I14 >= I13 - I12
4.86/4.83	  0 <= I19 - 1 /\ I19 <= I18  ==>  I18 - I19 > -I19 + I18 - I19  with  I18 - I19 >= 0
4.86/4.83	
4.86/4.83	We remove all the strictly oriented dependency pairs.
4.86/4.83	
4.86/4.83	DP problem for innermost termination.
4.86/4.83	P =
4.86/4.83	  f3#(I0, I1, I2, I3) -> f3#(I0, I1 - 1, I2 - 1, I2 - 1) [I2 = I3 /\ 0 <= I2 - 1]
4.86/4.83	  f3#(I12, I13, I14, I15) -> f2#(I13, I12, I16, I17) [0 = I15 /\ 0 = I14]
4.86/4.83	R = 
4.86/4.83	  init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 
4.86/4.83	  f3(I0, I1, I2, I3) -> f3(I0, I1 - 1, I2 - 1, I2 - 1) [I2 = I3 /\ 0 <= I2 - 1]
4.86/4.83	  f3(I4, I5, I6, I7) -> f3(I4, I5 + 1, I6 + 1, I6 + 1) [I6 = I7 /\ I6 <= 0 /\ 0 <= I6 - 1]
4.86/4.83	  f3(I8, I9, I10, I11) -> f3(I8, I9 + 1, I10 + 1, I10 + 1) [I10 = I11 /\ I10 <= 0 /\ I10 <= -1]
4.86/4.83	  f3(I12, I13, I14, I15) -> f2(I13, I12, I16, I17) [0 = I15 /\ 0 = I14]
4.86/4.83	  f2(I18, I19, I20, I21) -> f3(I19, I18, I19, I19) [0 <= I19 - 1 /\ I19 <= I18]
4.86/4.83	  f1(I22, I23, I24, I25) -> f2(I26, I27, I28, I29) [0 <= I22 - 1 /\ -1 <= I26 - 1 /\ 1 <= I23 - 1 /\ -1 <= I27 - 1]
4.86/4.83	
4.86/4.83	The dependency graph for this problem is:
4.86/4.83	  1 -> 1, 4
4.86/4.83	  4 -> 
4.86/4.83	Where:
4.86/4.83	  1) f3#(I0, I1, I2, I3) -> f3#(I0, I1 - 1, I2 - 1, I2 - 1) [I2 = I3 /\ 0 <= I2 - 1]
4.86/4.83	  4) f3#(I12, I13, I14, I15) -> f2#(I13, I12, I16, I17) [0 = I15 /\ 0 = I14]
4.86/4.83	
4.86/4.83	We have the following SCCs.
4.86/4.83	  { 1 }
4.86/4.83	
4.86/4.83	DP problem for innermost termination.
4.86/4.83	P =
4.86/4.83	  f3#(I0, I1, I2, I3) -> f3#(I0, I1 - 1, I2 - 1, I2 - 1) [I2 = I3 /\ 0 <= I2 - 1]
4.86/4.83	R = 
4.86/4.83	  init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 
4.86/4.83	  f3(I0, I1, I2, I3) -> f3(I0, I1 - 1, I2 - 1, I2 - 1) [I2 = I3 /\ 0 <= I2 - 1]
4.86/4.83	  f3(I4, I5, I6, I7) -> f3(I4, I5 + 1, I6 + 1, I6 + 1) [I6 = I7 /\ I6 <= 0 /\ 0 <= I6 - 1]
4.86/4.83	  f3(I8, I9, I10, I11) -> f3(I8, I9 + 1, I10 + 1, I10 + 1) [I10 = I11 /\ I10 <= 0 /\ I10 <= -1]
4.86/4.83	  f3(I12, I13, I14, I15) -> f2(I13, I12, I16, I17) [0 = I15 /\ 0 = I14]
4.86/4.83	  f2(I18, I19, I20, I21) -> f3(I19, I18, I19, I19) [0 <= I19 - 1 /\ I19 <= I18]
4.86/4.83	  f1(I22, I23, I24, I25) -> f2(I26, I27, I28, I29) [0 <= I22 - 1 /\ -1 <= I26 - 1 /\ 1 <= I23 - 1 /\ -1 <= I27 - 1]
4.86/4.83	
4.86/4.83	We use the basic value criterion with the projection function NU:
4.86/4.83	  NU[f3#(z1,z2,z3,z4)] = z4
4.86/4.83	
4.86/4.83	This gives the following inequalities:
4.86/4.83	  I2 = I3 /\ 0 <= I2 - 1  ==>  I3 >! I2 - 1
4.86/4.83	
4.86/4.83	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
4.86/4.83	
4.86/4.83	DP problem for innermost termination.
4.86/4.83	P =
4.86/4.83	  f3#(I8, I9, I10, I11) -> f3#(I8, I9 + 1, I10 + 1, I10 + 1) [I10 = I11 /\ I10 <= 0 /\ I10 <= -1]
4.86/4.83	R = 
4.86/4.83	  init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 
4.86/4.83	  f3(I0, I1, I2, I3) -> f3(I0, I1 - 1, I2 - 1, I2 - 1) [I2 = I3 /\ 0 <= I2 - 1]
4.86/4.83	  f3(I4, I5, I6, I7) -> f3(I4, I5 + 1, I6 + 1, I6 + 1) [I6 = I7 /\ I6 <= 0 /\ 0 <= I6 - 1]
4.86/4.83	  f3(I8, I9, I10, I11) -> f3(I8, I9 + 1, I10 + 1, I10 + 1) [I10 = I11 /\ I10 <= 0 /\ I10 <= -1]
4.86/4.83	  f3(I12, I13, I14, I15) -> f2(I13, I12, I16, I17) [0 = I15 /\ 0 = I14]
4.86/4.83	  f2(I18, I19, I20, I21) -> f3(I19, I18, I19, I19) [0 <= I19 - 1 /\ I19 <= I18]
4.86/4.83	  f1(I22, I23, I24, I25) -> f2(I26, I27, I28, I29) [0 <= I22 - 1 /\ -1 <= I26 - 1 /\ 1 <= I23 - 1 /\ -1 <= I27 - 1]
4.86/4.83	
4.86/4.83	We use the reverse value criterion with the projection function NU:
4.86/4.83	  NU[f3#(z1,z2,z3,z4)] = 0 + -1 * z3
4.86/4.83	
4.86/4.83	This gives the following inequalities:
4.86/4.83	  I10 = I11 /\ I10 <= 0 /\ I10 <= -1  ==>  0 + -1 * I10 > 0 + -1 * (I10 + 1)  with  0 + -1 * I10 >= 0
4.86/4.83	
4.86/4.83	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
4.86/7.81	EOF