7.59/7.81	YES
7.59/7.81	
7.59/7.81	DP problem for innermost termination.
7.59/7.81	P =
7.59/7.81	  init#(x1, x2, x3, x4, x5) -> f1#(rnd1, rnd2, rnd3, rnd4, rnd5) 
7.59/7.81	  f3#(I0, I1, I2, I3, I4) -> f2#(I1 + 1, I2, I0, I5, I6) [0 = I4 /\ 0 = I3 /\ -1 <= I1 - 1]
7.59/7.81	  f3#(I7, I8, I9, I10, I11) -> f3#(I7, I8, I9 - 1, I10 - 1, I10 - 1) [I10 = I11 /\ 0 <= I10 - 1]
7.59/7.81	  f3#(I12, I13, I14, I15, I16) -> f3#(I12, I13, I14 + 1, I15 + 1, I15 + 1) [I15 = I16 /\ I15 <= 0 /\ 0 <= I15 - 1]
7.59/7.81	  f3#(I17, I18, I19, I20, I21) -> f3#(I17, I18, I19 + 1, I20 + 1, I20 + 1) [I20 = I21 /\ I20 <= 0 /\ I20 <= -1]
7.59/7.81	  f2#(I22, I23, I24, I25, I26) -> f3#(I24, I22, I23, I24, I24) [0 <= I24 - 1 /\ I24 <= I23]
7.59/7.81	  f1#(I27, I28, I29, I30, I31) -> f2#(0, I32, I33, I34, I35) [0 <= I27 - 1 /\ -1 <= I32 - 1 /\ -1 <= I28 - 1 /\ -1 <= I33 - 1]
7.59/7.81	R = 
7.59/7.81	  init(x1, x2, x3, x4, x5) -> f1(rnd1, rnd2, rnd3, rnd4, rnd5) 
7.59/7.81	  f3(I0, I1, I2, I3, I4) -> f2(I1 + 1, I2, I0, I5, I6) [0 = I4 /\ 0 = I3 /\ -1 <= I1 - 1]
7.59/7.81	  f3(I7, I8, I9, I10, I11) -> f3(I7, I8, I9 - 1, I10 - 1, I10 - 1) [I10 = I11 /\ 0 <= I10 - 1]
7.59/7.81	  f3(I12, I13, I14, I15, I16) -> f3(I12, I13, I14 + 1, I15 + 1, I15 + 1) [I15 = I16 /\ I15 <= 0 /\ 0 <= I15 - 1]
7.59/7.81	  f3(I17, I18, I19, I20, I21) -> f3(I17, I18, I19 + 1, I20 + 1, I20 + 1) [I20 = I21 /\ I20 <= 0 /\ I20 <= -1]
7.59/7.81	  f2(I22, I23, I24, I25, I26) -> f3(I24, I22, I23, I24, I24) [0 <= I24 - 1 /\ I24 <= I23]
7.59/7.81	  f1(I27, I28, I29, I30, I31) -> f2(0, I32, I33, I34, I35) [0 <= I27 - 1 /\ -1 <= I32 - 1 /\ -1 <= I28 - 1 /\ -1 <= I33 - 1]
7.59/7.81	
7.59/7.81	The dependency graph for this problem is:
7.59/7.81	  0 -> 6
7.59/7.81	  1 -> 5
7.59/7.81	  2 -> 1, 2
7.59/7.81	  3 -> 
7.59/7.81	  4 -> 1, 4
7.59/7.81	  5 -> 2
7.59/7.81	  6 -> 5
7.59/7.81	Where:
7.59/7.81	  0) init#(x1, x2, x3, x4, x5) -> f1#(rnd1, rnd2, rnd3, rnd4, rnd5) 
7.59/7.81	  1) f3#(I0, I1, I2, I3, I4) -> f2#(I1 + 1, I2, I0, I5, I6) [0 = I4 /\ 0 = I3 /\ -1 <= I1 - 1]
7.59/7.81	  2) f3#(I7, I8, I9, I10, I11) -> f3#(I7, I8, I9 - 1, I10 - 1, I10 - 1) [I10 = I11 /\ 0 <= I10 - 1]
7.59/7.81	  3) f3#(I12, I13, I14, I15, I16) -> f3#(I12, I13, I14 + 1, I15 + 1, I15 + 1) [I15 = I16 /\ I15 <= 0 /\ 0 <= I15 - 1]
7.59/7.81	  4) f3#(I17, I18, I19, I20, I21) -> f3#(I17, I18, I19 + 1, I20 + 1, I20 + 1) [I20 = I21 /\ I20 <= 0 /\ I20 <= -1]
7.59/7.81	  5) f2#(I22, I23, I24, I25, I26) -> f3#(I24, I22, I23, I24, I24) [0 <= I24 - 1 /\ I24 <= I23]
7.59/7.81	  6) f1#(I27, I28, I29, I30, I31) -> f2#(0, I32, I33, I34, I35) [0 <= I27 - 1 /\ -1 <= I32 - 1 /\ -1 <= I28 - 1 /\ -1 <= I33 - 1]
7.59/7.81	
7.59/7.81	We have the following SCCs.
7.59/7.81	  { 4 }
7.59/7.81	  { 1, 2, 5 }
7.59/7.81	
7.59/7.81	DP problem for innermost termination.
7.59/7.81	P =
7.59/7.81	  f3#(I0, I1, I2, I3, I4) -> f2#(I1 + 1, I2, I0, I5, I6) [0 = I4 /\ 0 = I3 /\ -1 <= I1 - 1]
7.59/7.81	  f3#(I7, I8, I9, I10, I11) -> f3#(I7, I8, I9 - 1, I10 - 1, I10 - 1) [I10 = I11 /\ 0 <= I10 - 1]
7.59/7.81	  f2#(I22, I23, I24, I25, I26) -> f3#(I24, I22, I23, I24, I24) [0 <= I24 - 1 /\ I24 <= I23]
7.59/7.81	R = 
7.59/7.81	  init(x1, x2, x3, x4, x5) -> f1(rnd1, rnd2, rnd3, rnd4, rnd5) 
7.59/7.81	  f3(I0, I1, I2, I3, I4) -> f2(I1 + 1, I2, I0, I5, I6) [0 = I4 /\ 0 = I3 /\ -1 <= I1 - 1]
7.59/7.81	  f3(I7, I8, I9, I10, I11) -> f3(I7, I8, I9 - 1, I10 - 1, I10 - 1) [I10 = I11 /\ 0 <= I10 - 1]
7.59/7.81	  f3(I12, I13, I14, I15, I16) -> f3(I12, I13, I14 + 1, I15 + 1, I15 + 1) [I15 = I16 /\ I15 <= 0 /\ 0 <= I15 - 1]
7.59/7.81	  f3(I17, I18, I19, I20, I21) -> f3(I17, I18, I19 + 1, I20 + 1, I20 + 1) [I20 = I21 /\ I20 <= 0 /\ I20 <= -1]
7.59/7.81	  f2(I22, I23, I24, I25, I26) -> f3(I24, I22, I23, I24, I24) [0 <= I24 - 1 /\ I24 <= I23]
7.59/7.81	  f1(I27, I28, I29, I30, I31) -> f2(0, I32, I33, I34, I35) [0 <= I27 - 1 /\ -1 <= I32 - 1 /\ -1 <= I28 - 1 /\ -1 <= I33 - 1]
7.59/7.81	
7.59/7.81	We use the extended value criterion with the projection function NU:
7.59/7.81	  NU[f2#(x0,x1,x2,x3,x4)] = x1 - x2
7.59/7.81	  NU[f3#(x0,x1,x2,x3,x4)] = -x0 + x2 - x3
7.59/7.81	
7.59/7.81	This gives the following inequalities:
7.59/7.81	  0 = I4 /\ 0 = I3 /\ -1 <= I1 - 1  ==>  -I0 + I2 - I3 >= I2 - I0
7.59/7.81	  I10 = I11 /\ 0 <= I10 - 1  ==>  -I7 + I9 - I10 >= -I7 + (I9 - 1) - (I10 - 1)
7.59/7.81	  0 <= I24 - 1 /\ I24 <= I23  ==>  I23 - I24 > -I24 + I23 - I24  with  I23 - I24 >= 0
7.59/7.81	
7.59/7.81	We remove all the strictly oriented dependency pairs.
7.59/7.81	
7.59/7.81	DP problem for innermost termination.
7.59/7.81	P =
7.59/7.81	  f3#(I0, I1, I2, I3, I4) -> f2#(I1 + 1, I2, I0, I5, I6) [0 = I4 /\ 0 = I3 /\ -1 <= I1 - 1]
7.59/7.81	  f3#(I7, I8, I9, I10, I11) -> f3#(I7, I8, I9 - 1, I10 - 1, I10 - 1) [I10 = I11 /\ 0 <= I10 - 1]
7.59/7.81	R = 
7.59/7.81	  init(x1, x2, x3, x4, x5) -> f1(rnd1, rnd2, rnd3, rnd4, rnd5) 
7.59/7.81	  f3(I0, I1, I2, I3, I4) -> f2(I1 + 1, I2, I0, I5, I6) [0 = I4 /\ 0 = I3 /\ -1 <= I1 - 1]
7.59/7.81	  f3(I7, I8, I9, I10, I11) -> f3(I7, I8, I9 - 1, I10 - 1, I10 - 1) [I10 = I11 /\ 0 <= I10 - 1]
7.59/7.81	  f3(I12, I13, I14, I15, I16) -> f3(I12, I13, I14 + 1, I15 + 1, I15 + 1) [I15 = I16 /\ I15 <= 0 /\ 0 <= I15 - 1]
7.59/7.81	  f3(I17, I18, I19, I20, I21) -> f3(I17, I18, I19 + 1, I20 + 1, I20 + 1) [I20 = I21 /\ I20 <= 0 /\ I20 <= -1]
7.59/7.81	  f2(I22, I23, I24, I25, I26) -> f3(I24, I22, I23, I24, I24) [0 <= I24 - 1 /\ I24 <= I23]
7.59/7.81	  f1(I27, I28, I29, I30, I31) -> f2(0, I32, I33, I34, I35) [0 <= I27 - 1 /\ -1 <= I32 - 1 /\ -1 <= I28 - 1 /\ -1 <= I33 - 1]
7.59/7.81	
7.59/7.81	The dependency graph for this problem is:
7.59/7.81	  1 -> 
7.59/7.81	  2 -> 1, 2
7.59/7.81	Where:
7.59/7.81	  1) f3#(I0, I1, I2, I3, I4) -> f2#(I1 + 1, I2, I0, I5, I6) [0 = I4 /\ 0 = I3 /\ -1 <= I1 - 1]
7.59/7.81	  2) f3#(I7, I8, I9, I10, I11) -> f3#(I7, I8, I9 - 1, I10 - 1, I10 - 1) [I10 = I11 /\ 0 <= I10 - 1]
7.59/7.81	
7.59/7.81	We have the following SCCs.
7.59/7.81	  { 2 }
7.59/7.81	
7.59/7.81	DP problem for innermost termination.
7.59/7.81	P =
7.59/7.81	  f3#(I7, I8, I9, I10, I11) -> f3#(I7, I8, I9 - 1, I10 - 1, I10 - 1) [I10 = I11 /\ 0 <= I10 - 1]
7.59/7.81	R = 
7.59/7.81	  init(x1, x2, x3, x4, x5) -> f1(rnd1, rnd2, rnd3, rnd4, rnd5) 
7.59/7.81	  f3(I0, I1, I2, I3, I4) -> f2(I1 + 1, I2, I0, I5, I6) [0 = I4 /\ 0 = I3 /\ -1 <= I1 - 1]
7.59/7.81	  f3(I7, I8, I9, I10, I11) -> f3(I7, I8, I9 - 1, I10 - 1, I10 - 1) [I10 = I11 /\ 0 <= I10 - 1]
7.59/7.81	  f3(I12, I13, I14, I15, I16) -> f3(I12, I13, I14 + 1, I15 + 1, I15 + 1) [I15 = I16 /\ I15 <= 0 /\ 0 <= I15 - 1]
7.59/7.81	  f3(I17, I18, I19, I20, I21) -> f3(I17, I18, I19 + 1, I20 + 1, I20 + 1) [I20 = I21 /\ I20 <= 0 /\ I20 <= -1]
7.59/7.81	  f2(I22, I23, I24, I25, I26) -> f3(I24, I22, I23, I24, I24) [0 <= I24 - 1 /\ I24 <= I23]
7.59/7.81	  f1(I27, I28, I29, I30, I31) -> f2(0, I32, I33, I34, I35) [0 <= I27 - 1 /\ -1 <= I32 - 1 /\ -1 <= I28 - 1 /\ -1 <= I33 - 1]
7.59/7.81	
7.59/7.81	We use the basic value criterion with the projection function NU:
7.59/7.81	  NU[f3#(z1,z2,z3,z4,z5)] = z5
7.59/7.81	
7.59/7.81	This gives the following inequalities:
7.59/7.81	  I10 = I11 /\ 0 <= I10 - 1  ==>  I11 >! I10 - 1
7.59/7.81	
7.59/7.81	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
7.59/7.81	
7.59/7.81	DP problem for innermost termination.
7.59/7.81	P =
7.59/7.81	  f3#(I17, I18, I19, I20, I21) -> f3#(I17, I18, I19 + 1, I20 + 1, I20 + 1) [I20 = I21 /\ I20 <= 0 /\ I20 <= -1]
7.59/7.81	R = 
7.59/7.81	  init(x1, x2, x3, x4, x5) -> f1(rnd1, rnd2, rnd3, rnd4, rnd5) 
7.59/7.81	  f3(I0, I1, I2, I3, I4) -> f2(I1 + 1, I2, I0, I5, I6) [0 = I4 /\ 0 = I3 /\ -1 <= I1 - 1]
7.59/7.81	  f3(I7, I8, I9, I10, I11) -> f3(I7, I8, I9 - 1, I10 - 1, I10 - 1) [I10 = I11 /\ 0 <= I10 - 1]
7.59/7.81	  f3(I12, I13, I14, I15, I16) -> f3(I12, I13, I14 + 1, I15 + 1, I15 + 1) [I15 = I16 /\ I15 <= 0 /\ 0 <= I15 - 1]
7.59/7.81	  f3(I17, I18, I19, I20, I21) -> f3(I17, I18, I19 + 1, I20 + 1, I20 + 1) [I20 = I21 /\ I20 <= 0 /\ I20 <= -1]
7.59/7.81	  f2(I22, I23, I24, I25, I26) -> f3(I24, I22, I23, I24, I24) [0 <= I24 - 1 /\ I24 <= I23]
7.59/7.81	  f1(I27, I28, I29, I30, I31) -> f2(0, I32, I33, I34, I35) [0 <= I27 - 1 /\ -1 <= I32 - 1 /\ -1 <= I28 - 1 /\ -1 <= I33 - 1]
7.59/7.81	
7.59/7.81	We use the reverse value criterion with the projection function NU:
7.59/7.81	  NU[f3#(z1,z2,z3,z4,z5)] = -1 + -1 * z4
7.59/7.81	
7.59/7.81	This gives the following inequalities:
7.59/7.81	  I20 = I21 /\ I20 <= 0 /\ I20 <= -1  ==>  -1 + -1 * I20 > -1 + -1 * (I20 + 1)  with  -1 + -1 * I20 >= 0
7.59/7.81	
7.59/7.81	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
7.59/10.79	EOF