19.38/19.53	YES
19.38/19.53	
19.38/19.53	DP problem for innermost termination.
19.38/19.53	P =
19.38/19.53	  f6#(x1, x2, x3, x4, x5, x6) -> f1#(x1, x2, x3, x4, x5, x6) 
19.38/19.53	  f5#(I0, I1, I2, I3, I4, I5) -> f3#(I0, I1, I2, I3, I4, I5) 
19.38/19.53	  f3#(I6, I7, I8, I9, I10, I11) -> f5#(-1 + I6, I7, rnd3, -1 + I9, I10, rnd6) [1 <= rnd6 /\ -1 + rnd6 <= -1 + I9 /\ -1 + I9 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 <= I9 /\ rnd6 = rnd6 /\ rnd3 = rnd3]
19.38/19.53	  f3#(I12, I13, I14, I15, I16, I17) -> f2#(I12, I13, I14, I15, I16, I17) [I15 <= 0 /\ I15 <= 0]
19.38/19.53	  f4#(I18, I19, I20, I21, I22, I23) -> f3#(-1 + I18, rnd2, I20, -1 + I21, rnd5, I23) [1 <= rnd5 /\ 1 <= rnd2 /\ -1 + rnd5 <= -1 + I21 /\ -1 + I21 <= -1 + rnd5 /\ -1 + rnd2 <= -1 + I18 /\ -1 + I18 <= -1 + rnd2 /\ 1 <= I21 /\ rnd5 = rnd5 /\ rnd2 = rnd2]
19.38/19.53	  f2#(I24, I25, I26, I27, I28, I29) -> f3#(I24, I25, I26, 5000, I28, I29) [1 <= I24 /\ 1 <= 5000 /\ 1 <= I24]
19.38/19.53	  f1#(I30, I31, I32, I33, I34, I35) -> f2#(rnd1, I31, I32, rnd4, I34, I35) [rnd4 = rnd4 /\ rnd1 = rnd1]
19.38/19.53	R = 
19.38/19.53	  f6(x1, x2, x3, x4, x5, x6) -> f1(x1, x2, x3, x4, x5, x6) 
19.38/19.53	  f5(I0, I1, I2, I3, I4, I5) -> f3(I0, I1, I2, I3, I4, I5) 
19.38/19.53	  f3(I6, I7, I8, I9, I10, I11) -> f5(-1 + I6, I7, rnd3, -1 + I9, I10, rnd6) [1 <= rnd6 /\ -1 + rnd6 <= -1 + I9 /\ -1 + I9 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 <= I9 /\ rnd6 = rnd6 /\ rnd3 = rnd3]
19.38/19.53	  f3(I12, I13, I14, I15, I16, I17) -> f2(I12, I13, I14, I15, I16, I17) [I15 <= 0 /\ I15 <= 0]
19.38/19.53	  f4(I18, I19, I20, I21, I22, I23) -> f3(-1 + I18, rnd2, I20, -1 + I21, rnd5, I23) [1 <= rnd5 /\ 1 <= rnd2 /\ -1 + rnd5 <= -1 + I21 /\ -1 + I21 <= -1 + rnd5 /\ -1 + rnd2 <= -1 + I18 /\ -1 + I18 <= -1 + rnd2 /\ 1 <= I21 /\ rnd5 = rnd5 /\ rnd2 = rnd2]
19.38/19.53	  f2(I24, I25, I26, I27, I28, I29) -> f3(I24, I25, I26, 5000, I28, I29) [1 <= I24 /\ 1 <= 5000 /\ 1 <= I24]
19.38/19.53	  f1(I30, I31, I32, I33, I34, I35) -> f2(rnd1, I31, I32, rnd4, I34, I35) [rnd4 = rnd4 /\ rnd1 = rnd1]
19.38/19.53	
19.38/19.53	The dependency graph for this problem is:
19.38/19.53	  0 -> 6
19.38/19.53	  1 -> 2, 3
19.38/19.53	  2 -> 1
19.38/19.53	  3 -> 5
19.38/19.53	  4 -> 2, 3
19.38/19.53	  5 -> 2
19.38/19.53	  6 -> 5
19.38/19.53	Where:
19.38/19.53	  0) f6#(x1, x2, x3, x4, x5, x6) -> f1#(x1, x2, x3, x4, x5, x6) 
19.38/19.53	  1) f5#(I0, I1, I2, I3, I4, I5) -> f3#(I0, I1, I2, I3, I4, I5) 
19.38/19.53	  2) f3#(I6, I7, I8, I9, I10, I11) -> f5#(-1 + I6, I7, rnd3, -1 + I9, I10, rnd6) [1 <= rnd6 /\ -1 + rnd6 <= -1 + I9 /\ -1 + I9 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 <= I9 /\ rnd6 = rnd6 /\ rnd3 = rnd3]
19.38/19.53	  3) f3#(I12, I13, I14, I15, I16, I17) -> f2#(I12, I13, I14, I15, I16, I17) [I15 <= 0 /\ I15 <= 0]
19.38/19.53	  4) f4#(I18, I19, I20, I21, I22, I23) -> f3#(-1 + I18, rnd2, I20, -1 + I21, rnd5, I23) [1 <= rnd5 /\ 1 <= rnd2 /\ -1 + rnd5 <= -1 + I21 /\ -1 + I21 <= -1 + rnd5 /\ -1 + rnd2 <= -1 + I18 /\ -1 + I18 <= -1 + rnd2 /\ 1 <= I21 /\ rnd5 = rnd5 /\ rnd2 = rnd2]
19.38/19.53	  5) f2#(I24, I25, I26, I27, I28, I29) -> f3#(I24, I25, I26, 5000, I28, I29) [1 <= I24 /\ 1 <= 5000 /\ 1 <= I24]
19.38/19.53	  6) f1#(I30, I31, I32, I33, I34, I35) -> f2#(rnd1, I31, I32, rnd4, I34, I35) [rnd4 = rnd4 /\ rnd1 = rnd1]
19.38/19.53	
19.38/19.53	We have the following SCCs.
19.38/19.53	  { 1, 2, 3, 5 }
19.38/19.53	
19.38/19.53	DP problem for innermost termination.
19.38/19.53	P =
19.38/19.53	  f5#(I0, I1, I2, I3, I4, I5) -> f3#(I0, I1, I2, I3, I4, I5) 
19.38/19.53	  f3#(I6, I7, I8, I9, I10, I11) -> f5#(-1 + I6, I7, rnd3, -1 + I9, I10, rnd6) [1 <= rnd6 /\ -1 + rnd6 <= -1 + I9 /\ -1 + I9 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 <= I9 /\ rnd6 = rnd6 /\ rnd3 = rnd3]
19.38/19.53	  f3#(I12, I13, I14, I15, I16, I17) -> f2#(I12, I13, I14, I15, I16, I17) [I15 <= 0 /\ I15 <= 0]
19.38/19.53	  f2#(I24, I25, I26, I27, I28, I29) -> f3#(I24, I25, I26, 5000, I28, I29) [1 <= I24 /\ 1 <= 5000 /\ 1 <= I24]
19.38/19.53	R = 
19.38/19.53	  f6(x1, x2, x3, x4, x5, x6) -> f1(x1, x2, x3, x4, x5, x6) 
19.38/19.53	  f5(I0, I1, I2, I3, I4, I5) -> f3(I0, I1, I2, I3, I4, I5) 
19.38/19.53	  f3(I6, I7, I8, I9, I10, I11) -> f5(-1 + I6, I7, rnd3, -1 + I9, I10, rnd6) [1 <= rnd6 /\ -1 + rnd6 <= -1 + I9 /\ -1 + I9 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 <= I9 /\ rnd6 = rnd6 /\ rnd3 = rnd3]
19.38/19.53	  f3(I12, I13, I14, I15, I16, I17) -> f2(I12, I13, I14, I15, I16, I17) [I15 <= 0 /\ I15 <= 0]
19.38/19.53	  f4(I18, I19, I20, I21, I22, I23) -> f3(-1 + I18, rnd2, I20, -1 + I21, rnd5, I23) [1 <= rnd5 /\ 1 <= rnd2 /\ -1 + rnd5 <= -1 + I21 /\ -1 + I21 <= -1 + rnd5 /\ -1 + rnd2 <= -1 + I18 /\ -1 + I18 <= -1 + rnd2 /\ 1 <= I21 /\ rnd5 = rnd5 /\ rnd2 = rnd2]
19.38/19.53	  f2(I24, I25, I26, I27, I28, I29) -> f3(I24, I25, I26, 5000, I28, I29) [1 <= I24 /\ 1 <= 5000 /\ 1 <= I24]
19.38/19.53	  f1(I30, I31, I32, I33, I34, I35) -> f2(rnd1, I31, I32, rnd4, I34, I35) [rnd4 = rnd4 /\ rnd1 = rnd1]
19.38/19.53	
19.38/19.53	We use the extended value criterion with the projection function NU:
19.38/19.53	  NU[f2#(x0,x1,x2,x3,x4,x5)] = x0 - 1
19.38/19.53	  NU[f3#(x0,x1,x2,x3,x4,x5)] = x0 - x3 - 1
19.38/19.53	  NU[f5#(x0,x1,x2,x3,x4,x5)] = x0 - x3 - 1
19.38/19.53	
19.38/19.53	This gives the following inequalities:
19.38/19.53	    ==>  I0 - I3 - 1 >= I0 - I3 - 1
19.38/19.53	  1 <= rnd6 /\ -1 + rnd6 <= -1 + I9 /\ -1 + I9 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 <= I9 /\ rnd6 = rnd6 /\ rnd3 = rnd3  ==>  I6 - I9 - 1 >= (-1 + I6) - (-1 + I9) - 1
19.38/19.53	  I15 <= 0 /\ I15 <= 0  ==>  I12 - I15 - 1 >= I12 - 1
19.38/19.53	  1 <= I24 /\ 1 <= 5000 /\ 1 <= I24  ==>  I24 - 1 > I24 - 5000 - 1  with  I24 - 1 >= 0
19.38/19.53	
19.38/19.53	We remove all the strictly oriented dependency pairs.
19.38/19.53	
19.38/19.53	DP problem for innermost termination.
19.38/19.53	P =
19.38/19.53	  f5#(I0, I1, I2, I3, I4, I5) -> f3#(I0, I1, I2, I3, I4, I5) 
19.38/19.53	  f3#(I6, I7, I8, I9, I10, I11) -> f5#(-1 + I6, I7, rnd3, -1 + I9, I10, rnd6) [1 <= rnd6 /\ -1 + rnd6 <= -1 + I9 /\ -1 + I9 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 <= I9 /\ rnd6 = rnd6 /\ rnd3 = rnd3]
19.38/19.53	  f3#(I12, I13, I14, I15, I16, I17) -> f2#(I12, I13, I14, I15, I16, I17) [I15 <= 0 /\ I15 <= 0]
19.38/19.53	R = 
19.38/19.53	  f6(x1, x2, x3, x4, x5, x6) -> f1(x1, x2, x3, x4, x5, x6) 
19.38/19.53	  f5(I0, I1, I2, I3, I4, I5) -> f3(I0, I1, I2, I3, I4, I5) 
19.38/19.53	  f3(I6, I7, I8, I9, I10, I11) -> f5(-1 + I6, I7, rnd3, -1 + I9, I10, rnd6) [1 <= rnd6 /\ -1 + rnd6 <= -1 + I9 /\ -1 + I9 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 <= I9 /\ rnd6 = rnd6 /\ rnd3 = rnd3]
19.38/19.53	  f3(I12, I13, I14, I15, I16, I17) -> f2(I12, I13, I14, I15, I16, I17) [I15 <= 0 /\ I15 <= 0]
19.38/19.53	  f4(I18, I19, I20, I21, I22, I23) -> f3(-1 + I18, rnd2, I20, -1 + I21, rnd5, I23) [1 <= rnd5 /\ 1 <= rnd2 /\ -1 + rnd5 <= -1 + I21 /\ -1 + I21 <= -1 + rnd5 /\ -1 + rnd2 <= -1 + I18 /\ -1 + I18 <= -1 + rnd2 /\ 1 <= I21 /\ rnd5 = rnd5 /\ rnd2 = rnd2]
19.38/19.53	  f2(I24, I25, I26, I27, I28, I29) -> f3(I24, I25, I26, 5000, I28, I29) [1 <= I24 /\ 1 <= 5000 /\ 1 <= I24]
19.38/19.53	  f1(I30, I31, I32, I33, I34, I35) -> f2(rnd1, I31, I32, rnd4, I34, I35) [rnd4 = rnd4 /\ rnd1 = rnd1]
19.38/19.53	
19.38/19.53	The dependency graph for this problem is:
19.38/19.53	  1 -> 2, 3
19.38/19.53	  2 -> 1
19.38/19.53	  3 -> 
19.38/19.53	Where:
19.38/19.53	  1) f5#(I0, I1, I2, I3, I4, I5) -> f3#(I0, I1, I2, I3, I4, I5) 
19.38/19.53	  2) f3#(I6, I7, I8, I9, I10, I11) -> f5#(-1 + I6, I7, rnd3, -1 + I9, I10, rnd6) [1 <= rnd6 /\ -1 + rnd6 <= -1 + I9 /\ -1 + I9 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 <= I9 /\ rnd6 = rnd6 /\ rnd3 = rnd3]
19.38/19.53	  3) f3#(I12, I13, I14, I15, I16, I17) -> f2#(I12, I13, I14, I15, I16, I17) [I15 <= 0 /\ I15 <= 0]
19.38/19.53	
19.38/19.53	We have the following SCCs.
19.38/19.53	  { 1, 2 }
19.38/19.53	
19.38/19.53	DP problem for innermost termination.
19.38/19.53	P =
19.38/19.53	  f5#(I0, I1, I2, I3, I4, I5) -> f3#(I0, I1, I2, I3, I4, I5) 
19.38/19.53	  f3#(I6, I7, I8, I9, I10, I11) -> f5#(-1 + I6, I7, rnd3, -1 + I9, I10, rnd6) [1 <= rnd6 /\ -1 + rnd6 <= -1 + I9 /\ -1 + I9 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 <= I9 /\ rnd6 = rnd6 /\ rnd3 = rnd3]
19.38/19.53	R = 
19.38/19.53	  f6(x1, x2, x3, x4, x5, x6) -> f1(x1, x2, x3, x4, x5, x6) 
19.38/19.53	  f5(I0, I1, I2, I3, I4, I5) -> f3(I0, I1, I2, I3, I4, I5) 
19.38/19.53	  f3(I6, I7, I8, I9, I10, I11) -> f5(-1 + I6, I7, rnd3, -1 + I9, I10, rnd6) [1 <= rnd6 /\ -1 + rnd6 <= -1 + I9 /\ -1 + I9 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 <= I9 /\ rnd6 = rnd6 /\ rnd3 = rnd3]
19.38/19.53	  f3(I12, I13, I14, I15, I16, I17) -> f2(I12, I13, I14, I15, I16, I17) [I15 <= 0 /\ I15 <= 0]
19.38/19.53	  f4(I18, I19, I20, I21, I22, I23) -> f3(-1 + I18, rnd2, I20, -1 + I21, rnd5, I23) [1 <= rnd5 /\ 1 <= rnd2 /\ -1 + rnd5 <= -1 + I21 /\ -1 + I21 <= -1 + rnd5 /\ -1 + rnd2 <= -1 + I18 /\ -1 + I18 <= -1 + rnd2 /\ 1 <= I21 /\ rnd5 = rnd5 /\ rnd2 = rnd2]
19.38/19.53	  f2(I24, I25, I26, I27, I28, I29) -> f3(I24, I25, I26, 5000, I28, I29) [1 <= I24 /\ 1 <= 5000 /\ 1 <= I24]
19.38/19.53	  f1(I30, I31, I32, I33, I34, I35) -> f2(rnd1, I31, I32, rnd4, I34, I35) [rnd4 = rnd4 /\ rnd1 = rnd1]
19.38/19.53	
19.38/19.53	We use the basic value criterion with the projection function NU:
19.38/19.53	  NU[f3#(z1,z2,z3,z4,z5,z6)] = z4
19.38/19.53	  NU[f5#(z1,z2,z3,z4,z5,z6)] = z4
19.38/19.53	
19.38/19.53	This gives the following inequalities:
19.38/19.53	    ==>  I3 (>! \union =) I3
19.38/19.53	  1 <= rnd6 /\ -1 + rnd6 <= -1 + I9 /\ -1 + I9 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 <= I9 /\ rnd6 = rnd6 /\ rnd3 = rnd3  ==>  I9 >! -1 + I9
19.38/19.53	
19.38/19.53	We remove all the strictly oriented dependency pairs.
19.38/19.53	
19.38/19.53	DP problem for innermost termination.
19.38/19.53	P =
19.38/19.53	  f5#(I0, I1, I2, I3, I4, I5) -> f3#(I0, I1, I2, I3, I4, I5) 
19.38/19.53	R = 
19.38/19.53	  f6(x1, x2, x3, x4, x5, x6) -> f1(x1, x2, x3, x4, x5, x6) 
19.38/19.53	  f5(I0, I1, I2, I3, I4, I5) -> f3(I0, I1, I2, I3, I4, I5) 
19.38/19.53	  f3(I6, I7, I8, I9, I10, I11) -> f5(-1 + I6, I7, rnd3, -1 + I9, I10, rnd6) [1 <= rnd6 /\ -1 + rnd6 <= -1 + I9 /\ -1 + I9 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 <= I9 /\ rnd6 = rnd6 /\ rnd3 = rnd3]
19.38/19.53	  f3(I12, I13, I14, I15, I16, I17) -> f2(I12, I13, I14, I15, I16, I17) [I15 <= 0 /\ I15 <= 0]
19.38/19.53	  f4(I18, I19, I20, I21, I22, I23) -> f3(-1 + I18, rnd2, I20, -1 + I21, rnd5, I23) [1 <= rnd5 /\ 1 <= rnd2 /\ -1 + rnd5 <= -1 + I21 /\ -1 + I21 <= -1 + rnd5 /\ -1 + rnd2 <= -1 + I18 /\ -1 + I18 <= -1 + rnd2 /\ 1 <= I21 /\ rnd5 = rnd5 /\ rnd2 = rnd2]
19.38/19.53	  f2(I24, I25, I26, I27, I28, I29) -> f3(I24, I25, I26, 5000, I28, I29) [1 <= I24 /\ 1 <= 5000 /\ 1 <= I24]
19.38/19.53	  f1(I30, I31, I32, I33, I34, I35) -> f2(rnd1, I31, I32, rnd4, I34, I35) [rnd4 = rnd4 /\ rnd1 = rnd1]
19.38/19.53	
19.38/19.53	The dependency graph for this problem is:
19.38/19.53	  1 -> 
19.38/19.53	Where:
19.38/19.53	  1) f5#(I0, I1, I2, I3, I4, I5) -> f3#(I0, I1, I2, I3, I4, I5) 
19.38/19.53	
19.38/19.53	We have the following SCCs.
19.38/19.53	  
19.38/22.51	EOF