7.04/7.08	YES
7.04/7.08	
7.04/7.08	DP problem for innermost termination.
7.04/7.08	P =
7.04/7.08	  f6#(x1, x2, x3, x4, x5, x6) -> f4#(x1, x2, x3, x4, x5, x6) 
7.04/7.08	  f5#(I0, I1, I2, I3, I4, I5) -> f1#(I0, I1, I2, I3, 1 + I4, I5) [0 <= -1 + I5 /\ 0 <= I2 - I4]
7.04/7.08	  f4#(I12, I13, I14, I15, I16, I17) -> f5#(I12, I13, I14, I15, 0, I17) 
7.04/7.08	  f3#(I18, I19, I20, I21, I22, I23) -> f1#(I18, I19, I20, I21, I22, I23) 
7.04/7.08	  f1#(I24, I25, I26, I27, I28, I29) -> f3#(I24, I25, I26, I27, 1 + I28, I29) [0 <= -1 + I29 /\ 0 <= I26 - I28]
7.04/7.08	R = 
7.04/7.08	  f6(x1, x2, x3, x4, x5, x6) -> f4(x1, x2, x3, x4, x5, x6) 
7.04/7.08	  f5(I0, I1, I2, I3, I4, I5) -> f1(I0, I1, I2, I3, 1 + I4, I5) [0 <= -1 + I5 /\ 0 <= I2 - I4]
7.04/7.08	  f5(I6, I7, I8, I9, I10, I11) -> f2(rnd1, rnd2, I8, 0, I10, I11) [rnd1 = rnd2 /\ rnd2 = 0 /\ I11 <= 0 /\ 0 <= I8 - I10]
7.04/7.08	  f4(I12, I13, I14, I15, I16, I17) -> f5(I12, I13, I14, I15, 0, I17) 
7.04/7.08	  f3(I18, I19, I20, I21, I22, I23) -> f1(I18, I19, I20, I21, I22, I23) 
7.04/7.08	  f1(I24, I25, I26, I27, I28, I29) -> f3(I24, I25, I26, I27, 1 + I28, I29) [0 <= -1 + I29 /\ 0 <= I26 - I28]
7.04/7.08	  f1(I30, I31, I32, I33, I34, I35) -> f2(I36, I37, I32, 0, I34, I35) [I36 = I37 /\ I37 = 0 /\ 1 + I32 - I34 <= 0]
7.04/7.08	
7.04/7.08	The dependency graph for this problem is:
7.04/7.08	  0 -> 2
7.04/7.08	  1 -> 4
7.04/7.08	  2 -> 1
7.04/7.08	  3 -> 4
7.04/7.08	  4 -> 3
7.04/7.08	Where:
7.04/7.08	  0) f6#(x1, x2, x3, x4, x5, x6) -> f4#(x1, x2, x3, x4, x5, x6) 
7.04/7.08	  1) f5#(I0, I1, I2, I3, I4, I5) -> f1#(I0, I1, I2, I3, 1 + I4, I5) [0 <= -1 + I5 /\ 0 <= I2 - I4]
7.04/7.08	  2) f4#(I12, I13, I14, I15, I16, I17) -> f5#(I12, I13, I14, I15, 0, I17) 
7.04/7.08	  3) f3#(I18, I19, I20, I21, I22, I23) -> f1#(I18, I19, I20, I21, I22, I23) 
7.04/7.08	  4) f1#(I24, I25, I26, I27, I28, I29) -> f3#(I24, I25, I26, I27, 1 + I28, I29) [0 <= -1 + I29 /\ 0 <= I26 - I28]
7.04/7.08	
7.04/7.08	We have the following SCCs.
7.04/7.08	  { 3, 4 }
7.04/7.08	
7.04/7.08	DP problem for innermost termination.
7.04/7.08	P =
7.04/7.08	  f3#(I18, I19, I20, I21, I22, I23) -> f1#(I18, I19, I20, I21, I22, I23) 
7.04/7.08	  f1#(I24, I25, I26, I27, I28, I29) -> f3#(I24, I25, I26, I27, 1 + I28, I29) [0 <= -1 + I29 /\ 0 <= I26 - I28]
7.04/7.08	R = 
7.04/7.08	  f6(x1, x2, x3, x4, x5, x6) -> f4(x1, x2, x3, x4, x5, x6) 
7.04/7.08	  f5(I0, I1, I2, I3, I4, I5) -> f1(I0, I1, I2, I3, 1 + I4, I5) [0 <= -1 + I5 /\ 0 <= I2 - I4]
7.04/7.08	  f5(I6, I7, I8, I9, I10, I11) -> f2(rnd1, rnd2, I8, 0, I10, I11) [rnd1 = rnd2 /\ rnd2 = 0 /\ I11 <= 0 /\ 0 <= I8 - I10]
7.04/7.08	  f4(I12, I13, I14, I15, I16, I17) -> f5(I12, I13, I14, I15, 0, I17) 
7.04/7.08	  f3(I18, I19, I20, I21, I22, I23) -> f1(I18, I19, I20, I21, I22, I23) 
7.04/7.08	  f1(I24, I25, I26, I27, I28, I29) -> f3(I24, I25, I26, I27, 1 + I28, I29) [0 <= -1 + I29 /\ 0 <= I26 - I28]
7.04/7.08	  f1(I30, I31, I32, I33, I34, I35) -> f2(I36, I37, I32, 0, I34, I35) [I36 = I37 /\ I37 = 0 /\ 1 + I32 - I34 <= 0]
7.04/7.08	
7.04/7.08	We use the reverse value criterion with the projection function NU:
7.04/7.08	  NU[f1#(z1,z2,z3,z4,z5,z6)] = z3 - z5 + -1 * 0
7.04/7.08	  NU[f3#(z1,z2,z3,z4,z5,z6)] = z3 - z5 + -1 * 0
7.04/7.08	
7.04/7.08	This gives the following inequalities:
7.04/7.08	    ==>  I20 - I22 + -1 * 0 >= I20 - I22 + -1 * 0
7.04/7.08	  0 <= -1 + I29 /\ 0 <= I26 - I28  ==>  I26 - I28 + -1 * 0 > I26 - (1 + I28) + -1 * 0  with  I26 - I28 + -1 * 0 >= 0
7.04/7.08	
7.04/7.08	We remove all the strictly oriented dependency pairs.
7.04/7.08	
7.04/7.08	DP problem for innermost termination.
7.04/7.08	P =
7.04/7.08	  f3#(I18, I19, I20, I21, I22, I23) -> f1#(I18, I19, I20, I21, I22, I23) 
7.04/7.08	R = 
7.04/7.08	  f6(x1, x2, x3, x4, x5, x6) -> f4(x1, x2, x3, x4, x5, x6) 
7.04/7.08	  f5(I0, I1, I2, I3, I4, I5) -> f1(I0, I1, I2, I3, 1 + I4, I5) [0 <= -1 + I5 /\ 0 <= I2 - I4]
7.04/7.08	  f5(I6, I7, I8, I9, I10, I11) -> f2(rnd1, rnd2, I8, 0, I10, I11) [rnd1 = rnd2 /\ rnd2 = 0 /\ I11 <= 0 /\ 0 <= I8 - I10]
7.04/7.08	  f4(I12, I13, I14, I15, I16, I17) -> f5(I12, I13, I14, I15, 0, I17) 
7.04/7.08	  f3(I18, I19, I20, I21, I22, I23) -> f1(I18, I19, I20, I21, I22, I23) 
7.04/7.08	  f1(I24, I25, I26, I27, I28, I29) -> f3(I24, I25, I26, I27, 1 + I28, I29) [0 <= -1 + I29 /\ 0 <= I26 - I28]
7.04/7.08	  f1(I30, I31, I32, I33, I34, I35) -> f2(I36, I37, I32, 0, I34, I35) [I36 = I37 /\ I37 = 0 /\ 1 + I32 - I34 <= 0]
7.04/7.08	
7.04/7.08	The dependency graph for this problem is:
7.04/7.08	  3 -> 
7.04/7.08	Where:
7.04/7.08	  3) f3#(I18, I19, I20, I21, I22, I23) -> f1#(I18, I19, I20, I21, I22, I23) 
7.04/7.08	
7.04/7.08	We have the following SCCs.
7.04/7.08	  
7.04/10.06	EOF