6.36/6.31	YES
6.36/6.31	
6.36/6.31	DP problem for innermost termination.
6.36/6.31	P =
6.36/6.31	  f7#(x1, x2, x3, x4, x5, x6) -> f1#(x1, x2, x3, x4, x5, x6) 
6.36/6.31	  f6#(I0, I1, I2, I3, I4, I5) -> f2#(I0, I1, I2, I3, I4, I5) 
6.36/6.31	  f2#(I6, I7, I8, I9, I10, I11) -> f6#(I6, I7, 1 + I8, rnd4, I10, I11) [1 + rnd4 <= I6 /\ 1 + rnd4 <= 1 + I8 /\ 1 + I8 <= 1 + rnd4 /\ 1 + I8 <= I6 /\ rnd4 = rnd4]
6.36/6.31	  f4#(I18, I19, I20, I21, I22, I23) -> f2#(I18, I19, 1 + I20, I21, I22, I23) [1 + I20 <= 1 /\ 1 <= 1 + I20 /\ 1 + I20 <= I18]
6.36/6.31	  f3#(I24, I25, I26, I27, I28, I29) -> f2#(I24, rnd2, rnd3, I27, I28, I29) [1 + I26 <= I24 /\ y1 = 1 + I26 /\ rnd3 = rnd3 /\ 2 <= rnd3 /\ rnd3 <= 2 /\ rnd2 = rnd2 /\ rnd2 <= rnd3 /\ rnd3 <= rnd2]
6.36/6.31	  f1#(I30, I31, I32, I33, I34, I35) -> f2#(I30, I31, 0, I33, I34, I35) [0 <= 0 /\ 0 <= 0]
6.36/6.31	R = 
6.36/6.31	  f7(x1, x2, x3, x4, x5, x6) -> f1(x1, x2, x3, x4, x5, x6) 
6.36/6.31	  f6(I0, I1, I2, I3, I4, I5) -> f2(I0, I1, I2, I3, I4, I5) 
6.36/6.31	  f2(I6, I7, I8, I9, I10, I11) -> f6(I6, I7, 1 + I8, rnd4, I10, I11) [1 + rnd4 <= I6 /\ 1 + rnd4 <= 1 + I8 /\ 1 + I8 <= 1 + rnd4 /\ 1 + I8 <= I6 /\ rnd4 = rnd4]
6.36/6.31	  f2(I12, I13, I14, I15, I16, I17) -> f5(I12, I13, I14, I15, I17, I17) [I12 <= I14]
6.36/6.31	  f4(I18, I19, I20, I21, I22, I23) -> f2(I18, I19, 1 + I20, I21, I22, I23) [1 + I20 <= 1 /\ 1 <= 1 + I20 /\ 1 + I20 <= I18]
6.36/6.31	  f3(I24, I25, I26, I27, I28, I29) -> f2(I24, rnd2, rnd3, I27, I28, I29) [1 + I26 <= I24 /\ y1 = 1 + I26 /\ rnd3 = rnd3 /\ 2 <= rnd3 /\ rnd3 <= 2 /\ rnd2 = rnd2 /\ rnd2 <= rnd3 /\ rnd3 <= rnd2]
6.36/6.31	  f1(I30, I31, I32, I33, I34, I35) -> f2(I30, I31, 0, I33, I34, I35) [0 <= 0 /\ 0 <= 0]
6.36/6.31	
6.36/6.31	The dependency graph for this problem is:
6.36/6.31	  0 -> 5
6.36/6.31	  1 -> 2
6.36/6.31	  2 -> 1
6.36/6.31	  3 -> 2
6.36/6.31	  4 -> 2
6.36/6.31	  5 -> 2
6.36/6.31	Where:
6.36/6.31	  0) f7#(x1, x2, x3, x4, x5, x6) -> f1#(x1, x2, x3, x4, x5, x6) 
6.36/6.31	  1) f6#(I0, I1, I2, I3, I4, I5) -> f2#(I0, I1, I2, I3, I4, I5) 
6.36/6.31	  2) f2#(I6, I7, I8, I9, I10, I11) -> f6#(I6, I7, 1 + I8, rnd4, I10, I11) [1 + rnd4 <= I6 /\ 1 + rnd4 <= 1 + I8 /\ 1 + I8 <= 1 + rnd4 /\ 1 + I8 <= I6 /\ rnd4 = rnd4]
6.36/6.31	  3) f4#(I18, I19, I20, I21, I22, I23) -> f2#(I18, I19, 1 + I20, I21, I22, I23) [1 + I20 <= 1 /\ 1 <= 1 + I20 /\ 1 + I20 <= I18]
6.36/6.31	  4) f3#(I24, I25, I26, I27, I28, I29) -> f2#(I24, rnd2, rnd3, I27, I28, I29) [1 + I26 <= I24 /\ y1 = 1 + I26 /\ rnd3 = rnd3 /\ 2 <= rnd3 /\ rnd3 <= 2 /\ rnd2 = rnd2 /\ rnd2 <= rnd3 /\ rnd3 <= rnd2]
6.36/6.31	  5) f1#(I30, I31, I32, I33, I34, I35) -> f2#(I30, I31, 0, I33, I34, I35) [0 <= 0 /\ 0 <= 0]
6.36/6.31	
6.36/6.31	We have the following SCCs.
6.36/6.31	  { 1, 2 }
6.36/6.31	
6.36/6.31	DP problem for innermost termination.
6.36/6.31	P =
6.36/6.31	  f6#(I0, I1, I2, I3, I4, I5) -> f2#(I0, I1, I2, I3, I4, I5) 
6.36/6.31	  f2#(I6, I7, I8, I9, I10, I11) -> f6#(I6, I7, 1 + I8, rnd4, I10, I11) [1 + rnd4 <= I6 /\ 1 + rnd4 <= 1 + I8 /\ 1 + I8 <= 1 + rnd4 /\ 1 + I8 <= I6 /\ rnd4 = rnd4]
6.36/6.31	R = 
6.36/6.31	  f7(x1, x2, x3, x4, x5, x6) -> f1(x1, x2, x3, x4, x5, x6) 
6.36/6.31	  f6(I0, I1, I2, I3, I4, I5) -> f2(I0, I1, I2, I3, I4, I5) 
6.36/6.31	  f2(I6, I7, I8, I9, I10, I11) -> f6(I6, I7, 1 + I8, rnd4, I10, I11) [1 + rnd4 <= I6 /\ 1 + rnd4 <= 1 + I8 /\ 1 + I8 <= 1 + rnd4 /\ 1 + I8 <= I6 /\ rnd4 = rnd4]
6.36/6.31	  f2(I12, I13, I14, I15, I16, I17) -> f5(I12, I13, I14, I15, I17, I17) [I12 <= I14]
6.36/6.31	  f4(I18, I19, I20, I21, I22, I23) -> f2(I18, I19, 1 + I20, I21, I22, I23) [1 + I20 <= 1 /\ 1 <= 1 + I20 /\ 1 + I20 <= I18]
6.36/6.31	  f3(I24, I25, I26, I27, I28, I29) -> f2(I24, rnd2, rnd3, I27, I28, I29) [1 + I26 <= I24 /\ y1 = 1 + I26 /\ rnd3 = rnd3 /\ 2 <= rnd3 /\ rnd3 <= 2 /\ rnd2 = rnd2 /\ rnd2 <= rnd3 /\ rnd3 <= rnd2]
6.36/6.31	  f1(I30, I31, I32, I33, I34, I35) -> f2(I30, I31, 0, I33, I34, I35) [0 <= 0 /\ 0 <= 0]
6.36/6.31	
6.36/6.31	We use the reverse value criterion with the projection function NU:
6.36/6.31	  NU[f2#(z1,z2,z3,z4,z5,z6)] = z1 + -1 * (1 + z3)
6.36/6.31	  NU[f6#(z1,z2,z3,z4,z5,z6)] = z1 + -1 * (1 + z3)
6.36/6.31	
6.36/6.31	This gives the following inequalities:
6.36/6.31	    ==>  I0 + -1 * (1 + I2) >= I0 + -1 * (1 + I2)
6.36/6.31	  1 + rnd4 <= I6 /\ 1 + rnd4 <= 1 + I8 /\ 1 + I8 <= 1 + rnd4 /\ 1 + I8 <= I6 /\ rnd4 = rnd4  ==>  I6 + -1 * (1 + I8) > I6 + -1 * (1 + (1 + I8))  with  I6 + -1 * (1 + I8) >= 0
6.36/6.31	
6.36/6.31	We remove all the strictly oriented dependency pairs.
6.36/6.31	
6.36/6.31	DP problem for innermost termination.
6.36/6.31	P =
6.36/6.31	  f6#(I0, I1, I2, I3, I4, I5) -> f2#(I0, I1, I2, I3, I4, I5) 
6.36/6.31	R = 
6.36/6.31	  f7(x1, x2, x3, x4, x5, x6) -> f1(x1, x2, x3, x4, x5, x6) 
6.36/6.31	  f6(I0, I1, I2, I3, I4, I5) -> f2(I0, I1, I2, I3, I4, I5) 
6.36/6.31	  f2(I6, I7, I8, I9, I10, I11) -> f6(I6, I7, 1 + I8, rnd4, I10, I11) [1 + rnd4 <= I6 /\ 1 + rnd4 <= 1 + I8 /\ 1 + I8 <= 1 + rnd4 /\ 1 + I8 <= I6 /\ rnd4 = rnd4]
6.36/6.31	  f2(I12, I13, I14, I15, I16, I17) -> f5(I12, I13, I14, I15, I17, I17) [I12 <= I14]
6.36/6.31	  f4(I18, I19, I20, I21, I22, I23) -> f2(I18, I19, 1 + I20, I21, I22, I23) [1 + I20 <= 1 /\ 1 <= 1 + I20 /\ 1 + I20 <= I18]
6.36/6.31	  f3(I24, I25, I26, I27, I28, I29) -> f2(I24, rnd2, rnd3, I27, I28, I29) [1 + I26 <= I24 /\ y1 = 1 + I26 /\ rnd3 = rnd3 /\ 2 <= rnd3 /\ rnd3 <= 2 /\ rnd2 = rnd2 /\ rnd2 <= rnd3 /\ rnd3 <= rnd2]
6.36/6.31	  f1(I30, I31, I32, I33, I34, I35) -> f2(I30, I31, 0, I33, I34, I35) [0 <= 0 /\ 0 <= 0]
6.36/6.31	
6.36/6.31	The dependency graph for this problem is:
6.36/6.31	  1 -> 
6.36/6.31	Where:
6.36/6.31	  1) f6#(I0, I1, I2, I3, I4, I5) -> f2#(I0, I1, I2, I3, I4, I5) 
6.36/6.31	
6.36/6.31	We have the following SCCs.
6.36/6.31	  
6.36/9.28	EOF