1.66/1.74	YES
1.66/1.74	
1.66/1.74	DP problem for innermost termination.
1.66/1.74	P =
1.66/1.74	  f7#(x1, x2, x3) -> f6#(x1, x2, x3) 
1.66/1.74	  f6#(I0, I1, I2) -> f5#(I0, I1, I2) 
1.66/1.74	  f6#(I3, I4, I5) -> f2#(I3, I4, I5) 
1.66/1.74	  f6#(I6, I7, I8) -> f3#(I6, I7, I8) 
1.66/1.74	  f6#(I9, I10, I11) -> f1#(I9, I10, I11) 
1.66/1.74	  f6#(I15, I16, I17) -> f5#(I17, rnd2, rnd3) [rnd3 = rnd2 /\ rnd2 = rnd2]
1.66/1.74	  f5#(I18, I19, I20) -> f2#(I20, I19, 0) 
1.66/1.74	  f2#(I21, I22, I23) -> f3#(I23, I22, I23) [11 <= I23]
1.66/1.74	  f2#(I24, I25, I26) -> f1#(I26, I25, I26) [I26 <= 10]
1.66/1.74	  f1#(I32, I33, I34) -> f2#(I34, I33, 1 + I34) 
1.66/1.74	R = 
1.66/1.74	  f7(x1, x2, x3) -> f6(x1, x2, x3) 
1.66/1.74	  f6(I0, I1, I2) -> f5(I0, I1, I2) 
1.66/1.74	  f6(I3, I4, I5) -> f2(I3, I4, I5) 
1.66/1.74	  f6(I6, I7, I8) -> f3(I6, I7, I8) 
1.66/1.74	  f6(I9, I10, I11) -> f1(I9, I10, I11) 
1.66/1.74	  f6(I12, I13, I14) -> f4(I12, I13, I14) 
1.66/1.74	  f6(I15, I16, I17) -> f5(I17, rnd2, rnd3) [rnd3 = rnd2 /\ rnd2 = rnd2]
1.66/1.74	  f5(I18, I19, I20) -> f2(I20, I19, 0) 
1.66/1.74	  f2(I21, I22, I23) -> f3(I23, I22, I23) [11 <= I23]
1.66/1.74	  f2(I24, I25, I26) -> f1(I26, I25, I26) [I26 <= 10]
1.66/1.74	  f3(I27, I28, I29) -> f4(I29, I30, I31) [I31 = I30 /\ I30 = I30]
1.66/1.74	  f1(I32, I33, I34) -> f2(I34, I33, 1 + I34) 
1.66/1.74	
1.66/1.74	The dependency graph for this problem is:
1.66/1.74	  0 -> 1, 2, 3, 4, 5
1.66/1.74	  1 -> 6
1.66/1.74	  2 -> 7, 8
1.66/1.74	  3 -> 
1.66/1.74	  4 -> 9
1.66/1.74	  5 -> 6
1.66/1.74	  6 -> 8
1.66/1.74	  7 -> 
1.66/1.74	  8 -> 9
1.66/1.74	  9 -> 7, 8
1.66/1.74	Where:
1.66/1.74	  0) f7#(x1, x2, x3) -> f6#(x1, x2, x3) 
1.66/1.74	  1) f6#(I0, I1, I2) -> f5#(I0, I1, I2) 
1.66/1.74	  2) f6#(I3, I4, I5) -> f2#(I3, I4, I5) 
1.66/1.74	  3) f6#(I6, I7, I8) -> f3#(I6, I7, I8) 
1.66/1.74	  4) f6#(I9, I10, I11) -> f1#(I9, I10, I11) 
1.66/1.74	  5) f6#(I15, I16, I17) -> f5#(I17, rnd2, rnd3) [rnd3 = rnd2 /\ rnd2 = rnd2]
1.66/1.74	  6) f5#(I18, I19, I20) -> f2#(I20, I19, 0) 
1.66/1.74	  7) f2#(I21, I22, I23) -> f3#(I23, I22, I23) [11 <= I23]
1.66/1.74	  8) f2#(I24, I25, I26) -> f1#(I26, I25, I26) [I26 <= 10]
1.66/1.74	  9) f1#(I32, I33, I34) -> f2#(I34, I33, 1 + I34) 
1.66/1.74	
1.66/1.74	We have the following SCCs.
1.66/1.74	  { 8, 9 }
1.66/1.74	
1.66/1.74	DP problem for innermost termination.
1.66/1.74	P =
1.66/1.74	  f2#(I24, I25, I26) -> f1#(I26, I25, I26) [I26 <= 10]
1.66/1.74	  f1#(I32, I33, I34) -> f2#(I34, I33, 1 + I34) 
1.66/1.74	R = 
1.66/1.74	  f7(x1, x2, x3) -> f6(x1, x2, x3) 
1.66/1.74	  f6(I0, I1, I2) -> f5(I0, I1, I2) 
1.66/1.74	  f6(I3, I4, I5) -> f2(I3, I4, I5) 
1.66/1.74	  f6(I6, I7, I8) -> f3(I6, I7, I8) 
1.66/1.74	  f6(I9, I10, I11) -> f1(I9, I10, I11) 
1.66/1.74	  f6(I12, I13, I14) -> f4(I12, I13, I14) 
1.66/1.74	  f6(I15, I16, I17) -> f5(I17, rnd2, rnd3) [rnd3 = rnd2 /\ rnd2 = rnd2]
1.66/1.74	  f5(I18, I19, I20) -> f2(I20, I19, 0) 
1.66/1.74	  f2(I21, I22, I23) -> f3(I23, I22, I23) [11 <= I23]
1.66/1.74	  f2(I24, I25, I26) -> f1(I26, I25, I26) [I26 <= 10]
1.66/1.74	  f3(I27, I28, I29) -> f4(I29, I30, I31) [I31 = I30 /\ I30 = I30]
1.66/1.74	  f1(I32, I33, I34) -> f2(I34, I33, 1 + I34) 
1.66/1.74	
1.66/1.74	We use the reverse value criterion with the projection function NU:
1.66/1.74	  NU[f1#(z1,z2,z3)] = 10 + -1 * (1 + z3)
1.66/1.74	  NU[f2#(z1,z2,z3)] = 10 + -1 * z3
1.66/1.74	
1.66/1.74	This gives the following inequalities:
1.66/1.74	  I26 <= 10  ==>  10 + -1 * I26 > 10 + -1 * (1 + I26)  with  10 + -1 * I26 >= 0
1.66/1.74	    ==>  10 + -1 * (1 + I34) >= 10 + -1 * (1 + I34)
1.66/1.74	
1.66/1.74	We remove all the strictly oriented dependency pairs.
1.66/1.74	
1.66/1.74	DP problem for innermost termination.
1.66/1.74	P =
1.66/1.74	  f1#(I32, I33, I34) -> f2#(I34, I33, 1 + I34) 
1.66/1.74	R = 
1.66/1.74	  f7(x1, x2, x3) -> f6(x1, x2, x3) 
1.66/1.74	  f6(I0, I1, I2) -> f5(I0, I1, I2) 
1.66/1.74	  f6(I3, I4, I5) -> f2(I3, I4, I5) 
1.66/1.74	  f6(I6, I7, I8) -> f3(I6, I7, I8) 
1.66/1.74	  f6(I9, I10, I11) -> f1(I9, I10, I11) 
1.66/1.74	  f6(I12, I13, I14) -> f4(I12, I13, I14) 
1.66/1.74	  f6(I15, I16, I17) -> f5(I17, rnd2, rnd3) [rnd3 = rnd2 /\ rnd2 = rnd2]
1.66/1.74	  f5(I18, I19, I20) -> f2(I20, I19, 0) 
1.66/1.74	  f2(I21, I22, I23) -> f3(I23, I22, I23) [11 <= I23]
1.66/1.74	  f2(I24, I25, I26) -> f1(I26, I25, I26) [I26 <= 10]
1.66/1.74	  f3(I27, I28, I29) -> f4(I29, I30, I31) [I31 = I30 /\ I30 = I30]
1.66/1.74	  f1(I32, I33, I34) -> f2(I34, I33, 1 + I34) 
1.66/1.74	
1.66/1.74	The dependency graph for this problem is:
1.66/1.74	  9 -> 
1.66/1.74	Where:
1.66/1.74	  9) f1#(I32, I33, I34) -> f2#(I34, I33, 1 + I34) 
1.66/1.74	
1.66/1.74	We have the following SCCs.
1.66/1.74	  
1.66/4.71	EOF