8.80/8.70	YES
8.80/8.70	
8.80/8.70	DP problem for innermost termination.
8.80/8.70	P =
8.80/8.70	  f7#(x1, x2, x3, x4, x5) -> f6#(x1, x2, x3, x4, x5) 
8.80/8.70	  f6#(I0, I1, I2, I3, I4) -> f1#(0, I1, rnd3, rnd4, I4) [rnd4 = rnd4 /\ rnd3 = rnd3]
8.80/8.70	  f2#(I5, I6, I7, I8, I9) -> f1#(1 + I5, I6, I7, I8, I9) [1 + I5 <= 10]
8.80/8.70	  f2#(I10, I11, I12, I13, I14) -> f5#(I10, 0, I12, I13, rnd5) [rnd5 = rnd5 /\ 10 <= I10]
8.80/8.70	  f5#(I15, I16, I17, I18, I19) -> f3#(I15, I16, I17, I18, I19) 
8.80/8.70	  f3#(I20, I21, I22, I23, I24) -> f5#(I20, 1 + I21, I22, I23, I24) [1 + I21 <= 10]
8.80/8.70	  f1#(I30, I31, I32, I33, I34) -> f2#(I30, I31, I32, I33, I34) 
8.80/8.70	R = 
8.80/8.70	  f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) 
8.80/8.70	  f6(I0, I1, I2, I3, I4) -> f1(0, I1, rnd3, rnd4, I4) [rnd4 = rnd4 /\ rnd3 = rnd3]
8.80/8.70	  f2(I5, I6, I7, I8, I9) -> f1(1 + I5, I6, I7, I8, I9) [1 + I5 <= 10]
8.80/8.70	  f2(I10, I11, I12, I13, I14) -> f5(I10, 0, I12, I13, rnd5) [rnd5 = rnd5 /\ 10 <= I10]
8.80/8.70	  f5(I15, I16, I17, I18, I19) -> f3(I15, I16, I17, I18, I19) 
8.80/8.70	  f3(I20, I21, I22, I23, I24) -> f5(I20, 1 + I21, I22, I23, I24) [1 + I21 <= 10]
8.80/8.70	  f3(I25, I26, I27, I28, I29) -> f4(I25, I26, I27, I28, I29) [10 <= I26]
8.80/8.70	  f1(I30, I31, I32, I33, I34) -> f2(I30, I31, I32, I33, I34) 
8.80/8.70	
8.80/8.70	The dependency graph for this problem is:
8.80/8.70	  0 -> 1
8.80/8.70	  1 -> 6
8.80/8.70	  2 -> 6
8.80/8.70	  3 -> 4
8.80/8.70	  4 -> 5
8.80/8.70	  5 -> 4
8.80/8.70	  6 -> 2, 3
8.80/8.70	Where:
8.80/8.70	  0) f7#(x1, x2, x3, x4, x5) -> f6#(x1, x2, x3, x4, x5) 
8.80/8.70	  1) f6#(I0, I1, I2, I3, I4) -> f1#(0, I1, rnd3, rnd4, I4) [rnd4 = rnd4 /\ rnd3 = rnd3]
8.80/8.70	  2) f2#(I5, I6, I7, I8, I9) -> f1#(1 + I5, I6, I7, I8, I9) [1 + I5 <= 10]
8.80/8.70	  3) f2#(I10, I11, I12, I13, I14) -> f5#(I10, 0, I12, I13, rnd5) [rnd5 = rnd5 /\ 10 <= I10]
8.80/8.70	  4) f5#(I15, I16, I17, I18, I19) -> f3#(I15, I16, I17, I18, I19) 
8.80/8.70	  5) f3#(I20, I21, I22, I23, I24) -> f5#(I20, 1 + I21, I22, I23, I24) [1 + I21 <= 10]
8.80/8.70	  6) f1#(I30, I31, I32, I33, I34) -> f2#(I30, I31, I32, I33, I34) 
8.80/8.70	
8.80/8.70	We have the following SCCs.
8.80/8.70	  { 2, 6 }
8.80/8.70	  { 4, 5 }
8.80/8.70	
8.80/8.70	DP problem for innermost termination.
8.80/8.70	P =
8.80/8.70	  f5#(I15, I16, I17, I18, I19) -> f3#(I15, I16, I17, I18, I19) 
8.80/8.70	  f3#(I20, I21, I22, I23, I24) -> f5#(I20, 1 + I21, I22, I23, I24) [1 + I21 <= 10]
8.80/8.70	R = 
8.80/8.70	  f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) 
8.80/8.70	  f6(I0, I1, I2, I3, I4) -> f1(0, I1, rnd3, rnd4, I4) [rnd4 = rnd4 /\ rnd3 = rnd3]
8.80/8.70	  f2(I5, I6, I7, I8, I9) -> f1(1 + I5, I6, I7, I8, I9) [1 + I5 <= 10]
8.80/8.70	  f2(I10, I11, I12, I13, I14) -> f5(I10, 0, I12, I13, rnd5) [rnd5 = rnd5 /\ 10 <= I10]
8.80/8.70	  f5(I15, I16, I17, I18, I19) -> f3(I15, I16, I17, I18, I19) 
8.80/8.70	  f3(I20, I21, I22, I23, I24) -> f5(I20, 1 + I21, I22, I23, I24) [1 + I21 <= 10]
8.80/8.70	  f3(I25, I26, I27, I28, I29) -> f4(I25, I26, I27, I28, I29) [10 <= I26]
8.80/8.70	  f1(I30, I31, I32, I33, I34) -> f2(I30, I31, I32, I33, I34) 
8.80/8.70	
8.80/8.70	We use the reverse value criterion with the projection function NU:
8.80/8.70	  NU[f3#(z1,z2,z3,z4,z5)] = 10 + -1 * (1 + z2)
8.80/8.70	  NU[f5#(z1,z2,z3,z4,z5)] = 10 + -1 * (1 + z2)
8.80/8.70	
8.80/8.70	This gives the following inequalities:
8.80/8.70	    ==>  10 + -1 * (1 + I16) >= 10 + -1 * (1 + I16)
8.80/8.70	  1 + I21 <= 10  ==>  10 + -1 * (1 + I21) > 10 + -1 * (1 + (1 + I21))  with  10 + -1 * (1 + I21) >= 0
8.80/8.70	
8.80/8.70	We remove all the strictly oriented dependency pairs.
8.80/8.70	
8.80/8.70	DP problem for innermost termination.
8.80/8.70	P =
8.80/8.70	  f5#(I15, I16, I17, I18, I19) -> f3#(I15, I16, I17, I18, I19) 
8.80/8.70	R = 
8.80/8.70	  f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) 
8.80/8.70	  f6(I0, I1, I2, I3, I4) -> f1(0, I1, rnd3, rnd4, I4) [rnd4 = rnd4 /\ rnd3 = rnd3]
8.80/8.70	  f2(I5, I6, I7, I8, I9) -> f1(1 + I5, I6, I7, I8, I9) [1 + I5 <= 10]
8.80/8.70	  f2(I10, I11, I12, I13, I14) -> f5(I10, 0, I12, I13, rnd5) [rnd5 = rnd5 /\ 10 <= I10]
8.80/8.70	  f5(I15, I16, I17, I18, I19) -> f3(I15, I16, I17, I18, I19) 
8.80/8.70	  f3(I20, I21, I22, I23, I24) -> f5(I20, 1 + I21, I22, I23, I24) [1 + I21 <= 10]
8.80/8.70	  f3(I25, I26, I27, I28, I29) -> f4(I25, I26, I27, I28, I29) [10 <= I26]
8.80/8.70	  f1(I30, I31, I32, I33, I34) -> f2(I30, I31, I32, I33, I34) 
8.80/8.70	
8.80/8.70	The dependency graph for this problem is:
8.80/8.70	  4 -> 
8.80/8.70	Where:
8.80/8.70	  4) f5#(I15, I16, I17, I18, I19) -> f3#(I15, I16, I17, I18, I19) 
8.80/8.70	
8.80/8.70	We have the following SCCs.
8.80/8.70	  
8.80/8.70	
8.80/8.70	DP problem for innermost termination.
8.80/8.70	P =
8.80/8.70	  f2#(I5, I6, I7, I8, I9) -> f1#(1 + I5, I6, I7, I8, I9) [1 + I5 <= 10]
8.80/8.70	  f1#(I30, I31, I32, I33, I34) -> f2#(I30, I31, I32, I33, I34) 
8.80/8.70	R = 
8.80/8.70	  f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) 
8.80/8.70	  f6(I0, I1, I2, I3, I4) -> f1(0, I1, rnd3, rnd4, I4) [rnd4 = rnd4 /\ rnd3 = rnd3]
8.80/8.70	  f2(I5, I6, I7, I8, I9) -> f1(1 + I5, I6, I7, I8, I9) [1 + I5 <= 10]
8.80/8.70	  f2(I10, I11, I12, I13, I14) -> f5(I10, 0, I12, I13, rnd5) [rnd5 = rnd5 /\ 10 <= I10]
8.80/8.70	  f5(I15, I16, I17, I18, I19) -> f3(I15, I16, I17, I18, I19) 
8.80/8.70	  f3(I20, I21, I22, I23, I24) -> f5(I20, 1 + I21, I22, I23, I24) [1 + I21 <= 10]
8.80/8.70	  f3(I25, I26, I27, I28, I29) -> f4(I25, I26, I27, I28, I29) [10 <= I26]
8.80/8.70	  f1(I30, I31, I32, I33, I34) -> f2(I30, I31, I32, I33, I34) 
8.80/8.70	
8.80/8.70	We use the reverse value criterion with the projection function NU:
8.80/8.70	  NU[f1#(z1,z2,z3,z4,z5)] = 10 + -1 * (1 + z1)
8.80/8.70	  NU[f2#(z1,z2,z3,z4,z5)] = 10 + -1 * (1 + z1)
8.80/8.70	
8.80/8.70	This gives the following inequalities:
8.80/8.70	  1 + I5 <= 10  ==>  10 + -1 * (1 + I5) > 10 + -1 * (1 + (1 + I5))  with  10 + -1 * (1 + I5) >= 0
8.80/8.70	    ==>  10 + -1 * (1 + I30) >= 10 + -1 * (1 + I30)
8.80/8.70	
8.80/8.70	We remove all the strictly oriented dependency pairs.
8.80/8.70	
8.80/8.70	DP problem for innermost termination.
8.80/8.70	P =
8.80/8.70	  f1#(I30, I31, I32, I33, I34) -> f2#(I30, I31, I32, I33, I34) 
8.80/8.70	R = 
8.80/8.70	  f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) 
8.80/8.70	  f6(I0, I1, I2, I3, I4) -> f1(0, I1, rnd3, rnd4, I4) [rnd4 = rnd4 /\ rnd3 = rnd3]
8.80/8.70	  f2(I5, I6, I7, I8, I9) -> f1(1 + I5, I6, I7, I8, I9) [1 + I5 <= 10]
8.80/8.70	  f2(I10, I11, I12, I13, I14) -> f5(I10, 0, I12, I13, rnd5) [rnd5 = rnd5 /\ 10 <= I10]
8.80/8.70	  f5(I15, I16, I17, I18, I19) -> f3(I15, I16, I17, I18, I19) 
8.80/8.70	  f3(I20, I21, I22, I23, I24) -> f5(I20, 1 + I21, I22, I23, I24) [1 + I21 <= 10]
8.80/8.70	  f3(I25, I26, I27, I28, I29) -> f4(I25, I26, I27, I28, I29) [10 <= I26]
8.80/8.70	  f1(I30, I31, I32, I33, I34) -> f2(I30, I31, I32, I33, I34) 
8.80/8.70	
8.80/8.70	The dependency graph for this problem is:
8.80/8.70	  6 -> 
8.80/8.70	Where:
8.80/8.70	  6) f1#(I30, I31, I32, I33, I34) -> f2#(I30, I31, I32, I33, I34) 
8.80/8.70	
8.80/8.70	We have the following SCCs.
8.80/8.70	  
8.80/11.68	EOF