12.67/12.44	YES
12.67/12.44	
12.67/12.44	DP problem for innermost termination.
12.67/12.44	P =
12.67/12.44	  f5#(x1, x2, x3, x4, x5, x6, x7, x8, x9) -> f4#(x1, x2, x3, x4, x5, x6, x7, x8, x9) 
12.67/12.44	  f4#(I0, I1, I2, I3, I4, I5, I6, I7, I8) -> f3#(1, 0, 30, I3, 2, 30, I6, I7, I8) 
12.67/12.44	  f3#(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f1#(I9, I10, I11, I12, I13, I14, I15, I16, I17) 
12.67/12.44	  f1#(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f3#(I18 + I19, rnd2, I20, I21, 1 + I22, I23, I24, I18, I26) [rnd2 = I18 /\ I22 <= I23]
12.67/12.44	R = 
12.67/12.44	  f5(x1, x2, x3, x4, x5, x6, x7, x8, x9) -> f4(x1, x2, x3, x4, x5, x6, x7, x8, x9) 
12.67/12.44	  f4(I0, I1, I2, I3, I4, I5, I6, I7, I8) -> f3(1, 0, 30, I3, 2, 30, I6, I7, I8) 
12.67/12.44	  f3(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f1(I9, I10, I11, I12, I13, I14, I15, I16, I17) 
12.67/12.44	  f1(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f3(I18 + I19, rnd2, I20, I21, 1 + I22, I23, I24, I18, I26) [rnd2 = I18 /\ I22 <= I23]
12.67/12.44	  f1(I27, I28, I29, I30, I31, I32, I33, I34, I35) -> f2(I27, I28, I29, I27, I31, I32, I27, I34, I27) [1 + I32 <= I31]
12.67/12.44	
12.67/12.44	The dependency graph for this problem is:
12.67/12.44	  0 -> 1
12.67/12.44	  1 -> 2
12.67/12.44	  2 -> 3
12.67/12.44	  3 -> 2
12.67/12.44	Where:
12.67/12.44	  0) f5#(x1, x2, x3, x4, x5, x6, x7, x8, x9) -> f4#(x1, x2, x3, x4, x5, x6, x7, x8, x9) 
12.67/12.44	  1) f4#(I0, I1, I2, I3, I4, I5, I6, I7, I8) -> f3#(1, 0, 30, I3, 2, 30, I6, I7, I8) 
12.67/12.44	  2) f3#(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f1#(I9, I10, I11, I12, I13, I14, I15, I16, I17) 
12.67/12.44	  3) f1#(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f3#(I18 + I19, rnd2, I20, I21, 1 + I22, I23, I24, I18, I26) [rnd2 = I18 /\ I22 <= I23]
12.67/12.44	
12.67/12.44	We have the following SCCs.
12.67/12.44	  { 2, 3 }
12.67/12.44	
12.67/12.44	DP problem for innermost termination.
12.67/12.44	P =
12.67/12.44	  f3#(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f1#(I9, I10, I11, I12, I13, I14, I15, I16, I17) 
12.67/12.44	  f1#(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f3#(I18 + I19, rnd2, I20, I21, 1 + I22, I23, I24, I18, I26) [rnd2 = I18 /\ I22 <= I23]
12.67/12.44	R = 
12.67/12.44	  f5(x1, x2, x3, x4, x5, x6, x7, x8, x9) -> f4(x1, x2, x3, x4, x5, x6, x7, x8, x9) 
12.67/12.44	  f4(I0, I1, I2, I3, I4, I5, I6, I7, I8) -> f3(1, 0, 30, I3, 2, 30, I6, I7, I8) 
12.67/12.44	  f3(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f1(I9, I10, I11, I12, I13, I14, I15, I16, I17) 
12.67/12.44	  f1(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f3(I18 + I19, rnd2, I20, I21, 1 + I22, I23, I24, I18, I26) [rnd2 = I18 /\ I22 <= I23]
12.67/12.44	  f1(I27, I28, I29, I30, I31, I32, I33, I34, I35) -> f2(I27, I28, I29, I27, I31, I32, I27, I34, I27) [1 + I32 <= I31]
12.67/12.44	
12.67/12.44	We use the reverse value criterion with the projection function NU:
12.67/12.44	  NU[f1#(z1,z2,z3,z4,z5,z6,z7,z8,z9)] = z6 + -1 * z5
12.67/12.44	  NU[f3#(z1,z2,z3,z4,z5,z6,z7,z8,z9)] = z6 + -1 * z5
12.67/12.44	
12.67/12.44	This gives the following inequalities:
12.67/12.44	    ==>  I14 + -1 * I13 >= I14 + -1 * I13
12.67/12.44	  rnd2 = I18 /\ I22 <= I23  ==>  I23 + -1 * I22 > I23 + -1 * (1 + I22)  with  I23 + -1 * I22 >= 0
12.67/12.44	
12.67/12.44	We remove all the strictly oriented dependency pairs.
12.67/12.44	
12.67/12.44	DP problem for innermost termination.
12.67/12.44	P =
12.67/12.44	  f3#(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f1#(I9, I10, I11, I12, I13, I14, I15, I16, I17) 
12.67/12.44	R = 
12.67/12.44	  f5(x1, x2, x3, x4, x5, x6, x7, x8, x9) -> f4(x1, x2, x3, x4, x5, x6, x7, x8, x9) 
12.67/12.44	  f4(I0, I1, I2, I3, I4, I5, I6, I7, I8) -> f3(1, 0, 30, I3, 2, 30, I6, I7, I8) 
12.67/12.44	  f3(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f1(I9, I10, I11, I12, I13, I14, I15, I16, I17) 
12.67/12.44	  f1(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f3(I18 + I19, rnd2, I20, I21, 1 + I22, I23, I24, I18, I26) [rnd2 = I18 /\ I22 <= I23]
12.67/12.44	  f1(I27, I28, I29, I30, I31, I32, I33, I34, I35) -> f2(I27, I28, I29, I27, I31, I32, I27, I34, I27) [1 + I32 <= I31]
12.67/12.44	
12.67/12.44	The dependency graph for this problem is:
12.67/12.44	  2 -> 
12.67/12.44	Where:
12.67/12.44	  2) f3#(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f1#(I9, I10, I11, I12, I13, I14, I15, I16, I17) 
12.67/12.44	
12.67/12.44	We have the following SCCs.
12.67/12.44	  
12.67/15.42	EOF