4.12/4.12	YES
4.12/4.12	
4.12/4.12	DP problem for innermost termination.
4.12/4.12	P =
4.12/4.12	  f6#(x1, x2, x3, x4, x5) -> f5#(x1, x2, x3, x4, x5) 
4.12/4.12	  f5#(I0, I1, I2, I3, I4) -> f4#(I0, I1, 0, I3, I4) 
4.12/4.12	  f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
4.12/4.12	  f4#(I15, I16, I17, I18, I19) -> f3#(I15, I16, I17, I18, rnd5) [rnd5 = rnd5 /\ I18 <= I15]
4.12/4.12	  f1#(I20, I21, I22, I23, I24) -> f3#(I20, I21, 1 + I22, I23, 1 + I24) [1 + I22 <= I21]
4.12/4.12	R = 
4.12/4.12	  f6(x1, x2, x3, x4, x5) -> f5(x1, x2, x3, x4, x5) 
4.12/4.12	  f5(I0, I1, I2, I3, I4) -> f4(I0, I1, 0, I3, I4) 
4.12/4.12	  f3(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) 
4.12/4.12	  f4(I10, I11, I12, I13, I14) -> f2(I10, I11, I12, I13, I14) [1 + I10 <= I13]
4.12/4.12	  f4(I15, I16, I17, I18, I19) -> f3(I15, I16, I17, I18, rnd5) [rnd5 = rnd5 /\ I18 <= I15]
4.12/4.12	  f1(I20, I21, I22, I23, I24) -> f3(I20, I21, 1 + I22, I23, 1 + I24) [1 + I22 <= I21]
4.12/4.12	  f1(I25, I26, I27, I28, I29) -> f2(I25, I26, I27, I28, I29) [I26 <= I27]
4.12/4.12	
4.12/4.12	The dependency graph for this problem is:
4.12/4.12	  0 -> 1
4.12/4.12	  1 -> 3
4.12/4.12	  2 -> 4
4.12/4.12	  3 -> 2
4.12/4.12	  4 -> 2
4.12/4.12	Where:
4.12/4.12	  0) f6#(x1, x2, x3, x4, x5) -> f5#(x1, x2, x3, x4, x5) 
4.12/4.12	  1) f5#(I0, I1, I2, I3, I4) -> f4#(I0, I1, 0, I3, I4) 
4.12/4.12	  2) f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
4.12/4.12	  3) f4#(I15, I16, I17, I18, I19) -> f3#(I15, I16, I17, I18, rnd5) [rnd5 = rnd5 /\ I18 <= I15]
4.12/4.12	  4) f1#(I20, I21, I22, I23, I24) -> f3#(I20, I21, 1 + I22, I23, 1 + I24) [1 + I22 <= I21]
4.12/4.12	
4.12/4.12	We have the following SCCs.
4.12/4.12	  { 2, 4 }
4.12/4.12	
4.12/4.12	DP problem for innermost termination.
4.12/4.12	P =
4.12/4.12	  f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
4.12/4.12	  f1#(I20, I21, I22, I23, I24) -> f3#(I20, I21, 1 + I22, I23, 1 + I24) [1 + I22 <= I21]
4.12/4.12	R = 
4.12/4.12	  f6(x1, x2, x3, x4, x5) -> f5(x1, x2, x3, x4, x5) 
4.12/4.12	  f5(I0, I1, I2, I3, I4) -> f4(I0, I1, 0, I3, I4) 
4.12/4.12	  f3(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) 
4.12/4.12	  f4(I10, I11, I12, I13, I14) -> f2(I10, I11, I12, I13, I14) [1 + I10 <= I13]
4.12/4.12	  f4(I15, I16, I17, I18, I19) -> f3(I15, I16, I17, I18, rnd5) [rnd5 = rnd5 /\ I18 <= I15]
4.12/4.12	  f1(I20, I21, I22, I23, I24) -> f3(I20, I21, 1 + I22, I23, 1 + I24) [1 + I22 <= I21]
4.12/4.12	  f1(I25, I26, I27, I28, I29) -> f2(I25, I26, I27, I28, I29) [I26 <= I27]
4.12/4.12	
4.12/4.12	We use the reverse value criterion with the projection function NU:
4.12/4.12	  NU[f1#(z1,z2,z3,z4,z5)] = z2 + -1 * (1 + z3)
4.12/4.12	  NU[f3#(z1,z2,z3,z4,z5)] = z2 + -1 * (1 + z3)
4.12/4.12	
4.12/4.12	This gives the following inequalities:
4.12/4.12	    ==>  I6 + -1 * (1 + I7) >= I6 + -1 * (1 + I7)
4.12/4.12	  1 + I22 <= I21  ==>  I21 + -1 * (1 + I22) > I21 + -1 * (1 + (1 + I22))  with  I21 + -1 * (1 + I22) >= 0
4.12/4.12	
4.12/4.12	We remove all the strictly oriented dependency pairs.
4.12/4.12	
4.12/4.12	DP problem for innermost termination.
4.12/4.12	P =
4.12/4.12	  f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
4.12/4.12	R = 
4.12/4.12	  f6(x1, x2, x3, x4, x5) -> f5(x1, x2, x3, x4, x5) 
4.12/4.12	  f5(I0, I1, I2, I3, I4) -> f4(I0, I1, 0, I3, I4) 
4.12/4.12	  f3(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) 
4.12/4.12	  f4(I10, I11, I12, I13, I14) -> f2(I10, I11, I12, I13, I14) [1 + I10 <= I13]
4.12/4.12	  f4(I15, I16, I17, I18, I19) -> f3(I15, I16, I17, I18, rnd5) [rnd5 = rnd5 /\ I18 <= I15]
4.12/4.12	  f1(I20, I21, I22, I23, I24) -> f3(I20, I21, 1 + I22, I23, 1 + I24) [1 + I22 <= I21]
4.12/4.12	  f1(I25, I26, I27, I28, I29) -> f2(I25, I26, I27, I28, I29) [I26 <= I27]
4.12/4.12	
4.12/4.12	The dependency graph for this problem is:
4.12/4.12	  2 -> 
4.12/4.12	Where:
4.12/4.12	  2) f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
4.12/4.12	
4.12/4.12	We have the following SCCs.
4.12/4.12	  
4.12/7.09	EOF