5.07/5.57	YES
5.07/5.57	
5.07/5.57	DP problem for innermost termination.
5.07/5.57	P =
5.07/5.57	  f6#(x1, x2, x3, x4, x5) -> f4#(x1, x2, x3, x4, x5) 
5.07/5.57	  f5#(I0, I1, I2, I3, I4) -> f1#(I0, I1, I2, 1 + I3, I4) [0 <= -1 + I4 /\ 0 <= 100 - I3]
5.07/5.57	  f4#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, 0, I14) 
5.07/5.57	  f3#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) 
5.07/5.57	  f1#(I20, I21, I22, I23, I24) -> f3#(I20, I21, I22, 1 + I23, I24) [0 <= -1 + I24 /\ 0 <= 100 - I23]
5.07/5.57	R = 
5.07/5.57	  f6(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) 
5.07/5.57	  f5(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, 1 + I3, I4) [0 <= -1 + I4 /\ 0 <= 100 - I3]
5.07/5.57	  f5(I5, I6, I7, I8, I9) -> f2(rnd1, rnd2, 0, I8, I9) [rnd1 = rnd2 /\ rnd2 = 0 /\ I9 <= 0 /\ 0 <= 100 - I8]
5.07/5.57	  f4(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, 0, I14) 
5.07/5.57	  f3(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I18, I19) 
5.07/5.57	  f1(I20, I21, I22, I23, I24) -> f3(I20, I21, I22, 1 + I23, I24) [0 <= -1 + I24 /\ 0 <= 100 - I23]
5.07/5.57	  f1(I25, I26, I27, I28, I29) -> f2(I30, I31, 0, I28, I29) [I30 = I31 /\ I31 = 0 /\ 101 - I28 <= 0]
5.07/5.57	
5.07/5.57	The dependency graph for this problem is:
5.07/5.57	  0 -> 2
5.07/5.57	  1 -> 4
5.07/5.57	  2 -> 1
5.07/5.57	  3 -> 4
5.07/5.57	  4 -> 3
5.07/5.57	Where:
5.07/5.57	  0) f6#(x1, x2, x3, x4, x5) -> f4#(x1, x2, x3, x4, x5) 
5.07/5.57	  1) f5#(I0, I1, I2, I3, I4) -> f1#(I0, I1, I2, 1 + I3, I4) [0 <= -1 + I4 /\ 0 <= 100 - I3]
5.07/5.57	  2) f4#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, 0, I14) 
5.07/5.57	  3) f3#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) 
5.07/5.57	  4) f1#(I20, I21, I22, I23, I24) -> f3#(I20, I21, I22, 1 + I23, I24) [0 <= -1 + I24 /\ 0 <= 100 - I23]
5.07/5.57	
5.07/5.57	We have the following SCCs.
5.07/5.57	  { 3, 4 }
5.07/5.57	
5.07/5.57	DP problem for innermost termination.
5.07/5.57	P =
5.07/5.57	  f3#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) 
5.07/5.57	  f1#(I20, I21, I22, I23, I24) -> f3#(I20, I21, I22, 1 + I23, I24) [0 <= -1 + I24 /\ 0 <= 100 - I23]
5.07/5.57	R = 
5.07/5.57	  f6(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) 
5.07/5.57	  f5(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, 1 + I3, I4) [0 <= -1 + I4 /\ 0 <= 100 - I3]
5.07/5.57	  f5(I5, I6, I7, I8, I9) -> f2(rnd1, rnd2, 0, I8, I9) [rnd1 = rnd2 /\ rnd2 = 0 /\ I9 <= 0 /\ 0 <= 100 - I8]
5.07/5.57	  f4(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, 0, I14) 
5.07/5.57	  f3(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I18, I19) 
5.07/5.57	  f1(I20, I21, I22, I23, I24) -> f3(I20, I21, I22, 1 + I23, I24) [0 <= -1 + I24 /\ 0 <= 100 - I23]
5.07/5.57	  f1(I25, I26, I27, I28, I29) -> f2(I30, I31, 0, I28, I29) [I30 = I31 /\ I31 = 0 /\ 101 - I28 <= 0]
5.07/5.57	
5.07/5.57	We use the reverse value criterion with the projection function NU:
5.07/5.57	  NU[f1#(z1,z2,z3,z4,z5)] = 100 - z4 + -1 * 0
5.07/5.57	  NU[f3#(z1,z2,z3,z4,z5)] = 100 - z4 + -1 * 0
5.07/5.57	
5.07/5.57	This gives the following inequalities:
5.07/5.57	    ==>  100 - I18 + -1 * 0 >= 100 - I18 + -1 * 0
5.07/5.57	  0 <= -1 + I24 /\ 0 <= 100 - I23  ==>  100 - I23 + -1 * 0 > 100 - (1 + I23) + -1 * 0  with  100 - I23 + -1 * 0 >= 0
5.07/5.57	
5.07/5.57	We remove all the strictly oriented dependency pairs.
5.07/5.57	
5.07/5.57	DP problem for innermost termination.
5.07/5.57	P =
5.07/5.57	  f3#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) 
5.07/5.57	R = 
5.07/5.57	  f6(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) 
5.07/5.57	  f5(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, 1 + I3, I4) [0 <= -1 + I4 /\ 0 <= 100 - I3]
5.07/5.57	  f5(I5, I6, I7, I8, I9) -> f2(rnd1, rnd2, 0, I8, I9) [rnd1 = rnd2 /\ rnd2 = 0 /\ I9 <= 0 /\ 0 <= 100 - I8]
5.07/5.57	  f4(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, 0, I14) 
5.07/5.57	  f3(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I18, I19) 
5.07/5.57	  f1(I20, I21, I22, I23, I24) -> f3(I20, I21, I22, 1 + I23, I24) [0 <= -1 + I24 /\ 0 <= 100 - I23]
5.07/5.57	  f1(I25, I26, I27, I28, I29) -> f2(I30, I31, 0, I28, I29) [I30 = I31 /\ I31 = 0 /\ 101 - I28 <= 0]
5.07/5.57	
5.07/5.57	The dependency graph for this problem is:
5.07/5.57	  3 -> 
5.07/5.57	Where:
5.07/5.57	  3) f3#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) 
5.07/5.57	
5.07/5.57	We have the following SCCs.
5.07/5.57	  
5.07/8.54	EOF