5.92/6.24	YES
5.92/6.24	
5.92/6.24	DP problem for innermost termination.
5.92/6.24	P =
5.92/6.24	  f6#(x1, x2, x3, x4, x5, x6) -> f1#(x1, x2, x3, x4, x5, x6) 
5.92/6.24	  f2#(I0, I1, I2, I3, I4, I5) -> f3#(1 + I0, I1, I2, I3, I4, I5) [1 + I0 <= 500]
5.92/6.24	  f5#(I12, I13, I14, I15, I16, I17) -> f3#(I12, I13, I14, I15, I16, I17) 
5.92/6.24	  f3#(I18, I19, I20, I21, I22, I23) -> f5#(1 + I18, I19, I20, I21, I22, I23) [1 + I18 <= 500]
5.92/6.24	  f1#(I30, I31, I32, I33, I34, I35) -> f2#(rnd1, I33, I32, I33, I34, I35) [y1 = I34 /\ rnd1 = I33]
5.92/6.24	R = 
5.92/6.24	  f6(x1, x2, x3, x4, x5, x6) -> f1(x1, x2, x3, x4, x5, x6) 
5.92/6.24	  f2(I0, I1, I2, I3, I4, I5) -> f3(1 + I0, I1, I2, I3, I4, I5) [1 + I0 <= 500]
5.92/6.24	  f2(I6, I7, I8, I9, I10, I11) -> f4(I6, I7, I11, I9, I10, I11) [500 <= I6]
5.92/6.24	  f5(I12, I13, I14, I15, I16, I17) -> f3(I12, I13, I14, I15, I16, I17) 
5.92/6.24	  f3(I18, I19, I20, I21, I22, I23) -> f5(1 + I18, I19, I20, I21, I22, I23) [1 + I18 <= 500]
5.92/6.24	  f3(I24, I25, I26, I27, I28, I29) -> f4(I24, I25, I29, I27, I28, I29) [500 <= I24]
5.92/6.24	  f1(I30, I31, I32, I33, I34, I35) -> f2(rnd1, I33, I32, I33, I34, I35) [y1 = I34 /\ rnd1 = I33]
5.92/6.24	
5.92/6.24	The dependency graph for this problem is:
5.92/6.24	  0 -> 4
5.92/6.24	  1 -> 3
5.92/6.24	  2 -> 3
5.92/6.24	  3 -> 2
5.92/6.24	  4 -> 1
5.92/6.24	Where:
5.92/6.24	  0) f6#(x1, x2, x3, x4, x5, x6) -> f1#(x1, x2, x3, x4, x5, x6) 
5.92/6.24	  1) f2#(I0, I1, I2, I3, I4, I5) -> f3#(1 + I0, I1, I2, I3, I4, I5) [1 + I0 <= 500]
5.92/6.24	  2) f5#(I12, I13, I14, I15, I16, I17) -> f3#(I12, I13, I14, I15, I16, I17) 
5.92/6.24	  3) f3#(I18, I19, I20, I21, I22, I23) -> f5#(1 + I18, I19, I20, I21, I22, I23) [1 + I18 <= 500]
5.92/6.24	  4) f1#(I30, I31, I32, I33, I34, I35) -> f2#(rnd1, I33, I32, I33, I34, I35) [y1 = I34 /\ rnd1 = I33]
5.92/6.24	
5.92/6.24	We have the following SCCs.
5.92/6.24	  { 2, 3 }
5.92/6.24	
5.92/6.24	DP problem for innermost termination.
5.92/6.24	P =
5.92/6.24	  f5#(I12, I13, I14, I15, I16, I17) -> f3#(I12, I13, I14, I15, I16, I17) 
5.92/6.24	  f3#(I18, I19, I20, I21, I22, I23) -> f5#(1 + I18, I19, I20, I21, I22, I23) [1 + I18 <= 500]
5.92/6.24	R = 
5.92/6.24	  f6(x1, x2, x3, x4, x5, x6) -> f1(x1, x2, x3, x4, x5, x6) 
5.92/6.24	  f2(I0, I1, I2, I3, I4, I5) -> f3(1 + I0, I1, I2, I3, I4, I5) [1 + I0 <= 500]
5.92/6.24	  f2(I6, I7, I8, I9, I10, I11) -> f4(I6, I7, I11, I9, I10, I11) [500 <= I6]
5.92/6.24	  f5(I12, I13, I14, I15, I16, I17) -> f3(I12, I13, I14, I15, I16, I17) 
5.92/6.24	  f3(I18, I19, I20, I21, I22, I23) -> f5(1 + I18, I19, I20, I21, I22, I23) [1 + I18 <= 500]
5.92/6.24	  f3(I24, I25, I26, I27, I28, I29) -> f4(I24, I25, I29, I27, I28, I29) [500 <= I24]
5.92/6.24	  f1(I30, I31, I32, I33, I34, I35) -> f2(rnd1, I33, I32, I33, I34, I35) [y1 = I34 /\ rnd1 = I33]
5.92/6.24	
5.92/6.24	We use the reverse value criterion with the projection function NU:
5.92/6.24	  NU[f3#(z1,z2,z3,z4,z5,z6)] = 500 + -1 * (1 + z1)
5.92/6.24	  NU[f5#(z1,z2,z3,z4,z5,z6)] = 500 + -1 * (1 + z1)
5.92/6.24	
5.92/6.24	This gives the following inequalities:
5.92/6.24	    ==>  500 + -1 * (1 + I12) >= 500 + -1 * (1 + I12)
5.92/6.24	  1 + I18 <= 500  ==>  500 + -1 * (1 + I18) > 500 + -1 * (1 + (1 + I18))  with  500 + -1 * (1 + I18) >= 0
5.92/6.24	
5.92/6.24	We remove all the strictly oriented dependency pairs.
5.92/6.24	
5.92/6.24	DP problem for innermost termination.
5.92/6.24	P =
5.92/6.24	  f5#(I12, I13, I14, I15, I16, I17) -> f3#(I12, I13, I14, I15, I16, I17) 
5.92/6.24	R = 
5.92/6.24	  f6(x1, x2, x3, x4, x5, x6) -> f1(x1, x2, x3, x4, x5, x6) 
5.92/6.24	  f2(I0, I1, I2, I3, I4, I5) -> f3(1 + I0, I1, I2, I3, I4, I5) [1 + I0 <= 500]
5.92/6.24	  f2(I6, I7, I8, I9, I10, I11) -> f4(I6, I7, I11, I9, I10, I11) [500 <= I6]
5.92/6.24	  f5(I12, I13, I14, I15, I16, I17) -> f3(I12, I13, I14, I15, I16, I17) 
5.92/6.24	  f3(I18, I19, I20, I21, I22, I23) -> f5(1 + I18, I19, I20, I21, I22, I23) [1 + I18 <= 500]
5.92/6.24	  f3(I24, I25, I26, I27, I28, I29) -> f4(I24, I25, I29, I27, I28, I29) [500 <= I24]
5.92/6.24	  f1(I30, I31, I32, I33, I34, I35) -> f2(rnd1, I33, I32, I33, I34, I35) [y1 = I34 /\ rnd1 = I33]
5.92/6.24	
5.92/6.24	The dependency graph for this problem is:
5.92/6.24	  2 -> 
5.92/6.24	Where:
5.92/6.24	  2) f5#(I12, I13, I14, I15, I16, I17) -> f3#(I12, I13, I14, I15, I16, I17) 
5.92/6.24	
5.92/6.24	We have the following SCCs.
5.92/6.24	  
5.92/9.22	EOF