9.12/9.03	YES
9.12/9.03	
9.12/9.03	DP problem for innermost termination.
9.12/9.03	P =
9.12/9.03	  f7#(x1, x2, x3, x4, x5) -> f6#(x1, x2, x3, x4, x5) 
9.12/9.03	  f4#(I5, I6, I7, I8, I9) -> f1#(I5, I6, 1, 1 + I8, I9) [0 <= I7 /\ I7 <= 0 /\ 0 <= -1 - I8 + I9]
9.12/9.03	  f6#(I10, I11, I12, I13, I14) -> f4#(I10, I11, 0, I13, I14) 
9.12/9.03	  f3#(I21, I22, I23, I24, I25) -> f4#(I21, I22, 0, I24, -1 + I25) 
9.12/9.03	  f2#(I26, I27, I28, I29, I30) -> f3#(I26, I27, I28, I29, I30) [I27 = I27]
9.12/9.03	  f1#(I31, I32, I33, I34, I35) -> f2#(I31, I32, I33, I34, I35) [0 <= -1 - I34 + I35]
9.12/9.03	R = 
9.12/9.03	  f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) 
9.12/9.03	  f4(I0, I1, I2, I3, I4) -> f5(rnd1, I1, I2, I3, I4) [rnd1 = rnd1 /\ -1 * I3 + I4 <= 0]
9.12/9.03	  f4(I5, I6, I7, I8, I9) -> f1(I5, I6, 1, 1 + I8, I9) [0 <= I7 /\ I7 <= 0 /\ 0 <= -1 - I8 + I9]
9.12/9.03	  f6(I10, I11, I12, I13, I14) -> f4(I10, I11, 0, I13, I14) 
9.12/9.03	  f1(I15, I16, I17, I18, I19) -> f5(I20, I16, I17, I18, I19) [I20 = I20 /\ -1 * I18 + I19 <= 0]
9.12/9.03	  f3(I21, I22, I23, I24, I25) -> f4(I21, I22, 0, I24, -1 + I25) 
9.12/9.03	  f2(I26, I27, I28, I29, I30) -> f3(I26, I27, I28, I29, I30) [I27 = I27]
9.12/9.03	  f1(I31, I32, I33, I34, I35) -> f2(I31, I32, I33, I34, I35) [0 <= -1 - I34 + I35]
9.12/9.03	
9.12/9.03	The dependency graph for this problem is:
9.12/9.03	  0 -> 2
9.12/9.03	  1 -> 5
9.12/9.03	  2 -> 1
9.12/9.03	  3 -> 1
9.12/9.03	  4 -> 3
9.12/9.03	  5 -> 4
9.12/9.03	Where:
9.12/9.03	  0) f7#(x1, x2, x3, x4, x5) -> f6#(x1, x2, x3, x4, x5) 
9.12/9.03	  1) f4#(I5, I6, I7, I8, I9) -> f1#(I5, I6, 1, 1 + I8, I9) [0 <= I7 /\ I7 <= 0 /\ 0 <= -1 - I8 + I9]
9.12/9.03	  2) f6#(I10, I11, I12, I13, I14) -> f4#(I10, I11, 0, I13, I14) 
9.12/9.03	  3) f3#(I21, I22, I23, I24, I25) -> f4#(I21, I22, 0, I24, -1 + I25) 
9.12/9.03	  4) f2#(I26, I27, I28, I29, I30) -> f3#(I26, I27, I28, I29, I30) [I27 = I27]
9.12/9.03	  5) f1#(I31, I32, I33, I34, I35) -> f2#(I31, I32, I33, I34, I35) [0 <= -1 - I34 + I35]
9.12/9.03	
9.12/9.03	We have the following SCCs.
9.12/9.03	  { 1, 3, 4, 5 }
9.12/9.03	
9.12/9.03	DP problem for innermost termination.
9.12/9.03	P =
9.12/9.03	  f4#(I5, I6, I7, I8, I9) -> f1#(I5, I6, 1, 1 + I8, I9) [0 <= I7 /\ I7 <= 0 /\ 0 <= -1 - I8 + I9]
9.12/9.03	  f3#(I21, I22, I23, I24, I25) -> f4#(I21, I22, 0, I24, -1 + I25) 
9.12/9.03	  f2#(I26, I27, I28, I29, I30) -> f3#(I26, I27, I28, I29, I30) [I27 = I27]
9.12/9.03	  f1#(I31, I32, I33, I34, I35) -> f2#(I31, I32, I33, I34, I35) [0 <= -1 - I34 + I35]
9.12/9.03	R = 
9.12/9.03	  f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) 
9.12/9.03	  f4(I0, I1, I2, I3, I4) -> f5(rnd1, I1, I2, I3, I4) [rnd1 = rnd1 /\ -1 * I3 + I4 <= 0]
9.12/9.03	  f4(I5, I6, I7, I8, I9) -> f1(I5, I6, 1, 1 + I8, I9) [0 <= I7 /\ I7 <= 0 /\ 0 <= -1 - I8 + I9]
9.12/9.03	  f6(I10, I11, I12, I13, I14) -> f4(I10, I11, 0, I13, I14) 
9.12/9.03	  f1(I15, I16, I17, I18, I19) -> f5(I20, I16, I17, I18, I19) [I20 = I20 /\ -1 * I18 + I19 <= 0]
9.12/9.03	  f3(I21, I22, I23, I24, I25) -> f4(I21, I22, 0, I24, -1 + I25) 
9.12/9.03	  f2(I26, I27, I28, I29, I30) -> f3(I26, I27, I28, I29, I30) [I27 = I27]
9.12/9.03	  f1(I31, I32, I33, I34, I35) -> f2(I31, I32, I33, I34, I35) [0 <= -1 - I34 + I35]
9.12/9.03	
9.12/9.03	We use the extended value criterion with the projection function NU:
9.12/9.03	  NU[f2#(x0,x1,x2,x3,x4)] = -x3 + x4 - 2
9.12/9.03	  NU[f3#(x0,x1,x2,x3,x4)] = -x3 + x4 - 2
9.12/9.03	  NU[f1#(x0,x1,x2,x3,x4)] = -x3 + x4 - 1
9.12/9.03	  NU[f4#(x0,x1,x2,x3,x4)] = -x2 - x3 + x4 - 2
9.12/9.03	
9.12/9.03	This gives the following inequalities:
9.12/9.03	  0 <= I7 /\ I7 <= 0 /\ 0 <= -1 - I8 + I9  ==>  -I7 - I8 + I9 - 2 >= -(1 + I8) + I9 - 1
9.12/9.03	    ==>  -I24 + I25 - 2 >= -0 - I24 + (-1 + I25) - 2
9.12/9.03	  I27 = I27  ==>  -I29 + I30 - 2 >= -I29 + I30 - 2
9.12/9.03	  0 <= -1 - I34 + I35  ==>  -I34 + I35 - 1 > -I34 + I35 - 2  with  -I34 + I35 - 1 >= 0
9.12/9.03	
9.12/9.03	We remove all the strictly oriented dependency pairs.
9.12/9.03	
9.12/9.03	DP problem for innermost termination.
9.12/9.03	P =
9.12/9.03	  f4#(I5, I6, I7, I8, I9) -> f1#(I5, I6, 1, 1 + I8, I9) [0 <= I7 /\ I7 <= 0 /\ 0 <= -1 - I8 + I9]
9.12/9.03	  f3#(I21, I22, I23, I24, I25) -> f4#(I21, I22, 0, I24, -1 + I25) 
9.12/9.03	  f2#(I26, I27, I28, I29, I30) -> f3#(I26, I27, I28, I29, I30) [I27 = I27]
9.12/9.03	R = 
9.12/9.03	  f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) 
9.12/9.03	  f4(I0, I1, I2, I3, I4) -> f5(rnd1, I1, I2, I3, I4) [rnd1 = rnd1 /\ -1 * I3 + I4 <= 0]
9.12/9.03	  f4(I5, I6, I7, I8, I9) -> f1(I5, I6, 1, 1 + I8, I9) [0 <= I7 /\ I7 <= 0 /\ 0 <= -1 - I8 + I9]
9.12/9.03	  f6(I10, I11, I12, I13, I14) -> f4(I10, I11, 0, I13, I14) 
9.12/9.03	  f1(I15, I16, I17, I18, I19) -> f5(I20, I16, I17, I18, I19) [I20 = I20 /\ -1 * I18 + I19 <= 0]
9.12/9.03	  f3(I21, I22, I23, I24, I25) -> f4(I21, I22, 0, I24, -1 + I25) 
9.12/9.03	  f2(I26, I27, I28, I29, I30) -> f3(I26, I27, I28, I29, I30) [I27 = I27]
9.12/9.03	  f1(I31, I32, I33, I34, I35) -> f2(I31, I32, I33, I34, I35) [0 <= -1 - I34 + I35]
9.12/9.03	
9.12/9.03	The dependency graph for this problem is:
9.12/9.03	  1 -> 
9.12/9.03	  3 -> 1
9.12/9.03	  4 -> 3
9.12/9.03	Where:
9.12/9.03	  1) f4#(I5, I6, I7, I8, I9) -> f1#(I5, I6, 1, 1 + I8, I9) [0 <= I7 /\ I7 <= 0 /\ 0 <= -1 - I8 + I9]
9.12/9.03	  3) f3#(I21, I22, I23, I24, I25) -> f4#(I21, I22, 0, I24, -1 + I25) 
9.12/9.03	  4) f2#(I26, I27, I28, I29, I30) -> f3#(I26, I27, I28, I29, I30) [I27 = I27]
9.12/9.03	
9.12/9.03	We have the following SCCs.
9.12/9.03	  
9.12/12.01	EOF