15.45/15.21	YES
15.45/15.21	
15.45/15.21	DP problem for innermost termination.
15.45/15.21	P =
15.45/15.21	  f8#(x1, x2, x3, x4, x5) -> f7#(x1, x2, x3, x4, x5) 
15.45/15.21	  f7#(I0, I1, I2, I3, I4) -> f1#(I0, I1, I2, I3, I4) 
15.45/15.21	  f6#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
15.45/15.21	  f5#(I10, I11, I12, I13, I14) -> f6#(I10, I11, I12, 1 + I13, I14) 
15.45/15.21	  f4#(I15, I16, I17, I18, I19) -> f5#(I15, I16, I17, I18, I19) [I16 = I16]
15.45/15.21	  f1#(I20, I21, I22, I23, I24) -> f4#(I20, I21, rnd3, I23, I24) [rnd3 = rnd3 /\ 0 <= -1 - I23 + I24]
15.45/15.21	  f3#(I25, I26, I27, I28, I29) -> f1#(I25, I26, I27, I28, I29) 
15.45/15.21	  f1#(I30, I31, I32, I33, I34) -> f3#(I30, I31, I35, I33, -1 + I34) [0 <= I35 /\ I35 <= 0 /\ I35 = I35 /\ 0 <= -1 - I33 + I34]
15.45/15.21	R = 
15.45/15.21	  f8(x1, x2, x3, x4, x5) -> f7(x1, x2, x3, x4, x5) 
15.45/15.21	  f7(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, I3, I4) 
15.45/15.21	  f6(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) 
15.45/15.21	  f5(I10, I11, I12, I13, I14) -> f6(I10, I11, I12, 1 + I13, I14) 
15.45/15.21	  f4(I15, I16, I17, I18, I19) -> f5(I15, I16, I17, I18, I19) [I16 = I16]
15.45/15.21	  f1(I20, I21, I22, I23, I24) -> f4(I20, I21, rnd3, I23, I24) [rnd3 = rnd3 /\ 0 <= -1 - I23 + I24]
15.45/15.21	  f3(I25, I26, I27, I28, I29) -> f1(I25, I26, I27, I28, I29) 
15.45/15.21	  f1(I30, I31, I32, I33, I34) -> f3(I30, I31, I35, I33, -1 + I34) [0 <= I35 /\ I35 <= 0 /\ I35 = I35 /\ 0 <= -1 - I33 + I34]
15.45/15.21	  f1(I36, I37, I38, I39, I40) -> f2(rnd1, I37, I38, I39, I40) [rnd1 = rnd1 /\ -1 * I39 + I40 <= 0]
15.45/15.21	
15.45/15.21	The dependency graph for this problem is:
15.45/15.21	  0 -> 1
15.45/15.21	  1 -> 5, 7
15.45/15.21	  2 -> 5, 7
15.45/15.21	  3 -> 2
15.45/15.21	  4 -> 3
15.45/15.21	  5 -> 4
15.45/15.21	  6 -> 5, 7
15.45/15.21	  7 -> 6
15.45/15.21	Where:
15.45/15.21	  0) f8#(x1, x2, x3, x4, x5) -> f7#(x1, x2, x3, x4, x5) 
15.45/15.21	  1) f7#(I0, I1, I2, I3, I4) -> f1#(I0, I1, I2, I3, I4) 
15.45/15.21	  2) f6#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
15.45/15.21	  3) f5#(I10, I11, I12, I13, I14) -> f6#(I10, I11, I12, 1 + I13, I14) 
15.45/15.21	  4) f4#(I15, I16, I17, I18, I19) -> f5#(I15, I16, I17, I18, I19) [I16 = I16]
15.45/15.21	  5) f1#(I20, I21, I22, I23, I24) -> f4#(I20, I21, rnd3, I23, I24) [rnd3 = rnd3 /\ 0 <= -1 - I23 + I24]
15.45/15.21	  6) f3#(I25, I26, I27, I28, I29) -> f1#(I25, I26, I27, I28, I29) 
15.45/15.21	  7) f1#(I30, I31, I32, I33, I34) -> f3#(I30, I31, I35, I33, -1 + I34) [0 <= I35 /\ I35 <= 0 /\ I35 = I35 /\ 0 <= -1 - I33 + I34]
15.45/15.21	
15.45/15.21	We have the following SCCs.
15.45/15.21	  { 2, 3, 4, 5, 6, 7 }
15.45/15.21	
15.45/15.21	DP problem for innermost termination.
15.45/15.21	P =
15.45/15.21	  f6#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
15.45/15.21	  f5#(I10, I11, I12, I13, I14) -> f6#(I10, I11, I12, 1 + I13, I14) 
15.45/15.21	  f4#(I15, I16, I17, I18, I19) -> f5#(I15, I16, I17, I18, I19) [I16 = I16]
15.45/15.21	  f1#(I20, I21, I22, I23, I24) -> f4#(I20, I21, rnd3, I23, I24) [rnd3 = rnd3 /\ 0 <= -1 - I23 + I24]
15.45/15.21	  f3#(I25, I26, I27, I28, I29) -> f1#(I25, I26, I27, I28, I29) 
15.45/15.21	  f1#(I30, I31, I32, I33, I34) -> f3#(I30, I31, I35, I33, -1 + I34) [0 <= I35 /\ I35 <= 0 /\ I35 = I35 /\ 0 <= -1 - I33 + I34]
15.45/15.21	R = 
15.45/15.21	  f8(x1, x2, x3, x4, x5) -> f7(x1, x2, x3, x4, x5) 
15.45/15.21	  f7(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, I3, I4) 
15.45/15.21	  f6(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) 
15.45/15.21	  f5(I10, I11, I12, I13, I14) -> f6(I10, I11, I12, 1 + I13, I14) 
15.45/15.21	  f4(I15, I16, I17, I18, I19) -> f5(I15, I16, I17, I18, I19) [I16 = I16]
15.45/15.21	  f1(I20, I21, I22, I23, I24) -> f4(I20, I21, rnd3, I23, I24) [rnd3 = rnd3 /\ 0 <= -1 - I23 + I24]
15.45/15.21	  f3(I25, I26, I27, I28, I29) -> f1(I25, I26, I27, I28, I29) 
15.45/15.21	  f1(I30, I31, I32, I33, I34) -> f3(I30, I31, I35, I33, -1 + I34) [0 <= I35 /\ I35 <= 0 /\ I35 = I35 /\ 0 <= -1 - I33 + I34]
15.45/15.21	  f1(I36, I37, I38, I39, I40) -> f2(rnd1, I37, I38, I39, I40) [rnd1 = rnd1 /\ -1 * I39 + I40 <= 0]
15.45/15.21	
15.45/15.21	We use the extended value criterion with the projection function NU:
15.45/15.21	  NU[f3#(x0,x1,x2,x3,x4)] = -x3 + x4 - 1
15.45/15.21	  NU[f4#(x0,x1,x2,x3,x4)] = -x3 + x4 - 2
15.45/15.21	  NU[f5#(x0,x1,x2,x3,x4)] = -x3 + x4 - 2
15.45/15.21	  NU[f1#(x0,x1,x2,x3,x4)] = -x3 + x4 - 1
15.45/15.21	  NU[f6#(x0,x1,x2,x3,x4)] = -x3 + x4 - 1
15.45/15.21	
15.45/15.21	This gives the following inequalities:
15.45/15.21	    ==>  -I8 + I9 - 1 >= -I8 + I9 - 1
15.45/15.21	    ==>  -I13 + I14 - 2 >= -(1 + I13) + I14 - 1
15.45/15.21	  I16 = I16  ==>  -I18 + I19 - 2 >= -I18 + I19 - 2
15.45/15.21	  rnd3 = rnd3 /\ 0 <= -1 - I23 + I24  ==>  -I23 + I24 - 1 > -I23 + I24 - 2  with  -I23 + I24 - 1 >= 0
15.45/15.21	    ==>  -I28 + I29 - 1 >= -I28 + I29 - 1
15.45/15.21	  0 <= I35 /\ I35 <= 0 /\ I35 = I35 /\ 0 <= -1 - I33 + I34  ==>  -I33 + I34 - 1 >= -I33 + (-1 + I34) - 1
15.45/15.21	
15.45/15.21	We remove all the strictly oriented dependency pairs.
15.45/15.21	
15.45/15.21	DP problem for innermost termination.
15.45/15.21	P =
15.45/15.21	  f6#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
15.45/15.21	  f5#(I10, I11, I12, I13, I14) -> f6#(I10, I11, I12, 1 + I13, I14) 
15.45/15.21	  f4#(I15, I16, I17, I18, I19) -> f5#(I15, I16, I17, I18, I19) [I16 = I16]
15.45/15.21	  f3#(I25, I26, I27, I28, I29) -> f1#(I25, I26, I27, I28, I29) 
15.45/15.21	  f1#(I30, I31, I32, I33, I34) -> f3#(I30, I31, I35, I33, -1 + I34) [0 <= I35 /\ I35 <= 0 /\ I35 = I35 /\ 0 <= -1 - I33 + I34]
15.45/15.21	R = 
15.45/15.21	  f8(x1, x2, x3, x4, x5) -> f7(x1, x2, x3, x4, x5) 
15.45/15.21	  f7(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, I3, I4) 
15.45/15.21	  f6(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) 
15.45/15.21	  f5(I10, I11, I12, I13, I14) -> f6(I10, I11, I12, 1 + I13, I14) 
15.45/15.21	  f4(I15, I16, I17, I18, I19) -> f5(I15, I16, I17, I18, I19) [I16 = I16]
15.45/15.21	  f1(I20, I21, I22, I23, I24) -> f4(I20, I21, rnd3, I23, I24) [rnd3 = rnd3 /\ 0 <= -1 - I23 + I24]
15.45/15.21	  f3(I25, I26, I27, I28, I29) -> f1(I25, I26, I27, I28, I29) 
15.45/15.21	  f1(I30, I31, I32, I33, I34) -> f3(I30, I31, I35, I33, -1 + I34) [0 <= I35 /\ I35 <= 0 /\ I35 = I35 /\ 0 <= -1 - I33 + I34]
15.45/15.21	  f1(I36, I37, I38, I39, I40) -> f2(rnd1, I37, I38, I39, I40) [rnd1 = rnd1 /\ -1 * I39 + I40 <= 0]
15.45/15.21	
15.45/15.21	The dependency graph for this problem is:
15.45/15.21	  2 -> 7
15.45/15.21	  3 -> 2
15.45/15.21	  4 -> 3
15.45/15.21	  6 -> 7
15.45/15.21	  7 -> 6
15.45/15.21	Where:
15.45/15.21	  2) f6#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
15.45/15.21	  3) f5#(I10, I11, I12, I13, I14) -> f6#(I10, I11, I12, 1 + I13, I14) 
15.45/15.21	  4) f4#(I15, I16, I17, I18, I19) -> f5#(I15, I16, I17, I18, I19) [I16 = I16]
15.45/15.21	  6) f3#(I25, I26, I27, I28, I29) -> f1#(I25, I26, I27, I28, I29) 
15.45/15.21	  7) f1#(I30, I31, I32, I33, I34) -> f3#(I30, I31, I35, I33, -1 + I34) [0 <= I35 /\ I35 <= 0 /\ I35 = I35 /\ 0 <= -1 - I33 + I34]
15.45/15.21	
15.45/15.21	We have the following SCCs.
15.45/15.21	  { 6, 7 }
15.45/15.21	
15.45/15.21	DP problem for innermost termination.
15.45/15.21	P =
15.45/15.21	  f3#(I25, I26, I27, I28, I29) -> f1#(I25, I26, I27, I28, I29) 
15.45/15.21	  f1#(I30, I31, I32, I33, I34) -> f3#(I30, I31, I35, I33, -1 + I34) [0 <= I35 /\ I35 <= 0 /\ I35 = I35 /\ 0 <= -1 - I33 + I34]
15.45/15.21	R = 
15.45/15.21	  f8(x1, x2, x3, x4, x5) -> f7(x1, x2, x3, x4, x5) 
15.45/15.21	  f7(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, I3, I4) 
15.45/15.21	  f6(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) 
15.45/15.21	  f5(I10, I11, I12, I13, I14) -> f6(I10, I11, I12, 1 + I13, I14) 
15.45/15.21	  f4(I15, I16, I17, I18, I19) -> f5(I15, I16, I17, I18, I19) [I16 = I16]
15.45/15.21	  f1(I20, I21, I22, I23, I24) -> f4(I20, I21, rnd3, I23, I24) [rnd3 = rnd3 /\ 0 <= -1 - I23 + I24]
15.45/15.21	  f3(I25, I26, I27, I28, I29) -> f1(I25, I26, I27, I28, I29) 
15.45/15.21	  f1(I30, I31, I32, I33, I34) -> f3(I30, I31, I35, I33, -1 + I34) [0 <= I35 /\ I35 <= 0 /\ I35 = I35 /\ 0 <= -1 - I33 + I34]
15.45/15.21	  f1(I36, I37, I38, I39, I40) -> f2(rnd1, I37, I38, I39, I40) [rnd1 = rnd1 /\ -1 * I39 + I40 <= 0]
15.45/15.21	
15.45/15.21	We use the reverse value criterion with the projection function NU:
15.45/15.21	  NU[f1#(z1,z2,z3,z4,z5)] = -1 - z4 + z5 + -1 * 0
15.45/15.21	  NU[f3#(z1,z2,z3,z4,z5)] = -1 - z4 + z5 + -1 * 0
15.45/15.21	
15.45/15.21	This gives the following inequalities:
15.45/15.21	    ==>  -1 - I28 + I29 + -1 * 0 >= -1 - I28 + I29 + -1 * 0
15.45/15.21	  0 <= I35 /\ I35 <= 0 /\ I35 = I35 /\ 0 <= -1 - I33 + I34  ==>  -1 - I33 + I34 + -1 * 0 > -1 - I33 + (-1 + I34) + -1 * 0  with  -1 - I33 + I34 + -1 * 0 >= 0
15.45/15.21	
15.45/15.21	We remove all the strictly oriented dependency pairs.
15.45/15.21	
15.45/15.21	DP problem for innermost termination.
15.45/15.21	P =
15.45/15.21	  f3#(I25, I26, I27, I28, I29) -> f1#(I25, I26, I27, I28, I29) 
15.45/15.21	R = 
15.45/15.21	  f8(x1, x2, x3, x4, x5) -> f7(x1, x2, x3, x4, x5) 
15.45/15.21	  f7(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, I3, I4) 
15.45/15.21	  f6(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) 
15.45/15.21	  f5(I10, I11, I12, I13, I14) -> f6(I10, I11, I12, 1 + I13, I14) 
15.45/15.21	  f4(I15, I16, I17, I18, I19) -> f5(I15, I16, I17, I18, I19) [I16 = I16]
15.45/15.21	  f1(I20, I21, I22, I23, I24) -> f4(I20, I21, rnd3, I23, I24) [rnd3 = rnd3 /\ 0 <= -1 - I23 + I24]
15.45/15.21	  f3(I25, I26, I27, I28, I29) -> f1(I25, I26, I27, I28, I29) 
15.45/15.21	  f1(I30, I31, I32, I33, I34) -> f3(I30, I31, I35, I33, -1 + I34) [0 <= I35 /\ I35 <= 0 /\ I35 = I35 /\ 0 <= -1 - I33 + I34]
15.45/15.21	  f1(I36, I37, I38, I39, I40) -> f2(rnd1, I37, I38, I39, I40) [rnd1 = rnd1 /\ -1 * I39 + I40 <= 0]
15.45/15.21	
15.45/15.21	The dependency graph for this problem is:
15.45/15.21	  6 -> 
15.45/15.21	Where:
15.45/15.21	  6) f3#(I25, I26, I27, I28, I29) -> f1#(I25, I26, I27, I28, I29) 
15.45/15.21	
15.45/15.21	We have the following SCCs.
15.45/15.21	  
15.45/18.19	EOF