21.50/21.20	YES
21.50/21.20	
21.50/21.20	DP problem for innermost termination.
21.50/21.20	P =
21.50/21.20	  f8#(x1, x2, x3, x4, x5) -> f7#(x1, x2, x3, x4, x5) 
21.50/21.20	  f7#(I0, I1, I2, I3, I4) -> f1#(I0, I1, I2, I3, I4) 
21.50/21.20	  f6#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
21.50/21.20	  f1#(I10, I11, I12, I13, I14) -> f6#(I10, I11, I12, -1 * I11 + I13, 1 + I14) [-1 <= I14 /\ I14 <= -1 /\ 0 <= -1 - I13]
21.50/21.20	  f5#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) 
21.50/21.20	  f4#(I20, I21, I22, I23, I24) -> f5#(I20, I21, I22, 1 + I23, 1 + I24) 
21.50/21.20	  f3#(I25, I26, I27, I28, I29) -> f4#(I25, I26, I27, I28, I29) [I27 = I27]
21.50/21.20	  f1#(I30, I31, I32, I33, I34) -> f3#(I30, I31, I32, I33, I34) [0 <= -1 - I33]
21.50/21.20	R = 
21.50/21.20	  f8(x1, x2, x3, x4, x5) -> f7(x1, x2, x3, x4, x5) 
21.50/21.20	  f7(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, I3, I4) 
21.50/21.20	  f6(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) 
21.50/21.20	  f1(I10, I11, I12, I13, I14) -> f6(I10, I11, I12, -1 * I11 + I13, 1 + I14) [-1 <= I14 /\ I14 <= -1 /\ 0 <= -1 - I13]
21.50/21.20	  f5(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I18, I19) 
21.50/21.20	  f4(I20, I21, I22, I23, I24) -> f5(I20, I21, I22, 1 + I23, 1 + I24) 
21.50/21.20	  f3(I25, I26, I27, I28, I29) -> f4(I25, I26, I27, I28, I29) [I27 = I27]
21.50/21.20	  f1(I30, I31, I32, I33, I34) -> f3(I30, I31, I32, I33, I34) [0 <= -1 - I33]
21.50/21.20	  f1(I35, I36, I37, I38, I39) -> f2(rnd1, I36, I37, I38, I39) [rnd1 = rnd1 /\ -1 * I38 <= 0]
21.50/21.20	
21.50/21.20	The dependency graph for this problem is:
21.50/21.20	  0 -> 1
21.50/21.20	  1 -> 3, 7
21.50/21.20	  2 -> 3, 7
21.50/21.20	  3 -> 2
21.50/21.20	  4 -> 3, 7
21.50/21.20	  5 -> 4
21.50/21.20	  6 -> 5
21.50/21.20	  7 -> 6
21.50/21.20	Where:
21.50/21.20	  0) f8#(x1, x2, x3, x4, x5) -> f7#(x1, x2, x3, x4, x5) 
21.50/21.20	  1) f7#(I0, I1, I2, I3, I4) -> f1#(I0, I1, I2, I3, I4) 
21.50/21.20	  2) f6#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
21.50/21.20	  3) f1#(I10, I11, I12, I13, I14) -> f6#(I10, I11, I12, -1 * I11 + I13, 1 + I14) [-1 <= I14 /\ I14 <= -1 /\ 0 <= -1 - I13]
21.50/21.20	  4) f5#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) 
21.50/21.20	  5) f4#(I20, I21, I22, I23, I24) -> f5#(I20, I21, I22, 1 + I23, 1 + I24) 
21.50/21.20	  6) f3#(I25, I26, I27, I28, I29) -> f4#(I25, I26, I27, I28, I29) [I27 = I27]
21.50/21.20	  7) f1#(I30, I31, I32, I33, I34) -> f3#(I30, I31, I32, I33, I34) [0 <= -1 - I33]
21.50/21.20	
21.50/21.20	We have the following SCCs.
21.50/21.20	  { 2, 3, 4, 5, 6, 7 }
21.50/21.20	
21.50/21.20	DP problem for innermost termination.
21.50/21.20	P =
21.50/21.20	  f6#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
21.50/21.20	  f1#(I10, I11, I12, I13, I14) -> f6#(I10, I11, I12, -1 * I11 + I13, 1 + I14) [-1 <= I14 /\ I14 <= -1 /\ 0 <= -1 - I13]
21.50/21.20	  f5#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) 
21.50/21.20	  f4#(I20, I21, I22, I23, I24) -> f5#(I20, I21, I22, 1 + I23, 1 + I24) 
21.50/21.20	  f3#(I25, I26, I27, I28, I29) -> f4#(I25, I26, I27, I28, I29) [I27 = I27]
21.50/21.20	  f1#(I30, I31, I32, I33, I34) -> f3#(I30, I31, I32, I33, I34) [0 <= -1 - I33]
21.50/21.20	R = 
21.50/21.20	  f8(x1, x2, x3, x4, x5) -> f7(x1, x2, x3, x4, x5) 
21.50/21.20	  f7(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, I3, I4) 
21.50/21.20	  f6(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) 
21.50/21.20	  f1(I10, I11, I12, I13, I14) -> f6(I10, I11, I12, -1 * I11 + I13, 1 + I14) [-1 <= I14 /\ I14 <= -1 /\ 0 <= -1 - I13]
21.50/21.20	  f5(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I18, I19) 
21.50/21.20	  f4(I20, I21, I22, I23, I24) -> f5(I20, I21, I22, 1 + I23, 1 + I24) 
21.50/21.20	  f3(I25, I26, I27, I28, I29) -> f4(I25, I26, I27, I28, I29) [I27 = I27]
21.50/21.20	  f1(I30, I31, I32, I33, I34) -> f3(I30, I31, I32, I33, I34) [0 <= -1 - I33]
21.50/21.20	  f1(I35, I36, I37, I38, I39) -> f2(rnd1, I36, I37, I38, I39) [rnd1 = rnd1 /\ -1 * I38 <= 0]
21.50/21.20	
21.50/21.20	We use the extended value criterion with the projection function NU:
21.50/21.20	  NU[f3#(x0,x1,x2,x3,x4)] = -x4 - 2
21.50/21.20	  NU[f4#(x0,x1,x2,x3,x4)] = -x4 - 2
21.50/21.20	  NU[f5#(x0,x1,x2,x3,x4)] = -x4 - 1
21.50/21.20	  NU[f1#(x0,x1,x2,x3,x4)] = -x4 - 1
21.50/21.20	  NU[f6#(x0,x1,x2,x3,x4)] = -x4 - 1
21.50/21.20	
21.50/21.20	This gives the following inequalities:
21.50/21.20	    ==>  -I9 - 1 >= -I9 - 1
21.50/21.20	  -1 <= I14 /\ I14 <= -1 /\ 0 <= -1 - I13  ==>  -I14 - 1 > -(1 + I14) - 1  with  -I14 - 1 >= 0
21.50/21.20	    ==>  -I19 - 1 >= -I19 - 1
21.50/21.20	    ==>  -I24 - 2 >= -(1 + I24) - 1
21.50/21.20	  I27 = I27  ==>  -I29 - 2 >= -I29 - 2
21.50/21.20	  0 <= -1 - I33  ==>  -I34 - 1 >= -I34 - 2
21.50/21.20	
21.50/21.20	We remove all the strictly oriented dependency pairs.
21.50/21.20	
21.50/21.20	DP problem for innermost termination.
21.50/21.20	P =
21.50/21.20	  f6#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
21.50/21.20	  f5#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) 
21.50/21.20	  f4#(I20, I21, I22, I23, I24) -> f5#(I20, I21, I22, 1 + I23, 1 + I24) 
21.50/21.20	  f3#(I25, I26, I27, I28, I29) -> f4#(I25, I26, I27, I28, I29) [I27 = I27]
21.50/21.20	  f1#(I30, I31, I32, I33, I34) -> f3#(I30, I31, I32, I33, I34) [0 <= -1 - I33]
21.50/21.20	R = 
21.50/21.20	  f8(x1, x2, x3, x4, x5) -> f7(x1, x2, x3, x4, x5) 
21.50/21.20	  f7(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, I3, I4) 
21.50/21.20	  f6(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) 
21.50/21.20	  f1(I10, I11, I12, I13, I14) -> f6(I10, I11, I12, -1 * I11 + I13, 1 + I14) [-1 <= I14 /\ I14 <= -1 /\ 0 <= -1 - I13]
21.50/21.20	  f5(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I18, I19) 
21.50/21.20	  f4(I20, I21, I22, I23, I24) -> f5(I20, I21, I22, 1 + I23, 1 + I24) 
21.50/21.20	  f3(I25, I26, I27, I28, I29) -> f4(I25, I26, I27, I28, I29) [I27 = I27]
21.50/21.20	  f1(I30, I31, I32, I33, I34) -> f3(I30, I31, I32, I33, I34) [0 <= -1 - I33]
21.50/21.20	  f1(I35, I36, I37, I38, I39) -> f2(rnd1, I36, I37, I38, I39) [rnd1 = rnd1 /\ -1 * I38 <= 0]
21.50/21.20	
21.50/21.20	The dependency graph for this problem is:
21.50/21.20	  2 -> 7
21.50/21.20	  4 -> 7
21.50/21.20	  5 -> 4
21.50/21.20	  6 -> 5
21.50/21.20	  7 -> 6
21.50/21.20	Where:
21.50/21.20	  2) f6#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
21.50/21.20	  4) f5#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) 
21.50/21.20	  5) f4#(I20, I21, I22, I23, I24) -> f5#(I20, I21, I22, 1 + I23, 1 + I24) 
21.50/21.20	  6) f3#(I25, I26, I27, I28, I29) -> f4#(I25, I26, I27, I28, I29) [I27 = I27]
21.50/21.20	  7) f1#(I30, I31, I32, I33, I34) -> f3#(I30, I31, I32, I33, I34) [0 <= -1 - I33]
21.50/21.20	
21.50/21.20	We have the following SCCs.
21.50/21.20	  { 4, 5, 6, 7 }
21.50/21.20	
21.50/21.20	DP problem for innermost termination.
21.50/21.20	P =
21.50/21.20	  f5#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) 
21.50/21.20	  f4#(I20, I21, I22, I23, I24) -> f5#(I20, I21, I22, 1 + I23, 1 + I24) 
21.50/21.20	  f3#(I25, I26, I27, I28, I29) -> f4#(I25, I26, I27, I28, I29) [I27 = I27]
21.50/21.20	  f1#(I30, I31, I32, I33, I34) -> f3#(I30, I31, I32, I33, I34) [0 <= -1 - I33]
21.50/21.20	R = 
21.50/21.20	  f8(x1, x2, x3, x4, x5) -> f7(x1, x2, x3, x4, x5) 
21.50/21.20	  f7(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, I3, I4) 
21.50/21.20	  f6(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) 
21.50/21.20	  f1(I10, I11, I12, I13, I14) -> f6(I10, I11, I12, -1 * I11 + I13, 1 + I14) [-1 <= I14 /\ I14 <= -1 /\ 0 <= -1 - I13]
21.50/21.20	  f5(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I18, I19) 
21.50/21.20	  f4(I20, I21, I22, I23, I24) -> f5(I20, I21, I22, 1 + I23, 1 + I24) 
21.50/21.20	  f3(I25, I26, I27, I28, I29) -> f4(I25, I26, I27, I28, I29) [I27 = I27]
21.50/21.20	  f1(I30, I31, I32, I33, I34) -> f3(I30, I31, I32, I33, I34) [0 <= -1 - I33]
21.50/21.20	  f1(I35, I36, I37, I38, I39) -> f2(rnd1, I36, I37, I38, I39) [rnd1 = rnd1 /\ -1 * I38 <= 0]
21.50/21.20	
21.50/21.20	We use the extended value criterion with the projection function NU:
21.50/21.20	  NU[f3#(x0,x1,x2,x3,x4)] = -x3 - 2
21.50/21.20	  NU[f4#(x0,x1,x2,x3,x4)] = -x3 - 2
21.50/21.20	  NU[f1#(x0,x1,x2,x3,x4)] = -x3 - 1
21.50/21.20	  NU[f5#(x0,x1,x2,x3,x4)] = -x3 - 1
21.50/21.20	
21.50/21.20	This gives the following inequalities:
21.50/21.20	    ==>  -I18 - 1 >= -I18 - 1
21.50/21.20	    ==>  -I23 - 2 >= -(1 + I23) - 1
21.50/21.20	  I27 = I27  ==>  -I28 - 2 >= -I28 - 2
21.50/21.20	  0 <= -1 - I33  ==>  -I33 - 1 > -I33 - 2  with  -I33 - 1 >= 0
21.50/21.20	
21.50/21.20	We remove all the strictly oriented dependency pairs.
21.50/21.20	
21.50/21.20	DP problem for innermost termination.
21.50/21.20	P =
21.50/21.20	  f5#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) 
21.50/21.20	  f4#(I20, I21, I22, I23, I24) -> f5#(I20, I21, I22, 1 + I23, 1 + I24) 
21.50/21.20	  f3#(I25, I26, I27, I28, I29) -> f4#(I25, I26, I27, I28, I29) [I27 = I27]
21.50/21.20	R = 
21.50/21.20	  f8(x1, x2, x3, x4, x5) -> f7(x1, x2, x3, x4, x5) 
21.50/21.20	  f7(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, I3, I4) 
21.50/21.20	  f6(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) 
21.50/21.20	  f1(I10, I11, I12, I13, I14) -> f6(I10, I11, I12, -1 * I11 + I13, 1 + I14) [-1 <= I14 /\ I14 <= -1 /\ 0 <= -1 - I13]
21.50/21.20	  f5(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I18, I19) 
21.50/21.20	  f4(I20, I21, I22, I23, I24) -> f5(I20, I21, I22, 1 + I23, 1 + I24) 
21.50/21.20	  f3(I25, I26, I27, I28, I29) -> f4(I25, I26, I27, I28, I29) [I27 = I27]
21.50/21.20	  f1(I30, I31, I32, I33, I34) -> f3(I30, I31, I32, I33, I34) [0 <= -1 - I33]
21.50/21.20	  f1(I35, I36, I37, I38, I39) -> f2(rnd1, I36, I37, I38, I39) [rnd1 = rnd1 /\ -1 * I38 <= 0]
21.50/21.20	
21.50/21.20	The dependency graph for this problem is:
21.50/21.20	  4 -> 
21.50/21.20	  5 -> 4
21.50/21.20	  6 -> 5
21.50/21.20	Where:
21.50/21.20	  4) f5#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) 
21.50/21.20	  5) f4#(I20, I21, I22, I23, I24) -> f5#(I20, I21, I22, 1 + I23, 1 + I24) 
21.50/21.20	  6) f3#(I25, I26, I27, I28, I29) -> f4#(I25, I26, I27, I28, I29) [I27 = I27]
21.50/21.20	
21.50/21.20	We have the following SCCs.
21.50/21.20	  
21.50/24.17	EOF