13.16/13.08	YES
13.16/13.08	
13.16/13.08	DP problem for innermost termination.
13.16/13.08	P =
13.16/13.08	  f8#(x1, x2, x3) -> f7#(x1, x2, x3) 
13.16/13.08	  f7#(I0, I1, I2) -> f1#(I0, I1, I2) 
13.16/13.08	  f6#(I3, I4, I5) -> f1#(I3, I4, I5) 
13.16/13.08	  f1#(I6, I7, I8) -> f6#(I6, -99 + I7, 1 + I8) [-1 <= I8 /\ I8 <= -1 /\ 0 <= -1 - I7]
13.16/13.08	  f5#(I9, I10, I11) -> f1#(I9, I10, I11) 
13.16/13.08	  f4#(I12, I13, I14) -> f5#(I12, 1 + I13, 1 + I14) 
13.16/13.08	  f3#(I15, I16, I17) -> f4#(I15, I16, I17) [0 <= I17]
13.16/13.08	  f3#(I18, I19, I20) -> f4#(I18, I19, I20) [1 + I20 <= -1]
13.16/13.08	  f1#(I21, I22, I23) -> f3#(I21, I22, I23) [0 <= -1 - I22]
13.16/13.08	R = 
13.16/13.08	  f8(x1, x2, x3) -> f7(x1, x2, x3) 
13.16/13.08	  f7(I0, I1, I2) -> f1(I0, I1, I2) 
13.16/13.08	  f6(I3, I4, I5) -> f1(I3, I4, I5) 
13.16/13.08	  f1(I6, I7, I8) -> f6(I6, -99 + I7, 1 + I8) [-1 <= I8 /\ I8 <= -1 /\ 0 <= -1 - I7]
13.16/13.08	  f5(I9, I10, I11) -> f1(I9, I10, I11) 
13.16/13.08	  f4(I12, I13, I14) -> f5(I12, 1 + I13, 1 + I14) 
13.16/13.08	  f3(I15, I16, I17) -> f4(I15, I16, I17) [0 <= I17]
13.16/13.08	  f3(I18, I19, I20) -> f4(I18, I19, I20) [1 + I20 <= -1]
13.16/13.08	  f1(I21, I22, I23) -> f3(I21, I22, I23) [0 <= -1 - I22]
13.16/13.08	  f1(I24, I25, I26) -> f2(rnd1, I25, I26) [rnd1 = rnd1 /\ -1 * I25 <= 0]
13.16/13.08	
13.16/13.08	The dependency graph for this problem is:
13.16/13.08	  0 -> 1
13.16/13.08	  1 -> 3, 8
13.16/13.08	  2 -> 3, 8
13.16/13.08	  3 -> 2
13.16/13.08	  4 -> 3, 8
13.16/13.08	  5 -> 4
13.16/13.08	  6 -> 5
13.16/13.08	  7 -> 5
13.16/13.08	  8 -> 6, 7
13.16/13.08	Where:
13.16/13.08	  0) f8#(x1, x2, x3) -> f7#(x1, x2, x3) 
13.16/13.08	  1) f7#(I0, I1, I2) -> f1#(I0, I1, I2) 
13.16/13.08	  2) f6#(I3, I4, I5) -> f1#(I3, I4, I5) 
13.16/13.08	  3) f1#(I6, I7, I8) -> f6#(I6, -99 + I7, 1 + I8) [-1 <= I8 /\ I8 <= -1 /\ 0 <= -1 - I7]
13.16/13.08	  4) f5#(I9, I10, I11) -> f1#(I9, I10, I11) 
13.16/13.08	  5) f4#(I12, I13, I14) -> f5#(I12, 1 + I13, 1 + I14) 
13.16/13.08	  6) f3#(I15, I16, I17) -> f4#(I15, I16, I17) [0 <= I17]
13.16/13.08	  7) f3#(I18, I19, I20) -> f4#(I18, I19, I20) [1 + I20 <= -1]
13.16/13.08	  8) f1#(I21, I22, I23) -> f3#(I21, I22, I23) [0 <= -1 - I22]
13.16/13.08	
13.16/13.08	We have the following SCCs.
13.16/13.08	  { 2, 3, 4, 5, 6, 7, 8 }
13.16/13.08	
13.16/13.08	DP problem for innermost termination.
13.16/13.08	P =
13.16/13.08	  f6#(I3, I4, I5) -> f1#(I3, I4, I5) 
13.16/13.08	  f1#(I6, I7, I8) -> f6#(I6, -99 + I7, 1 + I8) [-1 <= I8 /\ I8 <= -1 /\ 0 <= -1 - I7]
13.16/13.08	  f5#(I9, I10, I11) -> f1#(I9, I10, I11) 
13.16/13.08	  f4#(I12, I13, I14) -> f5#(I12, 1 + I13, 1 + I14) 
13.16/13.08	  f3#(I15, I16, I17) -> f4#(I15, I16, I17) [0 <= I17]
13.16/13.08	  f3#(I18, I19, I20) -> f4#(I18, I19, I20) [1 + I20 <= -1]
13.16/13.08	  f1#(I21, I22, I23) -> f3#(I21, I22, I23) [0 <= -1 - I22]
13.16/13.08	R = 
13.16/13.08	  f8(x1, x2, x3) -> f7(x1, x2, x3) 
13.16/13.08	  f7(I0, I1, I2) -> f1(I0, I1, I2) 
13.16/13.08	  f6(I3, I4, I5) -> f1(I3, I4, I5) 
13.16/13.08	  f1(I6, I7, I8) -> f6(I6, -99 + I7, 1 + I8) [-1 <= I8 /\ I8 <= -1 /\ 0 <= -1 - I7]
13.16/13.08	  f5(I9, I10, I11) -> f1(I9, I10, I11) 
13.16/13.08	  f4(I12, I13, I14) -> f5(I12, 1 + I13, 1 + I14) 
13.16/13.08	  f3(I15, I16, I17) -> f4(I15, I16, I17) [0 <= I17]
13.16/13.08	  f3(I18, I19, I20) -> f4(I18, I19, I20) [1 + I20 <= -1]
13.16/13.08	  f1(I21, I22, I23) -> f3(I21, I22, I23) [0 <= -1 - I22]
13.16/13.08	  f1(I24, I25, I26) -> f2(rnd1, I25, I26) [rnd1 = rnd1 /\ -1 * I25 <= 0]
13.16/13.08	
13.16/13.08	We use the extended value criterion with the projection function NU:
13.16/13.08	  NU[f3#(x0,x1,x2)] = -x2 - 2
13.16/13.08	  NU[f4#(x0,x1,x2)] = -x2 - 2
13.16/13.08	  NU[f5#(x0,x1,x2)] = -x2 - 1
13.16/13.08	  NU[f1#(x0,x1,x2)] = -x2 - 1
13.16/13.08	  NU[f6#(x0,x1,x2)] = -x2 - 1
13.16/13.08	
13.16/13.08	This gives the following inequalities:
13.16/13.08	    ==>  -I5 - 1 >= -I5 - 1
13.16/13.08	  -1 <= I8 /\ I8 <= -1 /\ 0 <= -1 - I7  ==>  -I8 - 1 > -(1 + I8) - 1  with  -I8 - 1 >= 0
13.16/13.08	    ==>  -I11 - 1 >= -I11 - 1
13.16/13.08	    ==>  -I14 - 2 >= -(1 + I14) - 1
13.16/13.08	  0 <= I17  ==>  -I17 - 2 >= -I17 - 2
13.16/13.08	  1 + I20 <= -1  ==>  -I20 - 2 >= -I20 - 2
13.16/13.08	  0 <= -1 - I22  ==>  -I23 - 1 >= -I23 - 2
13.16/13.08	
13.16/13.08	We remove all the strictly oriented dependency pairs.
13.16/13.08	
13.16/13.08	DP problem for innermost termination.
13.16/13.08	P =
13.16/13.08	  f6#(I3, I4, I5) -> f1#(I3, I4, I5) 
13.16/13.08	  f5#(I9, I10, I11) -> f1#(I9, I10, I11) 
13.16/13.08	  f4#(I12, I13, I14) -> f5#(I12, 1 + I13, 1 + I14) 
13.16/13.08	  f3#(I15, I16, I17) -> f4#(I15, I16, I17) [0 <= I17]
13.16/13.08	  f3#(I18, I19, I20) -> f4#(I18, I19, I20) [1 + I20 <= -1]
13.16/13.08	  f1#(I21, I22, I23) -> f3#(I21, I22, I23) [0 <= -1 - I22]
13.16/13.08	R = 
13.16/13.08	  f8(x1, x2, x3) -> f7(x1, x2, x3) 
13.16/13.08	  f7(I0, I1, I2) -> f1(I0, I1, I2) 
13.16/13.08	  f6(I3, I4, I5) -> f1(I3, I4, I5) 
13.16/13.08	  f1(I6, I7, I8) -> f6(I6, -99 + I7, 1 + I8) [-1 <= I8 /\ I8 <= -1 /\ 0 <= -1 - I7]
13.16/13.08	  f5(I9, I10, I11) -> f1(I9, I10, I11) 
13.16/13.08	  f4(I12, I13, I14) -> f5(I12, 1 + I13, 1 + I14) 
13.16/13.08	  f3(I15, I16, I17) -> f4(I15, I16, I17) [0 <= I17]
13.16/13.08	  f3(I18, I19, I20) -> f4(I18, I19, I20) [1 + I20 <= -1]
13.16/13.08	  f1(I21, I22, I23) -> f3(I21, I22, I23) [0 <= -1 - I22]
13.16/13.08	  f1(I24, I25, I26) -> f2(rnd1, I25, I26) [rnd1 = rnd1 /\ -1 * I25 <= 0]
13.16/13.08	
13.16/13.08	The dependency graph for this problem is:
13.16/13.08	  2 -> 8
13.16/13.08	  4 -> 8
13.16/13.08	  5 -> 4
13.16/13.08	  6 -> 5
13.16/13.08	  7 -> 5
13.16/13.08	  8 -> 6, 7
13.16/13.08	Where:
13.16/13.08	  2) f6#(I3, I4, I5) -> f1#(I3, I4, I5) 
13.16/13.08	  4) f5#(I9, I10, I11) -> f1#(I9, I10, I11) 
13.16/13.08	  5) f4#(I12, I13, I14) -> f5#(I12, 1 + I13, 1 + I14) 
13.16/13.08	  6) f3#(I15, I16, I17) -> f4#(I15, I16, I17) [0 <= I17]
13.16/13.08	  7) f3#(I18, I19, I20) -> f4#(I18, I19, I20) [1 + I20 <= -1]
13.16/13.08	  8) f1#(I21, I22, I23) -> f3#(I21, I22, I23) [0 <= -1 - I22]
13.16/13.08	
13.16/13.08	We have the following SCCs.
13.16/13.08	  { 4, 5, 6, 7, 8 }
13.16/13.08	
13.16/13.08	DP problem for innermost termination.
13.16/13.08	P =
13.16/13.08	  f5#(I9, I10, I11) -> f1#(I9, I10, I11) 
13.16/13.08	  f4#(I12, I13, I14) -> f5#(I12, 1 + I13, 1 + I14) 
13.16/13.08	  f3#(I15, I16, I17) -> f4#(I15, I16, I17) [0 <= I17]
13.16/13.08	  f3#(I18, I19, I20) -> f4#(I18, I19, I20) [1 + I20 <= -1]
13.16/13.08	  f1#(I21, I22, I23) -> f3#(I21, I22, I23) [0 <= -1 - I22]
13.16/13.08	R = 
13.16/13.08	  f8(x1, x2, x3) -> f7(x1, x2, x3) 
13.16/13.08	  f7(I0, I1, I2) -> f1(I0, I1, I2) 
13.16/13.08	  f6(I3, I4, I5) -> f1(I3, I4, I5) 
13.16/13.08	  f1(I6, I7, I8) -> f6(I6, -99 + I7, 1 + I8) [-1 <= I8 /\ I8 <= -1 /\ 0 <= -1 - I7]
13.16/13.08	  f5(I9, I10, I11) -> f1(I9, I10, I11) 
13.16/13.08	  f4(I12, I13, I14) -> f5(I12, 1 + I13, 1 + I14) 
13.16/13.08	  f3(I15, I16, I17) -> f4(I15, I16, I17) [0 <= I17]
13.16/13.08	  f3(I18, I19, I20) -> f4(I18, I19, I20) [1 + I20 <= -1]
13.16/13.08	  f1(I21, I22, I23) -> f3(I21, I22, I23) [0 <= -1 - I22]
13.16/13.08	  f1(I24, I25, I26) -> f2(rnd1, I25, I26) [rnd1 = rnd1 /\ -1 * I25 <= 0]
13.16/13.08	
13.16/13.08	We use the extended value criterion with the projection function NU:
13.16/13.08	  NU[f3#(x0,x1,x2)] = -x1 - 2
13.16/13.08	  NU[f4#(x0,x1,x2)] = -x1 - 2
13.16/13.08	  NU[f1#(x0,x1,x2)] = -x1 - 1
13.16/13.08	  NU[f5#(x0,x1,x2)] = -x1 - 1
13.16/13.08	
13.16/13.08	This gives the following inequalities:
13.16/13.08	    ==>  -I10 - 1 >= -I10 - 1
13.16/13.08	    ==>  -I13 - 2 >= -(1 + I13) - 1
13.16/13.08	  0 <= I17  ==>  -I16 - 2 >= -I16 - 2
13.16/13.08	  1 + I20 <= -1  ==>  -I19 - 2 >= -I19 - 2
13.16/13.08	  0 <= -1 - I22  ==>  -I22 - 1 > -I22 - 2  with  -I22 - 1 >= 0
13.16/13.08	
13.16/13.08	We remove all the strictly oriented dependency pairs.
13.16/13.08	
13.16/13.08	DP problem for innermost termination.
13.16/13.08	P =
13.16/13.08	  f5#(I9, I10, I11) -> f1#(I9, I10, I11) 
13.16/13.08	  f4#(I12, I13, I14) -> f5#(I12, 1 + I13, 1 + I14) 
13.16/13.08	  f3#(I15, I16, I17) -> f4#(I15, I16, I17) [0 <= I17]
13.16/13.08	  f3#(I18, I19, I20) -> f4#(I18, I19, I20) [1 + I20 <= -1]
13.16/13.08	R = 
13.16/13.08	  f8(x1, x2, x3) -> f7(x1, x2, x3) 
13.16/13.08	  f7(I0, I1, I2) -> f1(I0, I1, I2) 
13.16/13.08	  f6(I3, I4, I5) -> f1(I3, I4, I5) 
13.16/13.08	  f1(I6, I7, I8) -> f6(I6, -99 + I7, 1 + I8) [-1 <= I8 /\ I8 <= -1 /\ 0 <= -1 - I7]
13.16/13.08	  f5(I9, I10, I11) -> f1(I9, I10, I11) 
13.16/13.08	  f4(I12, I13, I14) -> f5(I12, 1 + I13, 1 + I14) 
13.16/13.08	  f3(I15, I16, I17) -> f4(I15, I16, I17) [0 <= I17]
13.16/13.08	  f3(I18, I19, I20) -> f4(I18, I19, I20) [1 + I20 <= -1]
13.16/13.08	  f1(I21, I22, I23) -> f3(I21, I22, I23) [0 <= -1 - I22]
13.16/13.08	  f1(I24, I25, I26) -> f2(rnd1, I25, I26) [rnd1 = rnd1 /\ -1 * I25 <= 0]
13.16/13.08	
13.16/13.08	The dependency graph for this problem is:
13.16/13.08	  4 -> 
13.16/13.08	  5 -> 4
13.16/13.08	  6 -> 5
13.16/13.08	  7 -> 5
13.16/13.08	Where:
13.16/13.08	  4) f5#(I9, I10, I11) -> f1#(I9, I10, I11) 
13.16/13.08	  5) f4#(I12, I13, I14) -> f5#(I12, 1 + I13, 1 + I14) 
13.16/13.08	  6) f3#(I15, I16, I17) -> f4#(I15, I16, I17) [0 <= I17]
13.16/13.08	  7) f3#(I18, I19, I20) -> f4#(I18, I19, I20) [1 + I20 <= -1]
13.16/13.08	
13.16/13.08	We have the following SCCs.
13.16/13.08	  
13.16/16.06	EOF