2.07/2.10	YES
2.07/2.10	
2.07/2.10	DP problem for innermost termination.
2.07/2.10	P =
2.07/2.10	  init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
2.07/2.10	  f5#(I0, I1, I2) -> f5#(I3, I4, I5) [-1 <= I4 - 1 /\ 0 <= I3 - 1 /\ -1 <= I1 - 1 /\ 2 <= I0 - 1 /\ I4 + 3 <= I0 /\ I3 + 2 <= I0]
2.07/2.10	  f5#(I6, I7, I8) -> f5#(I9, I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1 /\ 0 <= I7 - 1 /\ 2 <= I6 - 1 /\ I10 + 1 <= I7 /\ I10 + 3 <= I6 /\ I9 <= I7 /\ I9 + 2 <= I6]
2.07/2.10	  f2#(I12, I13, I14) -> f5#(I15, I16, I17) [-1 <= I16 - 1 /\ 0 <= I15 - 1 /\ 0 <= I12 - 1 /\ I16 + 1 <= I12 /\ I15 <= I12]
2.07/2.10	  f4#(I18, I19, I20) -> f3#(I20, I20, I20) [2 <= I19 - 1 /\ 0 <= I18 - 1 /\ 0 <= I20 - 1]
2.07/2.10	  f3#(I21, I22, I23) -> f4#(I21, I24, I22 - 1) [I22 = I23 /\ 0 <= I21 - 1 /\ 3 <= I24 - 1 /\ 1 <= I22 - 1 /\ I22 - 1 <= I22 - 1]
2.07/2.10	  f3#(I25, I26, I27) -> f4#(I25, I28, I26 - 1) [I26 = I27 /\ 0 <= I25 - 1 /\ 4 <= I28 - 1 /\ 1 <= I26 - 1 /\ I26 - 1 <= I26 - 1]
2.07/2.10	  f3#(I29, I30, I31) -> f3#(I30 - 1, I30 - 1, I30 - 1) [I30 = I31 /\ 1 <= I30 - 1 /\ 0 <= I29 - 1 /\ I30 - 1 <= I30 - 1]
2.07/2.10	  f1#(I32, I33, I34) -> f3#(5, 5, 5) 
2.07/2.10	  f1#(I35, I36, I37) -> f2#(I38, I39, I40) [1 <= I38 - 1]
2.07/2.10	  f1#(I41, I42, I43) -> f2#(I44, I45, I46) [2 <= I44 - 1]
2.07/2.10	R = 
2.07/2.10	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
2.07/2.10	  f5(I0, I1, I2) -> f5(I3, I4, I5) [-1 <= I4 - 1 /\ 0 <= I3 - 1 /\ -1 <= I1 - 1 /\ 2 <= I0 - 1 /\ I4 + 3 <= I0 /\ I3 + 2 <= I0]
2.07/2.10	  f5(I6, I7, I8) -> f5(I9, I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1 /\ 0 <= I7 - 1 /\ 2 <= I6 - 1 /\ I10 + 1 <= I7 /\ I10 + 3 <= I6 /\ I9 <= I7 /\ I9 + 2 <= I6]
2.07/2.10	  f2(I12, I13, I14) -> f5(I15, I16, I17) [-1 <= I16 - 1 /\ 0 <= I15 - 1 /\ 0 <= I12 - 1 /\ I16 + 1 <= I12 /\ I15 <= I12]
2.07/2.10	  f4(I18, I19, I20) -> f3(I20, I20, I20) [2 <= I19 - 1 /\ 0 <= I18 - 1 /\ 0 <= I20 - 1]
2.07/2.10	  f3(I21, I22, I23) -> f4(I21, I24, I22 - 1) [I22 = I23 /\ 0 <= I21 - 1 /\ 3 <= I24 - 1 /\ 1 <= I22 - 1 /\ I22 - 1 <= I22 - 1]
2.07/2.10	  f3(I25, I26, I27) -> f4(I25, I28, I26 - 1) [I26 = I27 /\ 0 <= I25 - 1 /\ 4 <= I28 - 1 /\ 1 <= I26 - 1 /\ I26 - 1 <= I26 - 1]
2.07/2.10	  f3(I29, I30, I31) -> f3(I30 - 1, I30 - 1, I30 - 1) [I30 = I31 /\ 1 <= I30 - 1 /\ 0 <= I29 - 1 /\ I30 - 1 <= I30 - 1]
2.07/2.10	  f1(I32, I33, I34) -> f3(5, 5, 5) 
2.07/2.10	  f1(I35, I36, I37) -> f2(I38, I39, I40) [1 <= I38 - 1]
2.07/2.10	  f1(I41, I42, I43) -> f2(I44, I45, I46) [2 <= I44 - 1]
2.07/2.10	
2.07/2.10	The dependency graph for this problem is:
2.07/2.10	  0 -> 8, 9, 10
2.07/2.10	  1 -> 1, 2
2.07/2.10	  2 -> 1, 2
2.07/2.10	  3 -> 1, 2
2.07/2.10	  4 -> 5, 6, 7
2.07/2.10	  5 -> 4
2.07/2.10	  6 -> 4
2.07/2.10	  7 -> 5, 6, 7
2.07/2.10	  8 -> 5, 6, 7
2.07/2.10	  9 -> 3
2.07/2.10	  10 -> 3
2.07/2.10	Where:
2.07/2.10	  0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
2.07/2.10	  1) f5#(I0, I1, I2) -> f5#(I3, I4, I5) [-1 <= I4 - 1 /\ 0 <= I3 - 1 /\ -1 <= I1 - 1 /\ 2 <= I0 - 1 /\ I4 + 3 <= I0 /\ I3 + 2 <= I0]
2.07/2.10	  2) f5#(I6, I7, I8) -> f5#(I9, I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1 /\ 0 <= I7 - 1 /\ 2 <= I6 - 1 /\ I10 + 1 <= I7 /\ I10 + 3 <= I6 /\ I9 <= I7 /\ I9 + 2 <= I6]
2.07/2.10	  3) f2#(I12, I13, I14) -> f5#(I15, I16, I17) [-1 <= I16 - 1 /\ 0 <= I15 - 1 /\ 0 <= I12 - 1 /\ I16 + 1 <= I12 /\ I15 <= I12]
2.07/2.10	  4) f4#(I18, I19, I20) -> f3#(I20, I20, I20) [2 <= I19 - 1 /\ 0 <= I18 - 1 /\ 0 <= I20 - 1]
2.07/2.10	  5) f3#(I21, I22, I23) -> f4#(I21, I24, I22 - 1) [I22 = I23 /\ 0 <= I21 - 1 /\ 3 <= I24 - 1 /\ 1 <= I22 - 1 /\ I22 - 1 <= I22 - 1]
2.07/2.10	  6) f3#(I25, I26, I27) -> f4#(I25, I28, I26 - 1) [I26 = I27 /\ 0 <= I25 - 1 /\ 4 <= I28 - 1 /\ 1 <= I26 - 1 /\ I26 - 1 <= I26 - 1]
2.07/2.10	  7) f3#(I29, I30, I31) -> f3#(I30 - 1, I30 - 1, I30 - 1) [I30 = I31 /\ 1 <= I30 - 1 /\ 0 <= I29 - 1 /\ I30 - 1 <= I30 - 1]
2.07/2.10	  8) f1#(I32, I33, I34) -> f3#(5, 5, 5) 
2.07/2.10	  9) f1#(I35, I36, I37) -> f2#(I38, I39, I40) [1 <= I38 - 1]
2.07/2.10	  10) f1#(I41, I42, I43) -> f2#(I44, I45, I46) [2 <= I44 - 1]
2.07/2.10	
2.07/2.10	We have the following SCCs.
2.07/2.10	  { 4, 5, 6, 7 }
2.07/2.10	  { 1, 2 }
2.07/2.10	
2.07/2.10	DP problem for innermost termination.
2.07/2.10	P =
2.07/2.10	  f5#(I0, I1, I2) -> f5#(I3, I4, I5) [-1 <= I4 - 1 /\ 0 <= I3 - 1 /\ -1 <= I1 - 1 /\ 2 <= I0 - 1 /\ I4 + 3 <= I0 /\ I3 + 2 <= I0]
2.07/2.10	  f5#(I6, I7, I8) -> f5#(I9, I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1 /\ 0 <= I7 - 1 /\ 2 <= I6 - 1 /\ I10 + 1 <= I7 /\ I10 + 3 <= I6 /\ I9 <= I7 /\ I9 + 2 <= I6]
2.07/2.10	R = 
2.07/2.10	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
2.07/2.10	  f5(I0, I1, I2) -> f5(I3, I4, I5) [-1 <= I4 - 1 /\ 0 <= I3 - 1 /\ -1 <= I1 - 1 /\ 2 <= I0 - 1 /\ I4 + 3 <= I0 /\ I3 + 2 <= I0]
2.07/2.10	  f5(I6, I7, I8) -> f5(I9, I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1 /\ 0 <= I7 - 1 /\ 2 <= I6 - 1 /\ I10 + 1 <= I7 /\ I10 + 3 <= I6 /\ I9 <= I7 /\ I9 + 2 <= I6]
2.07/2.10	  f2(I12, I13, I14) -> f5(I15, I16, I17) [-1 <= I16 - 1 /\ 0 <= I15 - 1 /\ 0 <= I12 - 1 /\ I16 + 1 <= I12 /\ I15 <= I12]
2.07/2.10	  f4(I18, I19, I20) -> f3(I20, I20, I20) [2 <= I19 - 1 /\ 0 <= I18 - 1 /\ 0 <= I20 - 1]
2.07/2.10	  f3(I21, I22, I23) -> f4(I21, I24, I22 - 1) [I22 = I23 /\ 0 <= I21 - 1 /\ 3 <= I24 - 1 /\ 1 <= I22 - 1 /\ I22 - 1 <= I22 - 1]
2.07/2.10	  f3(I25, I26, I27) -> f4(I25, I28, I26 - 1) [I26 = I27 /\ 0 <= I25 - 1 /\ 4 <= I28 - 1 /\ 1 <= I26 - 1 /\ I26 - 1 <= I26 - 1]
2.07/2.10	  f3(I29, I30, I31) -> f3(I30 - 1, I30 - 1, I30 - 1) [I30 = I31 /\ 1 <= I30 - 1 /\ 0 <= I29 - 1 /\ I30 - 1 <= I30 - 1]
2.07/2.10	  f1(I32, I33, I34) -> f3(5, 5, 5) 
2.07/2.10	  f1(I35, I36, I37) -> f2(I38, I39, I40) [1 <= I38 - 1]
2.07/2.10	  f1(I41, I42, I43) -> f2(I44, I45, I46) [2 <= I44 - 1]
2.07/2.10	
2.07/2.10	We use the basic value criterion with the projection function NU:
2.07/2.10	  NU[f5#(z1,z2,z3)] = z1
2.07/2.10	
2.07/2.10	This gives the following inequalities:
2.07/2.10	  -1 <= I4 - 1 /\ 0 <= I3 - 1 /\ -1 <= I1 - 1 /\ 2 <= I0 - 1 /\ I4 + 3 <= I0 /\ I3 + 2 <= I0  ==>  I0 >! I3
2.07/2.10	  -1 <= I10 - 1 /\ 0 <= I9 - 1 /\ 0 <= I7 - 1 /\ 2 <= I6 - 1 /\ I10 + 1 <= I7 /\ I10 + 3 <= I6 /\ I9 <= I7 /\ I9 + 2 <= I6  ==>  I6 >! I9
2.07/2.10	
2.07/2.10	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
2.07/2.10	
2.07/2.10	DP problem for innermost termination.
2.07/2.10	P =
2.07/2.10	  f4#(I18, I19, I20) -> f3#(I20, I20, I20) [2 <= I19 - 1 /\ 0 <= I18 - 1 /\ 0 <= I20 - 1]
2.07/2.10	  f3#(I21, I22, I23) -> f4#(I21, I24, I22 - 1) [I22 = I23 /\ 0 <= I21 - 1 /\ 3 <= I24 - 1 /\ 1 <= I22 - 1 /\ I22 - 1 <= I22 - 1]
2.07/2.10	  f3#(I25, I26, I27) -> f4#(I25, I28, I26 - 1) [I26 = I27 /\ 0 <= I25 - 1 /\ 4 <= I28 - 1 /\ 1 <= I26 - 1 /\ I26 - 1 <= I26 - 1]
2.07/2.10	  f3#(I29, I30, I31) -> f3#(I30 - 1, I30 - 1, I30 - 1) [I30 = I31 /\ 1 <= I30 - 1 /\ 0 <= I29 - 1 /\ I30 - 1 <= I30 - 1]
2.07/2.10	R = 
2.07/2.10	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
2.07/2.10	  f5(I0, I1, I2) -> f5(I3, I4, I5) [-1 <= I4 - 1 /\ 0 <= I3 - 1 /\ -1 <= I1 - 1 /\ 2 <= I0 - 1 /\ I4 + 3 <= I0 /\ I3 + 2 <= I0]
2.07/2.10	  f5(I6, I7, I8) -> f5(I9, I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1 /\ 0 <= I7 - 1 /\ 2 <= I6 - 1 /\ I10 + 1 <= I7 /\ I10 + 3 <= I6 /\ I9 <= I7 /\ I9 + 2 <= I6]
2.07/2.10	  f2(I12, I13, I14) -> f5(I15, I16, I17) [-1 <= I16 - 1 /\ 0 <= I15 - 1 /\ 0 <= I12 - 1 /\ I16 + 1 <= I12 /\ I15 <= I12]
2.07/2.10	  f4(I18, I19, I20) -> f3(I20, I20, I20) [2 <= I19 - 1 /\ 0 <= I18 - 1 /\ 0 <= I20 - 1]
2.07/2.10	  f3(I21, I22, I23) -> f4(I21, I24, I22 - 1) [I22 = I23 /\ 0 <= I21 - 1 /\ 3 <= I24 - 1 /\ 1 <= I22 - 1 /\ I22 - 1 <= I22 - 1]
2.07/2.10	  f3(I25, I26, I27) -> f4(I25, I28, I26 - 1) [I26 = I27 /\ 0 <= I25 - 1 /\ 4 <= I28 - 1 /\ 1 <= I26 - 1 /\ I26 - 1 <= I26 - 1]
2.07/2.10	  f3(I29, I30, I31) -> f3(I30 - 1, I30 - 1, I30 - 1) [I30 = I31 /\ 1 <= I30 - 1 /\ 0 <= I29 - 1 /\ I30 - 1 <= I30 - 1]
2.07/2.10	  f1(I32, I33, I34) -> f3(5, 5, 5) 
2.07/2.10	  f1(I35, I36, I37) -> f2(I38, I39, I40) [1 <= I38 - 1]
2.07/2.10	  f1(I41, I42, I43) -> f2(I44, I45, I46) [2 <= I44 - 1]
2.07/2.10	
2.07/2.10	We use the basic value criterion with the projection function NU:
2.07/2.10	  NU[f3#(z1,z2,z3)] = z3
2.07/2.10	  NU[f4#(z1,z2,z3)] = z3
2.07/2.10	
2.07/2.10	This gives the following inequalities:
2.07/2.10	  2 <= I19 - 1 /\ 0 <= I18 - 1 /\ 0 <= I20 - 1  ==>  I20 (>! \union =) I20
2.07/2.10	  I22 = I23 /\ 0 <= I21 - 1 /\ 3 <= I24 - 1 /\ 1 <= I22 - 1 /\ I22 - 1 <= I22 - 1  ==>  I23 >! I22 - 1
2.07/2.10	  I26 = I27 /\ 0 <= I25 - 1 /\ 4 <= I28 - 1 /\ 1 <= I26 - 1 /\ I26 - 1 <= I26 - 1  ==>  I27 >! I26 - 1
2.07/2.10	  I30 = I31 /\ 1 <= I30 - 1 /\ 0 <= I29 - 1 /\ I30 - 1 <= I30 - 1  ==>  I31 >! I30 - 1
2.07/2.10	
2.07/2.10	We remove all the strictly oriented dependency pairs.
2.07/2.10	
2.07/2.10	DP problem for innermost termination.
2.07/2.10	P =
2.07/2.10	  f4#(I18, I19, I20) -> f3#(I20, I20, I20) [2 <= I19 - 1 /\ 0 <= I18 - 1 /\ 0 <= I20 - 1]
2.07/2.10	R = 
2.07/2.10	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
2.07/2.10	  f5(I0, I1, I2) -> f5(I3, I4, I5) [-1 <= I4 - 1 /\ 0 <= I3 - 1 /\ -1 <= I1 - 1 /\ 2 <= I0 - 1 /\ I4 + 3 <= I0 /\ I3 + 2 <= I0]
2.07/2.10	  f5(I6, I7, I8) -> f5(I9, I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1 /\ 0 <= I7 - 1 /\ 2 <= I6 - 1 /\ I10 + 1 <= I7 /\ I10 + 3 <= I6 /\ I9 <= I7 /\ I9 + 2 <= I6]
2.07/2.10	  f2(I12, I13, I14) -> f5(I15, I16, I17) [-1 <= I16 - 1 /\ 0 <= I15 - 1 /\ 0 <= I12 - 1 /\ I16 + 1 <= I12 /\ I15 <= I12]
2.07/2.10	  f4(I18, I19, I20) -> f3(I20, I20, I20) [2 <= I19 - 1 /\ 0 <= I18 - 1 /\ 0 <= I20 - 1]
2.07/2.10	  f3(I21, I22, I23) -> f4(I21, I24, I22 - 1) [I22 = I23 /\ 0 <= I21 - 1 /\ 3 <= I24 - 1 /\ 1 <= I22 - 1 /\ I22 - 1 <= I22 - 1]
2.07/2.10	  f3(I25, I26, I27) -> f4(I25, I28, I26 - 1) [I26 = I27 /\ 0 <= I25 - 1 /\ 4 <= I28 - 1 /\ 1 <= I26 - 1 /\ I26 - 1 <= I26 - 1]
2.07/2.10	  f3(I29, I30, I31) -> f3(I30 - 1, I30 - 1, I30 - 1) [I30 = I31 /\ 1 <= I30 - 1 /\ 0 <= I29 - 1 /\ I30 - 1 <= I30 - 1]
2.07/2.10	  f1(I32, I33, I34) -> f3(5, 5, 5) 
2.07/2.10	  f1(I35, I36, I37) -> f2(I38, I39, I40) [1 <= I38 - 1]
2.07/2.10	  f1(I41, I42, I43) -> f2(I44, I45, I46) [2 <= I44 - 1]
2.07/2.10	
2.07/2.10	The dependency graph for this problem is:
2.07/2.10	  4 -> 
2.07/2.10	Where:
2.07/2.10	  4) f4#(I18, I19, I20) -> f3#(I20, I20, I20) [2 <= I19 - 1 /\ 0 <= I18 - 1 /\ 0 <= I20 - 1]
2.07/2.10	
2.07/2.10	We have the following SCCs.
2.07/2.10	  
2.07/5.08	EOF