0.86/0.89	YES
0.86/0.89	
0.86/0.89	DP problem for innermost termination.
0.86/0.89	P =
0.86/0.89	  init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
0.86/0.89	  f5#(I0, I1, I2) -> f5#(I0 - 1, I1, I3) [0 <= I0 - 1 /\ I0 - 1 <= I0 - 1 /\ I1 <= 0]
0.86/0.89	  f5#(I4, I5, I6) -> f5#(I4, I5 - 1, I7) [I5 - 1 <= I5 - 1 /\ 0 <= I5 - 1]
0.86/0.89	  f3#(I8, I9, I10) -> f5#(I9, I10, I11) [0 <= I8 - 1]
0.86/0.89	  f4#(I12, I13, I14) -> f3#(I12, I14, I13) 
0.86/0.89	  f2#(I15, I16, I17) -> f3#(I16, I18, 0) [0 = I15 /\ I16 - 1 <= I16 - 1 /\ 0 <= I16 - 1]
0.86/0.89	  f2#(I19, I20, I21) -> f3#(I20, I22, I19) [I20 - 1 <= I20 - 1 /\ 0 <= I20 - 1]
0.86/0.89	  f2#(I23, I24, I25) -> f3#(I24, 0, 0) [0 = I23 /\ I24 - 1 <= I24 - 1 /\ 0 <= I24 - 1]
0.86/0.89	  f2#(I26, I27, I28) -> f3#(1, 0, I26) [1 = I27]
0.86/0.89	  f2#(I29, I30, I31) -> f2#(I29, I30 - 1, I32) [I30 - 1 <= I30 - 1 /\ 0 <= I30 - 1]
0.86/0.89	  f1#(I33, I34, I35) -> f2#(I36, I37, I38) [0 <= I33 - 1 /\ -1 <= I36 - 1 /\ 1 <= I34 - 1 /\ -1 <= I37 - 1]
0.86/0.89	R = 
0.86/0.89	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.86/0.89	  f5(I0, I1, I2) -> f5(I0 - 1, I1, I3) [0 <= I0 - 1 /\ I0 - 1 <= I0 - 1 /\ I1 <= 0]
0.86/0.89	  f5(I4, I5, I6) -> f5(I4, I5 - 1, I7) [I5 - 1 <= I5 - 1 /\ 0 <= I5 - 1]
0.86/0.89	  f3(I8, I9, I10) -> f5(I9, I10, I11) [0 <= I8 - 1]
0.86/0.89	  f4(I12, I13, I14) -> f3(I12, I14, I13) 
0.86/0.89	  f2(I15, I16, I17) -> f3(I16, I18, 0) [0 = I15 /\ I16 - 1 <= I16 - 1 /\ 0 <= I16 - 1]
0.86/0.89	  f2(I19, I20, I21) -> f3(I20, I22, I19) [I20 - 1 <= I20 - 1 /\ 0 <= I20 - 1]
0.86/0.89	  f2(I23, I24, I25) -> f3(I24, 0, 0) [0 = I23 /\ I24 - 1 <= I24 - 1 /\ 0 <= I24 - 1]
0.86/0.89	  f2(I26, I27, I28) -> f3(1, 0, I26) [1 = I27]
0.86/0.89	  f2(I29, I30, I31) -> f2(I29, I30 - 1, I32) [I30 - 1 <= I30 - 1 /\ 0 <= I30 - 1]
0.86/0.89	  f1(I33, I34, I35) -> f2(I36, I37, I38) [0 <= I33 - 1 /\ -1 <= I36 - 1 /\ 1 <= I34 - 1 /\ -1 <= I37 - 1]
0.86/0.89	
0.86/0.89	The dependency graph for this problem is:
0.86/0.89	  0 -> 10
0.86/0.89	  1 -> 1
0.86/0.89	  2 -> 1, 2
0.86/0.89	  3 -> 1, 2
0.86/0.89	  4 -> 3
0.86/0.89	  5 -> 3
0.86/0.89	  6 -> 3
0.86/0.89	  7 -> 3
0.86/0.89	  8 -> 3
0.86/0.89	  9 -> 5, 6, 7, 8, 9
0.86/0.89	  10 -> 5, 6, 7, 8, 9
0.86/0.89	Where:
0.86/0.89	  0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
0.86/0.89	  1) f5#(I0, I1, I2) -> f5#(I0 - 1, I1, I3) [0 <= I0 - 1 /\ I0 - 1 <= I0 - 1 /\ I1 <= 0]
0.86/0.89	  2) f5#(I4, I5, I6) -> f5#(I4, I5 - 1, I7) [I5 - 1 <= I5 - 1 /\ 0 <= I5 - 1]
0.86/0.89	  3) f3#(I8, I9, I10) -> f5#(I9, I10, I11) [0 <= I8 - 1]
0.86/0.89	  4) f4#(I12, I13, I14) -> f3#(I12, I14, I13) 
0.86/0.89	  5) f2#(I15, I16, I17) -> f3#(I16, I18, 0) [0 = I15 /\ I16 - 1 <= I16 - 1 /\ 0 <= I16 - 1]
0.86/0.89	  6) f2#(I19, I20, I21) -> f3#(I20, I22, I19) [I20 - 1 <= I20 - 1 /\ 0 <= I20 - 1]
0.86/0.89	  7) f2#(I23, I24, I25) -> f3#(I24, 0, 0) [0 = I23 /\ I24 - 1 <= I24 - 1 /\ 0 <= I24 - 1]
0.86/0.89	  8) f2#(I26, I27, I28) -> f3#(1, 0, I26) [1 = I27]
0.86/0.89	  9) f2#(I29, I30, I31) -> f2#(I29, I30 - 1, I32) [I30 - 1 <= I30 - 1 /\ 0 <= I30 - 1]
0.86/0.89	  10) f1#(I33, I34, I35) -> f2#(I36, I37, I38) [0 <= I33 - 1 /\ -1 <= I36 - 1 /\ 1 <= I34 - 1 /\ -1 <= I37 - 1]
0.86/0.89	
0.86/0.89	We have the following SCCs.
0.86/0.89	  { 9 }
0.86/0.89	  { 2 }
0.86/0.89	  { 1 }
0.86/0.89	
0.86/0.89	DP problem for innermost termination.
0.86/0.89	P =
0.86/0.89	  f5#(I0, I1, I2) -> f5#(I0 - 1, I1, I3) [0 <= I0 - 1 /\ I0 - 1 <= I0 - 1 /\ I1 <= 0]
0.86/0.89	R = 
0.86/0.89	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.86/0.89	  f5(I0, I1, I2) -> f5(I0 - 1, I1, I3) [0 <= I0 - 1 /\ I0 - 1 <= I0 - 1 /\ I1 <= 0]
0.86/0.89	  f5(I4, I5, I6) -> f5(I4, I5 - 1, I7) [I5 - 1 <= I5 - 1 /\ 0 <= I5 - 1]
0.86/0.89	  f3(I8, I9, I10) -> f5(I9, I10, I11) [0 <= I8 - 1]
0.86/0.89	  f4(I12, I13, I14) -> f3(I12, I14, I13) 
0.86/0.89	  f2(I15, I16, I17) -> f3(I16, I18, 0) [0 = I15 /\ I16 - 1 <= I16 - 1 /\ 0 <= I16 - 1]
0.86/0.89	  f2(I19, I20, I21) -> f3(I20, I22, I19) [I20 - 1 <= I20 - 1 /\ 0 <= I20 - 1]
0.86/0.89	  f2(I23, I24, I25) -> f3(I24, 0, 0) [0 = I23 /\ I24 - 1 <= I24 - 1 /\ 0 <= I24 - 1]
0.86/0.89	  f2(I26, I27, I28) -> f3(1, 0, I26) [1 = I27]
0.86/0.89	  f2(I29, I30, I31) -> f2(I29, I30 - 1, I32) [I30 - 1 <= I30 - 1 /\ 0 <= I30 - 1]
0.86/0.89	  f1(I33, I34, I35) -> f2(I36, I37, I38) [0 <= I33 - 1 /\ -1 <= I36 - 1 /\ 1 <= I34 - 1 /\ -1 <= I37 - 1]
0.86/0.89	
0.86/0.89	We use the basic value criterion with the projection function NU:
0.86/0.89	  NU[f5#(z1,z2,z3)] = z1
0.86/0.89	
0.86/0.89	This gives the following inequalities:
0.86/0.89	  0 <= I0 - 1 /\ I0 - 1 <= I0 - 1 /\ I1 <= 0  ==>  I0 >! I0 - 1
0.86/0.89	
0.86/0.89	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.86/0.89	
0.86/0.89	DP problem for innermost termination.
0.86/0.89	P =
0.86/0.89	  f5#(I4, I5, I6) -> f5#(I4, I5 - 1, I7) [I5 - 1 <= I5 - 1 /\ 0 <= I5 - 1]
0.86/0.89	R = 
0.86/0.89	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.86/0.89	  f5(I0, I1, I2) -> f5(I0 - 1, I1, I3) [0 <= I0 - 1 /\ I0 - 1 <= I0 - 1 /\ I1 <= 0]
0.86/0.89	  f5(I4, I5, I6) -> f5(I4, I5 - 1, I7) [I5 - 1 <= I5 - 1 /\ 0 <= I5 - 1]
0.86/0.89	  f3(I8, I9, I10) -> f5(I9, I10, I11) [0 <= I8 - 1]
0.86/0.89	  f4(I12, I13, I14) -> f3(I12, I14, I13) 
0.86/0.89	  f2(I15, I16, I17) -> f3(I16, I18, 0) [0 = I15 /\ I16 - 1 <= I16 - 1 /\ 0 <= I16 - 1]
0.86/0.89	  f2(I19, I20, I21) -> f3(I20, I22, I19) [I20 - 1 <= I20 - 1 /\ 0 <= I20 - 1]
0.86/0.89	  f2(I23, I24, I25) -> f3(I24, 0, 0) [0 = I23 /\ I24 - 1 <= I24 - 1 /\ 0 <= I24 - 1]
0.86/0.89	  f2(I26, I27, I28) -> f3(1, 0, I26) [1 = I27]
0.86/0.89	  f2(I29, I30, I31) -> f2(I29, I30 - 1, I32) [I30 - 1 <= I30 - 1 /\ 0 <= I30 - 1]
0.86/0.89	  f1(I33, I34, I35) -> f2(I36, I37, I38) [0 <= I33 - 1 /\ -1 <= I36 - 1 /\ 1 <= I34 - 1 /\ -1 <= I37 - 1]
0.86/0.89	
0.86/0.89	We use the basic value criterion with the projection function NU:
0.86/0.89	  NU[f5#(z1,z2,z3)] = z2
0.86/0.89	
0.86/0.89	This gives the following inequalities:
0.86/0.89	  I5 - 1 <= I5 - 1 /\ 0 <= I5 - 1  ==>  I5 >! I5 - 1
0.86/0.89	
0.86/0.89	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.86/0.89	
0.86/0.89	DP problem for innermost termination.
0.86/0.89	P =
0.86/0.89	  f2#(I29, I30, I31) -> f2#(I29, I30 - 1, I32) [I30 - 1 <= I30 - 1 /\ 0 <= I30 - 1]
0.86/0.89	R = 
0.86/0.89	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.86/0.89	  f5(I0, I1, I2) -> f5(I0 - 1, I1, I3) [0 <= I0 - 1 /\ I0 - 1 <= I0 - 1 /\ I1 <= 0]
0.86/0.89	  f5(I4, I5, I6) -> f5(I4, I5 - 1, I7) [I5 - 1 <= I5 - 1 /\ 0 <= I5 - 1]
0.86/0.89	  f3(I8, I9, I10) -> f5(I9, I10, I11) [0 <= I8 - 1]
0.86/0.89	  f4(I12, I13, I14) -> f3(I12, I14, I13) 
0.86/0.89	  f2(I15, I16, I17) -> f3(I16, I18, 0) [0 = I15 /\ I16 - 1 <= I16 - 1 /\ 0 <= I16 - 1]
0.86/0.89	  f2(I19, I20, I21) -> f3(I20, I22, I19) [I20 - 1 <= I20 - 1 /\ 0 <= I20 - 1]
0.86/0.89	  f2(I23, I24, I25) -> f3(I24, 0, 0) [0 = I23 /\ I24 - 1 <= I24 - 1 /\ 0 <= I24 - 1]
0.86/0.89	  f2(I26, I27, I28) -> f3(1, 0, I26) [1 = I27]
0.86/0.89	  f2(I29, I30, I31) -> f2(I29, I30 - 1, I32) [I30 - 1 <= I30 - 1 /\ 0 <= I30 - 1]
0.86/0.89	  f1(I33, I34, I35) -> f2(I36, I37, I38) [0 <= I33 - 1 /\ -1 <= I36 - 1 /\ 1 <= I34 - 1 /\ -1 <= I37 - 1]
0.86/0.89	
0.86/0.89	We use the basic value criterion with the projection function NU:
0.86/0.89	  NU[f2#(z1,z2,z3)] = z2
0.86/0.89	
0.86/0.89	This gives the following inequalities:
0.86/0.89	  I30 - 1 <= I30 - 1 /\ 0 <= I30 - 1  ==>  I30 >! I30 - 1
0.86/0.89	
0.86/0.89	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.86/3.88	EOF