1.16/1.19	YES
1.16/1.19	
1.16/1.19	DP problem for innermost termination.
1.16/1.19	P =
1.16/1.19	  init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
1.16/1.19	  f5#(I0, I1, I2) -> f4#(I0, I3, I0) [I0 = I2 /\ 0 <= I1 - I0 * I3 /\ I1 - I0 * I3 <= I0 - 1 /\ I3 <= I1 - 1 /\ I0 <= I1 /\ 1 <= I0 - 1 /\ 1 <= I1 - 1]
1.16/1.19	  f4#(I4, I5, I6) -> f5#(I4, I5, I4) [1 <= I5 - 1 /\ 1 <= I4 - 1 /\ y1 <= I5 - 1 /\ I4 <= I5 /\ I4 = I6]
1.16/1.19	  f3#(I7, I8, I9) -> f4#(I9, I8, I9) [0 <= I7 - 1 /\ 1 <= I10 - 1]
1.16/1.19	  f2#(I11, I12, I13) -> f4#(0, I12, 0) [0 <= I11 - 1]
1.16/1.19	  f1#(I14, I15, I16) -> f4#(0, 0, 0) [0 = I15 /\ 0 <= I14 - 1]
1.16/1.19	  f2#(I17, I18, I19) -> f3#(I20, I18, I21) [1 <= I22 - 1 /\ -1 <= I21 - 1 /\ I20 <= I17 /\ 0 <= I17 - 1 /\ 0 <= I20 - 1]
1.16/1.19	  f2#(I23, I24, I25) -> f3#(I26, I24, 0) [I26 <= I23 /\ 1 <= I27 - 1 /\ 0 <= I23 - 1 /\ 0 <= I26 - 1]
1.16/1.19	  f1#(I28, I29, I30) -> f2#(I31, I32, I33) [0 <= I31 - 1 /\ 0 <= I28 - 1 /\ I31 <= I28 /\ 0 <= I29 - 1 /\ -1 <= I32 - 1]
1.16/1.19	  f1#(I34, I35, I36) -> f2#(I37, 0, I38) [0 <= I37 - 1 /\ 0 <= I34 - 1 /\ 0 <= I35 - 1 /\ I37 <= I34]
1.16/1.19	R = 
1.16/1.19	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
1.16/1.19	  f5(I0, I1, I2) -> f4(I0, I3, I0) [I0 = I2 /\ 0 <= I1 - I0 * I3 /\ I1 - I0 * I3 <= I0 - 1 /\ I3 <= I1 - 1 /\ I0 <= I1 /\ 1 <= I0 - 1 /\ 1 <= I1 - 1]
1.16/1.19	  f4(I4, I5, I6) -> f5(I4, I5, I4) [1 <= I5 - 1 /\ 1 <= I4 - 1 /\ y1 <= I5 - 1 /\ I4 <= I5 /\ I4 = I6]
1.16/1.19	  f3(I7, I8, I9) -> f4(I9, I8, I9) [0 <= I7 - 1 /\ 1 <= I10 - 1]
1.16/1.19	  f2(I11, I12, I13) -> f4(0, I12, 0) [0 <= I11 - 1]
1.16/1.19	  f1(I14, I15, I16) -> f4(0, 0, 0) [0 = I15 /\ 0 <= I14 - 1]
1.16/1.19	  f2(I17, I18, I19) -> f3(I20, I18, I21) [1 <= I22 - 1 /\ -1 <= I21 - 1 /\ I20 <= I17 /\ 0 <= I17 - 1 /\ 0 <= I20 - 1]
1.16/1.19	  f2(I23, I24, I25) -> f3(I26, I24, 0) [I26 <= I23 /\ 1 <= I27 - 1 /\ 0 <= I23 - 1 /\ 0 <= I26 - 1]
1.16/1.19	  f1(I28, I29, I30) -> f2(I31, I32, I33) [0 <= I31 - 1 /\ 0 <= I28 - 1 /\ I31 <= I28 /\ 0 <= I29 - 1 /\ -1 <= I32 - 1]
1.16/1.19	  f1(I34, I35, I36) -> f2(I37, 0, I38) [0 <= I37 - 1 /\ 0 <= I34 - 1 /\ 0 <= I35 - 1 /\ I37 <= I34]
1.16/1.19	
1.16/1.19	The dependency graph for this problem is:
1.16/1.19	  0 -> 5, 8, 9
1.16/1.19	  1 -> 2
1.16/1.19	  2 -> 1
1.16/1.19	  3 -> 2
1.16/1.19	  4 -> 
1.16/1.19	  5 -> 
1.16/1.19	  6 -> 3
1.16/1.19	  7 -> 3
1.16/1.19	  8 -> 4, 6, 7
1.16/1.19	  9 -> 4, 6, 7
1.16/1.19	Where:
1.16/1.19	  0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
1.16/1.19	  1) f5#(I0, I1, I2) -> f4#(I0, I3, I0) [I0 = I2 /\ 0 <= I1 - I0 * I3 /\ I1 - I0 * I3 <= I0 - 1 /\ I3 <= I1 - 1 /\ I0 <= I1 /\ 1 <= I0 - 1 /\ 1 <= I1 - 1]
1.16/1.19	  2) f4#(I4, I5, I6) -> f5#(I4, I5, I4) [1 <= I5 - 1 /\ 1 <= I4 - 1 /\ y1 <= I5 - 1 /\ I4 <= I5 /\ I4 = I6]
1.16/1.19	  3) f3#(I7, I8, I9) -> f4#(I9, I8, I9) [0 <= I7 - 1 /\ 1 <= I10 - 1]
1.16/1.19	  4) f2#(I11, I12, I13) -> f4#(0, I12, 0) [0 <= I11 - 1]
1.16/1.19	  5) f1#(I14, I15, I16) -> f4#(0, 0, 0) [0 = I15 /\ 0 <= I14 - 1]
1.16/1.19	  6) f2#(I17, I18, I19) -> f3#(I20, I18, I21) [1 <= I22 - 1 /\ -1 <= I21 - 1 /\ I20 <= I17 /\ 0 <= I17 - 1 /\ 0 <= I20 - 1]
1.16/1.19	  7) f2#(I23, I24, I25) -> f3#(I26, I24, 0) [I26 <= I23 /\ 1 <= I27 - 1 /\ 0 <= I23 - 1 /\ 0 <= I26 - 1]
1.16/1.19	  8) f1#(I28, I29, I30) -> f2#(I31, I32, I33) [0 <= I31 - 1 /\ 0 <= I28 - 1 /\ I31 <= I28 /\ 0 <= I29 - 1 /\ -1 <= I32 - 1]
1.16/1.19	  9) f1#(I34, I35, I36) -> f2#(I37, 0, I38) [0 <= I37 - 1 /\ 0 <= I34 - 1 /\ 0 <= I35 - 1 /\ I37 <= I34]
1.16/1.19	
1.16/1.19	We have the following SCCs.
1.16/1.19	  { 1, 2 }
1.16/1.19	
1.16/1.19	DP problem for innermost termination.
1.16/1.19	P =
1.16/1.19	  f5#(I0, I1, I2) -> f4#(I0, I3, I0) [I0 = I2 /\ 0 <= I1 - I0 * I3 /\ I1 - I0 * I3 <= I0 - 1 /\ I3 <= I1 - 1 /\ I0 <= I1 /\ 1 <= I0 - 1 /\ 1 <= I1 - 1]
1.16/1.19	  f4#(I4, I5, I6) -> f5#(I4, I5, I4) [1 <= I5 - 1 /\ 1 <= I4 - 1 /\ y1 <= I5 - 1 /\ I4 <= I5 /\ I4 = I6]
1.16/1.19	R = 
1.16/1.19	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
1.16/1.19	  f5(I0, I1, I2) -> f4(I0, I3, I0) [I0 = I2 /\ 0 <= I1 - I0 * I3 /\ I1 - I0 * I3 <= I0 - 1 /\ I3 <= I1 - 1 /\ I0 <= I1 /\ 1 <= I0 - 1 /\ 1 <= I1 - 1]
1.16/1.19	  f4(I4, I5, I6) -> f5(I4, I5, I4) [1 <= I5 - 1 /\ 1 <= I4 - 1 /\ y1 <= I5 - 1 /\ I4 <= I5 /\ I4 = I6]
1.16/1.19	  f3(I7, I8, I9) -> f4(I9, I8, I9) [0 <= I7 - 1 /\ 1 <= I10 - 1]
1.16/1.19	  f2(I11, I12, I13) -> f4(0, I12, 0) [0 <= I11 - 1]
1.16/1.19	  f1(I14, I15, I16) -> f4(0, 0, 0) [0 = I15 /\ 0 <= I14 - 1]
1.16/1.19	  f2(I17, I18, I19) -> f3(I20, I18, I21) [1 <= I22 - 1 /\ -1 <= I21 - 1 /\ I20 <= I17 /\ 0 <= I17 - 1 /\ 0 <= I20 - 1]
1.16/1.19	  f2(I23, I24, I25) -> f3(I26, I24, 0) [I26 <= I23 /\ 1 <= I27 - 1 /\ 0 <= I23 - 1 /\ 0 <= I26 - 1]
1.16/1.19	  f1(I28, I29, I30) -> f2(I31, I32, I33) [0 <= I31 - 1 /\ 0 <= I28 - 1 /\ I31 <= I28 /\ 0 <= I29 - 1 /\ -1 <= I32 - 1]
1.16/1.19	  f1(I34, I35, I36) -> f2(I37, 0, I38) [0 <= I37 - 1 /\ 0 <= I34 - 1 /\ 0 <= I35 - 1 /\ I37 <= I34]
1.16/1.19	
1.16/1.19	We use the basic value criterion with the projection function NU:
1.16/1.19	  NU[f4#(z1,z2,z3)] = z2
1.16/1.19	  NU[f5#(z1,z2,z3)] = z2
1.16/1.19	
1.16/1.19	This gives the following inequalities:
1.16/1.19	  I0 = I2 /\ 0 <= I1 - I0 * I3 /\ I1 - I0 * I3 <= I0 - 1 /\ I3 <= I1 - 1 /\ I0 <= I1 /\ 1 <= I0 - 1 /\ 1 <= I1 - 1  ==>  I1 >! I3
1.16/1.19	  1 <= I5 - 1 /\ 1 <= I4 - 1 /\ y1 <= I5 - 1 /\ I4 <= I5 /\ I4 = I6  ==>  I5 (>! \union =) I5
1.16/1.19	
1.16/1.19	We remove all the strictly oriented dependency pairs.
1.16/1.19	
1.16/1.19	DP problem for innermost termination.
1.16/1.19	P =
1.16/1.19	  f4#(I4, I5, I6) -> f5#(I4, I5, I4) [1 <= I5 - 1 /\ 1 <= I4 - 1 /\ y1 <= I5 - 1 /\ I4 <= I5 /\ I4 = I6]
1.16/1.19	R = 
1.16/1.19	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
1.16/1.19	  f5(I0, I1, I2) -> f4(I0, I3, I0) [I0 = I2 /\ 0 <= I1 - I0 * I3 /\ I1 - I0 * I3 <= I0 - 1 /\ I3 <= I1 - 1 /\ I0 <= I1 /\ 1 <= I0 - 1 /\ 1 <= I1 - 1]
1.16/1.19	  f4(I4, I5, I6) -> f5(I4, I5, I4) [1 <= I5 - 1 /\ 1 <= I4 - 1 /\ y1 <= I5 - 1 /\ I4 <= I5 /\ I4 = I6]
1.16/1.20	  f3(I7, I8, I9) -> f4(I9, I8, I9) [0 <= I7 - 1 /\ 1 <= I10 - 1]
1.16/1.20	  f2(I11, I12, I13) -> f4(0, I12, 0) [0 <= I11 - 1]
1.16/1.20	  f1(I14, I15, I16) -> f4(0, 0, 0) [0 = I15 /\ 0 <= I14 - 1]
1.16/1.20	  f2(I17, I18, I19) -> f3(I20, I18, I21) [1 <= I22 - 1 /\ -1 <= I21 - 1 /\ I20 <= I17 /\ 0 <= I17 - 1 /\ 0 <= I20 - 1]
1.16/1.20	  f2(I23, I24, I25) -> f3(I26, I24, 0) [I26 <= I23 /\ 1 <= I27 - 1 /\ 0 <= I23 - 1 /\ 0 <= I26 - 1]
1.16/1.20	  f1(I28, I29, I30) -> f2(I31, I32, I33) [0 <= I31 - 1 /\ 0 <= I28 - 1 /\ I31 <= I28 /\ 0 <= I29 - 1 /\ -1 <= I32 - 1]
1.16/1.20	  f1(I34, I35, I36) -> f2(I37, 0, I38) [0 <= I37 - 1 /\ 0 <= I34 - 1 /\ 0 <= I35 - 1 /\ I37 <= I34]
1.16/1.20	
1.16/1.20	The dependency graph for this problem is:
1.16/1.20	  2 -> 
1.16/1.20	Where:
1.16/1.20	  2) f4#(I4, I5, I6) -> f5#(I4, I5, I4) [1 <= I5 - 1 /\ 1 <= I4 - 1 /\ y1 <= I5 - 1 /\ I4 <= I5 /\ I4 = I6]
1.16/1.20	
1.16/1.20	We have the following SCCs.
1.16/1.20	  
1.19/4.17	EOF