16.71/16.51	YES
16.71/16.51	
16.71/16.51	DP problem for innermost termination.
16.71/16.51	P =
16.71/16.51	  f6#(x1, x2, x3, x4, x5) -> f4#(x1, x2, x3, x4, x5) 
16.71/16.51	  f5#(I0, I1, I2, I3, I4) -> f3#(I0, I1, I2, I3, I4) 
16.71/16.51	  f3#(I5, I6, I7, I8, I9) -> f5#(I5, I6, I7, I8, -1 + I9) [0 <= -1 * I7 + I9]
16.71/16.51	  f3#(I10, I11, I12, I13, I14) -> f1#(I10, I11, I12, -1 + I13, I14) [1 - I12 + I14 <= 0]
16.71/16.51	  f4#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) 
16.71/16.51	  f1#(I20, I21, I22, I23, I24) -> f3#(I20, I21, I22, I23, I24) [0 <= -1 * I21 + I23]
16.71/16.51	R = 
16.71/16.51	  f6(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) 
16.71/16.51	  f5(I0, I1, I2, I3, I4) -> f3(I0, I1, I2, I3, I4) 
16.71/16.51	  f3(I5, I6, I7, I8, I9) -> f5(I5, I6, I7, I8, -1 + I9) [0 <= -1 * I7 + I9]
16.71/16.51	  f3(I10, I11, I12, I13, I14) -> f1(I10, I11, I12, -1 + I13, I14) [1 - I12 + I14 <= 0]
16.71/16.51	  f4(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I18, I19) 
16.71/16.51	  f1(I20, I21, I22, I23, I24) -> f3(I20, I21, I22, I23, I24) [0 <= -1 * I21 + I23]
16.71/16.51	  f1(I25, I26, I27, I28, I29) -> f2(rnd1, I26, I27, I28, I29) [rnd1 = rnd1 /\ 1 - I26 + I28 <= 0]
16.71/16.51	
16.71/16.51	The dependency graph for this problem is:
16.71/16.51	  0 -> 4
16.71/16.51	  1 -> 2, 3
16.71/16.51	  2 -> 1
16.71/16.51	  3 -> 5
16.71/16.51	  4 -> 5
16.71/16.51	  5 -> 2, 3
16.71/16.51	Where:
16.71/16.51	  0) f6#(x1, x2, x3, x4, x5) -> f4#(x1, x2, x3, x4, x5) 
16.71/16.51	  1) f5#(I0, I1, I2, I3, I4) -> f3#(I0, I1, I2, I3, I4) 
16.71/16.51	  2) f3#(I5, I6, I7, I8, I9) -> f5#(I5, I6, I7, I8, -1 + I9) [0 <= -1 * I7 + I9]
16.71/16.51	  3) f3#(I10, I11, I12, I13, I14) -> f1#(I10, I11, I12, -1 + I13, I14) [1 - I12 + I14 <= 0]
16.71/16.51	  4) f4#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) 
16.71/16.51	  5) f1#(I20, I21, I22, I23, I24) -> f3#(I20, I21, I22, I23, I24) [0 <= -1 * I21 + I23]
16.71/16.51	
16.71/16.51	We have the following SCCs.
16.71/16.51	  { 1, 2, 3, 5 }
16.71/16.51	
16.71/16.51	DP problem for innermost termination.
16.71/16.51	P =
16.71/16.51	  f5#(I0, I1, I2, I3, I4) -> f3#(I0, I1, I2, I3, I4) 
16.71/16.51	  f3#(I5, I6, I7, I8, I9) -> f5#(I5, I6, I7, I8, -1 + I9) [0 <= -1 * I7 + I9]
16.71/16.51	  f3#(I10, I11, I12, I13, I14) -> f1#(I10, I11, I12, -1 + I13, I14) [1 - I12 + I14 <= 0]
16.71/16.51	  f1#(I20, I21, I22, I23, I24) -> f3#(I20, I21, I22, I23, I24) [0 <= -1 * I21 + I23]
16.71/16.51	R = 
16.71/16.51	  f6(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) 
16.71/16.51	  f5(I0, I1, I2, I3, I4) -> f3(I0, I1, I2, I3, I4) 
16.71/16.51	  f3(I5, I6, I7, I8, I9) -> f5(I5, I6, I7, I8, -1 + I9) [0 <= -1 * I7 + I9]
16.71/16.51	  f3(I10, I11, I12, I13, I14) -> f1(I10, I11, I12, -1 + I13, I14) [1 - I12 + I14 <= 0]
16.71/16.51	  f4(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I18, I19) 
16.71/16.51	  f1(I20, I21, I22, I23, I24) -> f3(I20, I21, I22, I23, I24) [0 <= -1 * I21 + I23]
16.71/16.51	  f1(I25, I26, I27, I28, I29) -> f2(rnd1, I26, I27, I28, I29) [rnd1 = rnd1 /\ 1 - I26 + I28 <= 0]
16.71/16.51	
16.71/16.51	We use the reverse value criterion with the projection function NU:
16.71/16.51	  NU[f1#(z1,z2,z3,z4,z5)] = -1 * z3 + z5 + -1 * 0
16.71/16.51	  NU[f3#(z1,z2,z3,z4,z5)] = -1 * z3 + z5 + -1 * 0
16.71/16.51	  NU[f5#(z1,z2,z3,z4,z5)] = -1 * z3 + z5 + -1 * 0
16.71/16.51	
16.71/16.51	This gives the following inequalities:
16.71/16.51	    ==>  -1 * I2 + I4 + -1 * 0 >= -1 * I2 + I4 + -1 * 0
16.71/16.51	  0 <= -1 * I7 + I9  ==>  -1 * I7 + I9 + -1 * 0 > -1 * I7 + (-1 + I9) + -1 * 0  with  -1 * I7 + I9 + -1 * 0 >= 0
16.71/16.51	  1 - I12 + I14 <= 0  ==>  -1 * I12 + I14 + -1 * 0 >= -1 * I12 + I14 + -1 * 0
16.71/16.51	  0 <= -1 * I21 + I23  ==>  -1 * I22 + I24 + -1 * 0 >= -1 * I22 + I24 + -1 * 0
16.71/16.51	
16.71/16.51	We remove all the strictly oriented dependency pairs.
16.71/16.51	
16.71/16.51	DP problem for innermost termination.
16.71/16.51	P =
16.71/16.51	  f5#(I0, I1, I2, I3, I4) -> f3#(I0, I1, I2, I3, I4) 
16.71/16.51	  f3#(I10, I11, I12, I13, I14) -> f1#(I10, I11, I12, -1 + I13, I14) [1 - I12 + I14 <= 0]
16.71/16.51	  f1#(I20, I21, I22, I23, I24) -> f3#(I20, I21, I22, I23, I24) [0 <= -1 * I21 + I23]
16.71/16.51	R = 
16.71/16.51	  f6(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) 
16.71/16.51	  f5(I0, I1, I2, I3, I4) -> f3(I0, I1, I2, I3, I4) 
16.71/16.51	  f3(I5, I6, I7, I8, I9) -> f5(I5, I6, I7, I8, -1 + I9) [0 <= -1 * I7 + I9]
16.71/16.51	  f3(I10, I11, I12, I13, I14) -> f1(I10, I11, I12, -1 + I13, I14) [1 - I12 + I14 <= 0]
16.71/16.51	  f4(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I18, I19) 
16.71/16.51	  f1(I20, I21, I22, I23, I24) -> f3(I20, I21, I22, I23, I24) [0 <= -1 * I21 + I23]
16.71/16.51	  f1(I25, I26, I27, I28, I29) -> f2(rnd1, I26, I27, I28, I29) [rnd1 = rnd1 /\ 1 - I26 + I28 <= 0]
16.71/16.51	
16.71/16.51	The dependency graph for this problem is:
16.71/16.51	  1 -> 3
16.71/16.51	  3 -> 5
16.71/16.51	  5 -> 3
16.71/16.51	Where:
16.71/16.51	  1) f5#(I0, I1, I2, I3, I4) -> f3#(I0, I1, I2, I3, I4) 
16.71/16.51	  3) f3#(I10, I11, I12, I13, I14) -> f1#(I10, I11, I12, -1 + I13, I14) [1 - I12 + I14 <= 0]
16.71/16.51	  5) f1#(I20, I21, I22, I23, I24) -> f3#(I20, I21, I22, I23, I24) [0 <= -1 * I21 + I23]
16.71/16.51	
16.71/16.51	We have the following SCCs.
16.71/16.51	  { 3, 5 }
16.71/16.51	
16.71/16.51	DP problem for innermost termination.
16.71/16.51	P =
16.71/16.51	  f3#(I10, I11, I12, I13, I14) -> f1#(I10, I11, I12, -1 + I13, I14) [1 - I12 + I14 <= 0]
16.71/16.51	  f1#(I20, I21, I22, I23, I24) -> f3#(I20, I21, I22, I23, I24) [0 <= -1 * I21 + I23]
16.71/16.51	R = 
16.71/16.51	  f6(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) 
16.71/16.51	  f5(I0, I1, I2, I3, I4) -> f3(I0, I1, I2, I3, I4) 
16.71/16.51	  f3(I5, I6, I7, I8, I9) -> f5(I5, I6, I7, I8, -1 + I9) [0 <= -1 * I7 + I9]
16.71/16.51	  f3(I10, I11, I12, I13, I14) -> f1(I10, I11, I12, -1 + I13, I14) [1 - I12 + I14 <= 0]
16.71/16.51	  f4(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I18, I19) 
16.71/16.51	  f1(I20, I21, I22, I23, I24) -> f3(I20, I21, I22, I23, I24) [0 <= -1 * I21 + I23]
16.71/16.51	  f1(I25, I26, I27, I28, I29) -> f2(rnd1, I26, I27, I28, I29) [rnd1 = rnd1 /\ 1 - I26 + I28 <= 0]
16.71/16.51	
16.71/16.51	We use the reverse value criterion with the projection function NU:
16.71/16.51	  NU[f1#(z1,z2,z3,z4,z5)] = -1 * z2 + z4 + -1 * 0
16.71/16.51	  NU[f3#(z1,z2,z3,z4,z5)] = -1 * z2 + (-1 + z4) + -1 * 0
16.71/16.51	
16.71/16.51	This gives the following inequalities:
16.71/16.51	  1 - I12 + I14 <= 0  ==>  -1 * I11 + (-1 + I13) + -1 * 0 >= -1 * I11 + (-1 + I13) + -1 * 0
16.71/16.51	  0 <= -1 * I21 + I23  ==>  -1 * I21 + I23 + -1 * 0 > -1 * I21 + (-1 + I23) + -1 * 0  with  -1 * I21 + I23 + -1 * 0 >= 0
16.71/16.51	
16.71/16.51	We remove all the strictly oriented dependency pairs.
16.71/16.51	
16.71/16.51	DP problem for innermost termination.
16.71/16.51	P =
16.71/16.51	  f3#(I10, I11, I12, I13, I14) -> f1#(I10, I11, I12, -1 + I13, I14) [1 - I12 + I14 <= 0]
16.71/16.51	R = 
16.71/16.51	  f6(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) 
16.71/16.51	  f5(I0, I1, I2, I3, I4) -> f3(I0, I1, I2, I3, I4) 
16.71/16.51	  f3(I5, I6, I7, I8, I9) -> f5(I5, I6, I7, I8, -1 + I9) [0 <= -1 * I7 + I9]
16.71/16.51	  f3(I10, I11, I12, I13, I14) -> f1(I10, I11, I12, -1 + I13, I14) [1 - I12 + I14 <= 0]
16.71/16.51	  f4(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I18, I19) 
16.71/16.51	  f1(I20, I21, I22, I23, I24) -> f3(I20, I21, I22, I23, I24) [0 <= -1 * I21 + I23]
16.71/16.51	  f1(I25, I26, I27, I28, I29) -> f2(rnd1, I26, I27, I28, I29) [rnd1 = rnd1 /\ 1 - I26 + I28 <= 0]
16.71/16.51	
16.71/16.51	The dependency graph for this problem is:
16.71/16.51	  3 -> 
16.71/16.51	Where:
16.71/16.51	  3) f3#(I10, I11, I12, I13, I14) -> f1#(I10, I11, I12, -1 + I13, I14) [1 - I12 + I14 <= 0]
16.71/16.51	
16.71/16.51	We have the following SCCs.
16.71/16.51	  
16.71/19.48	EOF