3.29/3.24	YES
3.29/3.24	
3.29/3.24	DP problem for innermost termination.
3.29/3.24	P =
3.29/3.24	  f7#(x1, x2, x3) -> f6#(x1, x2, x3) 
3.29/3.24	  f6#(I0, I1, I2) -> f3#(0, I1, rnd3) [rnd3 = rnd3]
3.29/3.24	  f4#(I3, I4, I5) -> f2#(I3, 1 + I4, I5) [1 + I4 <= 10]
3.29/3.24	  f2#(I9, I10, I11) -> f4#(I9, I10, I11) 
3.29/3.24	  f3#(I12, I13, I14) -> f1#(I12, I13, I14) 
3.29/3.24	  f1#(I15, I16, I17) -> f3#(1 + I15, I16, I17) [1 + I15 <= 10]
3.29/3.24	  f1#(I18, I19, I20) -> f2#(I18, 0, I20) [10 <= I18]
3.29/3.24	R = 
3.29/3.24	  f7(x1, x2, x3) -> f6(x1, x2, x3) 
3.29/3.24	  f6(I0, I1, I2) -> f3(0, I1, rnd3) [rnd3 = rnd3]
3.29/3.24	  f4(I3, I4, I5) -> f2(I3, 1 + I4, I5) [1 + I4 <= 10]
3.29/3.24	  f4(I6, I7, I8) -> f5(I6, I7, I8) [10 <= I7]
3.29/3.24	  f2(I9, I10, I11) -> f4(I9, I10, I11) 
3.29/3.24	  f3(I12, I13, I14) -> f1(I12, I13, I14) 
3.29/3.24	  f1(I15, I16, I17) -> f3(1 + I15, I16, I17) [1 + I15 <= 10]
3.29/3.24	  f1(I18, I19, I20) -> f2(I18, 0, I20) [10 <= I18]
3.29/3.24	
3.29/3.24	The dependency graph for this problem is:
3.29/3.24	  0 -> 1
3.29/3.24	  1 -> 4
3.29/3.24	  2 -> 3
3.29/3.24	  3 -> 2
3.29/3.24	  4 -> 5, 6
3.29/3.24	  5 -> 4
3.29/3.24	  6 -> 3
3.29/3.24	Where:
3.29/3.24	  0) f7#(x1, x2, x3) -> f6#(x1, x2, x3) 
3.29/3.24	  1) f6#(I0, I1, I2) -> f3#(0, I1, rnd3) [rnd3 = rnd3]
3.29/3.24	  2) f4#(I3, I4, I5) -> f2#(I3, 1 + I4, I5) [1 + I4 <= 10]
3.29/3.24	  3) f2#(I9, I10, I11) -> f4#(I9, I10, I11) 
3.29/3.24	  4) f3#(I12, I13, I14) -> f1#(I12, I13, I14) 
3.29/3.24	  5) f1#(I15, I16, I17) -> f3#(1 + I15, I16, I17) [1 + I15 <= 10]
3.29/3.24	  6) f1#(I18, I19, I20) -> f2#(I18, 0, I20) [10 <= I18]
3.29/3.24	
3.29/3.24	We have the following SCCs.
3.29/3.24	  { 4, 5 }
3.29/3.24	  { 2, 3 }
3.29/3.24	
3.29/3.24	DP problem for innermost termination.
3.29/3.24	P =
3.29/3.24	  f4#(I3, I4, I5) -> f2#(I3, 1 + I4, I5) [1 + I4 <= 10]
3.29/3.24	  f2#(I9, I10, I11) -> f4#(I9, I10, I11) 
3.29/3.24	R = 
3.29/3.24	  f7(x1, x2, x3) -> f6(x1, x2, x3) 
3.29/3.24	  f6(I0, I1, I2) -> f3(0, I1, rnd3) [rnd3 = rnd3]
3.29/3.24	  f4(I3, I4, I5) -> f2(I3, 1 + I4, I5) [1 + I4 <= 10]
3.29/3.24	  f4(I6, I7, I8) -> f5(I6, I7, I8) [10 <= I7]
3.29/3.24	  f2(I9, I10, I11) -> f4(I9, I10, I11) 
3.29/3.24	  f3(I12, I13, I14) -> f1(I12, I13, I14) 
3.29/3.24	  f1(I15, I16, I17) -> f3(1 + I15, I16, I17) [1 + I15 <= 10]
3.29/3.24	  f1(I18, I19, I20) -> f2(I18, 0, I20) [10 <= I18]
3.29/3.24	
3.29/3.24	We use the reverse value criterion with the projection function NU:
3.29/3.24	  NU[f2#(z1,z2,z3)] = 10 + -1 * (1 + z2)
3.29/3.24	  NU[f4#(z1,z2,z3)] = 10 + -1 * (1 + z2)
3.29/3.24	
3.29/3.24	This gives the following inequalities:
3.29/3.24	  1 + I4 <= 10  ==>  10 + -1 * (1 + I4) > 10 + -1 * (1 + (1 + I4))  with  10 + -1 * (1 + I4) >= 0
3.29/3.24	    ==>  10 + -1 * (1 + I10) >= 10 + -1 * (1 + I10)
3.29/3.24	
3.29/3.24	We remove all the strictly oriented dependency pairs.
3.29/3.24	
3.29/3.24	DP problem for innermost termination.
3.29/3.24	P =
3.29/3.24	  f2#(I9, I10, I11) -> f4#(I9, I10, I11) 
3.29/3.24	R = 
3.29/3.24	  f7(x1, x2, x3) -> f6(x1, x2, x3) 
3.29/3.24	  f6(I0, I1, I2) -> f3(0, I1, rnd3) [rnd3 = rnd3]
3.29/3.24	  f4(I3, I4, I5) -> f2(I3, 1 + I4, I5) [1 + I4 <= 10]
3.29/3.24	  f4(I6, I7, I8) -> f5(I6, I7, I8) [10 <= I7]
3.29/3.24	  f2(I9, I10, I11) -> f4(I9, I10, I11) 
3.29/3.24	  f3(I12, I13, I14) -> f1(I12, I13, I14) 
3.29/3.24	  f1(I15, I16, I17) -> f3(1 + I15, I16, I17) [1 + I15 <= 10]
3.29/3.24	  f1(I18, I19, I20) -> f2(I18, 0, I20) [10 <= I18]
3.29/3.24	
3.29/3.24	The dependency graph for this problem is:
3.29/3.24	  3 -> 
3.29/3.24	Where:
3.29/3.24	  3) f2#(I9, I10, I11) -> f4#(I9, I10, I11) 
3.29/3.24	
3.29/3.24	We have the following SCCs.
3.29/3.24	  
3.29/3.24	
3.29/3.24	DP problem for innermost termination.
3.29/3.24	P =
3.29/3.24	  f3#(I12, I13, I14) -> f1#(I12, I13, I14) 
3.29/3.24	  f1#(I15, I16, I17) -> f3#(1 + I15, I16, I17) [1 + I15 <= 10]
3.29/3.24	R = 
3.29/3.24	  f7(x1, x2, x3) -> f6(x1, x2, x3) 
3.29/3.24	  f6(I0, I1, I2) -> f3(0, I1, rnd3) [rnd3 = rnd3]
3.29/3.24	  f4(I3, I4, I5) -> f2(I3, 1 + I4, I5) [1 + I4 <= 10]
3.29/3.24	  f4(I6, I7, I8) -> f5(I6, I7, I8) [10 <= I7]
3.29/3.24	  f2(I9, I10, I11) -> f4(I9, I10, I11) 
3.29/3.24	  f3(I12, I13, I14) -> f1(I12, I13, I14) 
3.29/3.24	  f1(I15, I16, I17) -> f3(1 + I15, I16, I17) [1 + I15 <= 10]
3.29/3.24	  f1(I18, I19, I20) -> f2(I18, 0, I20) [10 <= I18]
3.29/3.24	
3.29/3.24	We use the reverse value criterion with the projection function NU:
3.29/3.24	  NU[f1#(z1,z2,z3)] = 10 + -1 * (1 + z1)
3.29/3.24	  NU[f3#(z1,z2,z3)] = 10 + -1 * (1 + z1)
3.29/3.24	
3.29/3.24	This gives the following inequalities:
3.29/3.24	    ==>  10 + -1 * (1 + I12) >= 10 + -1 * (1 + I12)
3.29/3.24	  1 + I15 <= 10  ==>  10 + -1 * (1 + I15) > 10 + -1 * (1 + (1 + I15))  with  10 + -1 * (1 + I15) >= 0
3.29/3.24	
3.29/3.24	We remove all the strictly oriented dependency pairs.
3.29/3.24	
3.29/3.24	DP problem for innermost termination.
3.29/3.24	P =
3.29/3.24	  f3#(I12, I13, I14) -> f1#(I12, I13, I14) 
3.29/3.24	R = 
3.29/3.24	  f7(x1, x2, x3) -> f6(x1, x2, x3) 
3.29/3.24	  f6(I0, I1, I2) -> f3(0, I1, rnd3) [rnd3 = rnd3]
3.29/3.24	  f4(I3, I4, I5) -> f2(I3, 1 + I4, I5) [1 + I4 <= 10]
3.29/3.24	  f4(I6, I7, I8) -> f5(I6, I7, I8) [10 <= I7]
3.29/3.24	  f2(I9, I10, I11) -> f4(I9, I10, I11) 
3.29/3.24	  f3(I12, I13, I14) -> f1(I12, I13, I14) 
3.29/3.24	  f1(I15, I16, I17) -> f3(1 + I15, I16, I17) [1 + I15 <= 10]
3.29/3.24	  f1(I18, I19, I20) -> f2(I18, 0, I20) [10 <= I18]
3.29/3.24	
3.29/3.24	The dependency graph for this problem is:
3.29/3.24	  4 -> 
3.29/3.24	Where:
3.29/3.24	  4) f3#(I12, I13, I14) -> f1#(I12, I13, I14) 
3.29/3.24	
3.29/3.24	We have the following SCCs.
3.29/3.24	  
3.29/6.22	EOF