6.54/6.44	YES
6.54/6.44	
6.54/6.44	DP problem for innermost termination.
6.54/6.44	P =
6.54/6.44	  f5#(x1, x2, x3) -> f4#(x1, x2, x3) 
6.54/6.44	  f4#(I0, I1, I2) -> f2#(0, I1, I2) 
6.54/6.44	  f2#(I3, I4, I5) -> f1#(I3, 0, I5) [1 + I3 <= I5]
6.54/6.44	  f3#(I6, I7, I8) -> f1#(I6, I7, I8) 
6.54/6.44	  f1#(I9, I10, I11) -> f3#(I9, 1 + I10, I11) [1 + I10 <= I9]
6.54/6.44	  f1#(I12, I13, I14) -> f2#(1 + I12, I13, I14) [I12 <= I13]
6.54/6.44	R = 
6.54/6.44	  f5(x1, x2, x3) -> f4(x1, x2, x3) 
6.54/6.44	  f4(I0, I1, I2) -> f2(0, I1, I2) 
6.54/6.44	  f2(I3, I4, I5) -> f1(I3, 0, I5) [1 + I3 <= I5]
6.54/6.44	  f3(I6, I7, I8) -> f1(I6, I7, I8) 
6.54/6.44	  f1(I9, I10, I11) -> f3(I9, 1 + I10, I11) [1 + I10 <= I9]
6.54/6.44	  f1(I12, I13, I14) -> f2(1 + I12, I13, I14) [I12 <= I13]
6.54/6.44	
6.54/6.44	The dependency graph for this problem is:
6.54/6.44	  0 -> 1
6.54/6.44	  1 -> 2
6.54/6.44	  2 -> 4, 5
6.54/6.44	  3 -> 4, 5
6.54/6.44	  4 -> 3
6.54/6.44	  5 -> 2
6.54/6.44	Where:
6.54/6.44	  0) f5#(x1, x2, x3) -> f4#(x1, x2, x3) 
6.54/6.44	  1) f4#(I0, I1, I2) -> f2#(0, I1, I2) 
6.54/6.44	  2) f2#(I3, I4, I5) -> f1#(I3, 0, I5) [1 + I3 <= I5]
6.54/6.44	  3) f3#(I6, I7, I8) -> f1#(I6, I7, I8) 
6.54/6.44	  4) f1#(I9, I10, I11) -> f3#(I9, 1 + I10, I11) [1 + I10 <= I9]
6.54/6.44	  5) f1#(I12, I13, I14) -> f2#(1 + I12, I13, I14) [I12 <= I13]
6.54/6.44	
6.54/6.44	We have the following SCCs.
6.54/6.44	  { 2, 3, 4, 5 }
6.54/6.44	
6.54/6.44	DP problem for innermost termination.
6.54/6.44	P =
6.54/6.44	  f2#(I3, I4, I5) -> f1#(I3, 0, I5) [1 + I3 <= I5]
6.54/6.44	  f3#(I6, I7, I8) -> f1#(I6, I7, I8) 
6.54/6.44	  f1#(I9, I10, I11) -> f3#(I9, 1 + I10, I11) [1 + I10 <= I9]
6.54/6.44	  f1#(I12, I13, I14) -> f2#(1 + I12, I13, I14) [I12 <= I13]
6.54/6.44	R = 
6.54/6.44	  f5(x1, x2, x3) -> f4(x1, x2, x3) 
6.54/6.44	  f4(I0, I1, I2) -> f2(0, I1, I2) 
6.54/6.44	  f2(I3, I4, I5) -> f1(I3, 0, I5) [1 + I3 <= I5]
6.54/6.44	  f3(I6, I7, I8) -> f1(I6, I7, I8) 
6.54/6.44	  f1(I9, I10, I11) -> f3(I9, 1 + I10, I11) [1 + I10 <= I9]
6.54/6.44	  f1(I12, I13, I14) -> f2(1 + I12, I13, I14) [I12 <= I13]
6.54/6.44	
6.54/6.44	We use the extended value criterion with the projection function NU:
6.54/6.44	  NU[f3#(x0,x1,x2)] = -x0 + x2 - 2
6.54/6.44	  NU[f1#(x0,x1,x2)] = -x0 + x2 - 2
6.54/6.44	  NU[f2#(x0,x1,x2)] = -x0 + x2 - 1
6.54/6.44	
6.54/6.44	This gives the following inequalities:
6.54/6.44	  1 + I3 <= I5  ==>  -I3 + I5 - 1 > -I3 + I5 - 2  with  -I3 + I5 - 1 >= 0
6.54/6.44	    ==>  -I6 + I8 - 2 >= -I6 + I8 - 2
6.54/6.44	  1 + I10 <= I9  ==>  -I9 + I11 - 2 >= -I9 + I11 - 2
6.54/6.44	  I12 <= I13  ==>  -I12 + I14 - 2 >= -(1 + I12) + I14 - 1
6.54/6.44	
6.54/6.44	We remove all the strictly oriented dependency pairs.
6.54/6.44	
6.54/6.44	DP problem for innermost termination.
6.54/6.44	P =
6.54/6.44	  f3#(I6, I7, I8) -> f1#(I6, I7, I8) 
6.54/6.44	  f1#(I9, I10, I11) -> f3#(I9, 1 + I10, I11) [1 + I10 <= I9]
6.54/6.44	  f1#(I12, I13, I14) -> f2#(1 + I12, I13, I14) [I12 <= I13]
6.54/6.44	R = 
6.54/6.44	  f5(x1, x2, x3) -> f4(x1, x2, x3) 
6.54/6.44	  f4(I0, I1, I2) -> f2(0, I1, I2) 
6.54/6.44	  f2(I3, I4, I5) -> f1(I3, 0, I5) [1 + I3 <= I5]
6.54/6.44	  f3(I6, I7, I8) -> f1(I6, I7, I8) 
6.54/6.44	  f1(I9, I10, I11) -> f3(I9, 1 + I10, I11) [1 + I10 <= I9]
6.54/6.44	  f1(I12, I13, I14) -> f2(1 + I12, I13, I14) [I12 <= I13]
6.54/6.44	
6.54/6.44	The dependency graph for this problem is:
6.54/6.44	  3 -> 4, 5
6.54/6.44	  4 -> 3
6.54/6.44	  5 -> 
6.54/6.44	Where:
6.54/6.44	  3) f3#(I6, I7, I8) -> f1#(I6, I7, I8) 
6.54/6.44	  4) f1#(I9, I10, I11) -> f3#(I9, 1 + I10, I11) [1 + I10 <= I9]
6.54/6.44	  5) f1#(I12, I13, I14) -> f2#(1 + I12, I13, I14) [I12 <= I13]
6.54/6.44	
6.54/6.44	We have the following SCCs.
6.54/6.44	  { 3, 4 }
6.54/6.44	
6.54/6.44	DP problem for innermost termination.
6.54/6.44	P =
6.54/6.44	  f3#(I6, I7, I8) -> f1#(I6, I7, I8) 
6.54/6.44	  f1#(I9, I10, I11) -> f3#(I9, 1 + I10, I11) [1 + I10 <= I9]
6.54/6.44	R = 
6.54/6.44	  f5(x1, x2, x3) -> f4(x1, x2, x3) 
6.54/6.44	  f4(I0, I1, I2) -> f2(0, I1, I2) 
6.54/6.44	  f2(I3, I4, I5) -> f1(I3, 0, I5) [1 + I3 <= I5]
6.54/6.44	  f3(I6, I7, I8) -> f1(I6, I7, I8) 
6.54/6.44	  f1(I9, I10, I11) -> f3(I9, 1 + I10, I11) [1 + I10 <= I9]
6.54/6.44	  f1(I12, I13, I14) -> f2(1 + I12, I13, I14) [I12 <= I13]
6.54/6.44	
6.54/6.44	We use the reverse value criterion with the projection function NU:
6.54/6.44	  NU[f1#(z1,z2,z3)] = z1 + -1 * (1 + z2)
6.54/6.44	  NU[f3#(z1,z2,z3)] = z1 + -1 * (1 + z2)
6.54/6.44	
6.54/6.44	This gives the following inequalities:
6.54/6.44	    ==>  I6 + -1 * (1 + I7) >= I6 + -1 * (1 + I7)
6.54/6.44	  1 + I10 <= I9  ==>  I9 + -1 * (1 + I10) > I9 + -1 * (1 + (1 + I10))  with  I9 + -1 * (1 + I10) >= 0
6.54/6.44	
6.54/6.44	We remove all the strictly oriented dependency pairs.
6.54/6.44	
6.54/6.44	DP problem for innermost termination.
6.54/6.44	P =
6.54/6.44	  f3#(I6, I7, I8) -> f1#(I6, I7, I8) 
6.54/6.44	R = 
6.54/6.44	  f5(x1, x2, x3) -> f4(x1, x2, x3) 
6.54/6.44	  f4(I0, I1, I2) -> f2(0, I1, I2) 
6.54/6.44	  f2(I3, I4, I5) -> f1(I3, 0, I5) [1 + I3 <= I5]
6.54/6.44	  f3(I6, I7, I8) -> f1(I6, I7, I8) 
6.54/6.44	  f1(I9, I10, I11) -> f3(I9, 1 + I10, I11) [1 + I10 <= I9]
6.54/6.44	  f1(I12, I13, I14) -> f2(1 + I12, I13, I14) [I12 <= I13]
6.54/6.44	
6.54/6.44	The dependency graph for this problem is:
6.54/6.44	  3 -> 
6.54/6.44	Where:
6.54/6.44	  3) f3#(I6, I7, I8) -> f1#(I6, I7, I8) 
6.54/6.44	
6.54/6.44	We have the following SCCs.
6.54/6.44	  
6.54/9.42	EOF