2.32/2.33	YES
2.32/2.33	
2.32/2.33	DP problem for innermost termination.
2.32/2.33	P =
2.32/2.33	  f7#(x1, x2, x3) -> f6#(x1, x2, x3) 
2.32/2.33	  f6#(I0, I1, I2) -> f2#(I0, I1, I2) [I0 <= 0]
2.32/2.33	  f2#(I3, I4, I5) -> f5#(I3, I4, -1 + I5) [1 <= I5]
2.32/2.33	  f5#(I6, I7, I8) -> f1#(I6, I7, I8) [1 <= I6]
2.32/2.33	  f5#(I9, I10, I11) -> f4#(I9, I10, I11) [I9 <= 0]
2.32/2.33	  f4#(I12, I13, I14) -> f2#(1, I14, I14) 
2.32/2.33	  f4#(I15, I16, I17) -> f2#(I15, I16, I17) [I15 <= 0]
2.32/2.33	  f1#(I21, I22, I23) -> f2#(I21, I22, I23) [1 + I23 <= I22]
2.32/2.33	R = 
2.32/2.33	  f7(x1, x2, x3) -> f6(x1, x2, x3) 
2.32/2.33	  f6(I0, I1, I2) -> f2(I0, I1, I2) [I0 <= 0]
2.32/2.33	  f2(I3, I4, I5) -> f5(I3, I4, -1 + I5) [1 <= I5]
2.32/2.33	  f5(I6, I7, I8) -> f1(I6, I7, I8) [1 <= I6]
2.32/2.33	  f5(I9, I10, I11) -> f4(I9, I10, I11) [I9 <= 0]
2.32/2.33	  f4(I12, I13, I14) -> f2(1, I14, I14) 
2.32/2.33	  f4(I15, I16, I17) -> f2(I15, I16, I17) [I15 <= 0]
2.32/2.33	  f1(I18, I19, I20) -> f3(I18, I19, I20) [I19 <= I20]
2.32/2.33	  f1(I21, I22, I23) -> f2(I21, I22, I23) [1 + I23 <= I22]
2.32/2.33	
2.32/2.33	The dependency graph for this problem is:
2.32/2.33	  0 -> 1
2.32/2.33	  1 -> 2
2.32/2.33	  2 -> 3, 4
2.32/2.33	  3 -> 7
2.32/2.33	  4 -> 5, 6
2.32/2.33	  5 -> 2
2.32/2.33	  6 -> 2
2.32/2.33	  7 -> 2
2.32/2.33	Where:
2.32/2.33	  0) f7#(x1, x2, x3) -> f6#(x1, x2, x3) 
2.32/2.33	  1) f6#(I0, I1, I2) -> f2#(I0, I1, I2) [I0 <= 0]
2.32/2.33	  2) f2#(I3, I4, I5) -> f5#(I3, I4, -1 + I5) [1 <= I5]
2.32/2.33	  3) f5#(I6, I7, I8) -> f1#(I6, I7, I8) [1 <= I6]
2.32/2.33	  4) f5#(I9, I10, I11) -> f4#(I9, I10, I11) [I9 <= 0]
2.32/2.33	  5) f4#(I12, I13, I14) -> f2#(1, I14, I14) 
2.32/2.33	  6) f4#(I15, I16, I17) -> f2#(I15, I16, I17) [I15 <= 0]
2.32/2.33	  7) f1#(I21, I22, I23) -> f2#(I21, I22, I23) [1 + I23 <= I22]
2.32/2.33	
2.32/2.33	We have the following SCCs.
2.32/2.33	  { 2, 3, 4, 5, 6, 7 }
2.32/2.33	
2.32/2.33	DP problem for innermost termination.
2.32/2.33	P =
2.32/2.33	  f2#(I3, I4, I5) -> f5#(I3, I4, -1 + I5) [1 <= I5]
2.32/2.33	  f5#(I6, I7, I8) -> f1#(I6, I7, I8) [1 <= I6]
2.32/2.33	  f5#(I9, I10, I11) -> f4#(I9, I10, I11) [I9 <= 0]
2.32/2.33	  f4#(I12, I13, I14) -> f2#(1, I14, I14) 
2.32/2.33	  f4#(I15, I16, I17) -> f2#(I15, I16, I17) [I15 <= 0]
2.32/2.33	  f1#(I21, I22, I23) -> f2#(I21, I22, I23) [1 + I23 <= I22]
2.32/2.33	R = 
2.32/2.33	  f7(x1, x2, x3) -> f6(x1, x2, x3) 
2.32/2.33	  f6(I0, I1, I2) -> f2(I0, I1, I2) [I0 <= 0]
2.32/2.33	  f2(I3, I4, I5) -> f5(I3, I4, -1 + I5) [1 <= I5]
2.32/2.33	  f5(I6, I7, I8) -> f1(I6, I7, I8) [1 <= I6]
2.32/2.33	  f5(I9, I10, I11) -> f4(I9, I10, I11) [I9 <= 0]
2.32/2.33	  f4(I12, I13, I14) -> f2(1, I14, I14) 
2.32/2.33	  f4(I15, I16, I17) -> f2(I15, I16, I17) [I15 <= 0]
2.32/2.33	  f1(I18, I19, I20) -> f3(I18, I19, I20) [I19 <= I20]
2.32/2.33	  f1(I21, I22, I23) -> f2(I21, I22, I23) [1 + I23 <= I22]
2.32/2.33	
2.32/2.33	We use the basic value criterion with the projection function NU:
2.32/2.33	  NU[f4#(z1,z2,z3)] = z3
2.32/2.33	  NU[f1#(z1,z2,z3)] = z3
2.32/2.33	  NU[f5#(z1,z2,z3)] = z3
2.32/2.33	  NU[f2#(z1,z2,z3)] = z3
2.32/2.33	
2.32/2.33	This gives the following inequalities:
2.32/2.33	  1 <= I5  ==>  I5 >! -1 + I5
2.32/2.33	  1 <= I6  ==>  I8 (>! \union =) I8
2.32/2.33	  I9 <= 0  ==>  I11 (>! \union =) I11
2.32/2.33	    ==>  I14 (>! \union =) I14
2.32/2.33	  I15 <= 0  ==>  I17 (>! \union =) I17
2.32/2.33	  1 + I23 <= I22  ==>  I23 (>! \union =) I23
2.32/2.33	
2.32/2.33	We remove all the strictly oriented dependency pairs.
2.32/2.33	
2.32/2.33	DP problem for innermost termination.
2.32/2.33	P =
2.32/2.33	  f5#(I6, I7, I8) -> f1#(I6, I7, I8) [1 <= I6]
2.32/2.33	  f5#(I9, I10, I11) -> f4#(I9, I10, I11) [I9 <= 0]
2.32/2.33	  f4#(I12, I13, I14) -> f2#(1, I14, I14) 
2.32/2.33	  f4#(I15, I16, I17) -> f2#(I15, I16, I17) [I15 <= 0]
2.32/2.33	  f1#(I21, I22, I23) -> f2#(I21, I22, I23) [1 + I23 <= I22]
2.32/2.33	R = 
2.32/2.33	  f7(x1, x2, x3) -> f6(x1, x2, x3) 
2.32/2.33	  f6(I0, I1, I2) -> f2(I0, I1, I2) [I0 <= 0]
2.32/2.33	  f2(I3, I4, I5) -> f5(I3, I4, -1 + I5) [1 <= I5]
2.32/2.33	  f5(I6, I7, I8) -> f1(I6, I7, I8) [1 <= I6]
2.32/2.33	  f5(I9, I10, I11) -> f4(I9, I10, I11) [I9 <= 0]
2.32/2.33	  f4(I12, I13, I14) -> f2(1, I14, I14) 
2.32/2.33	  f4(I15, I16, I17) -> f2(I15, I16, I17) [I15 <= 0]
2.32/2.33	  f1(I18, I19, I20) -> f3(I18, I19, I20) [I19 <= I20]
2.32/2.33	  f1(I21, I22, I23) -> f2(I21, I22, I23) [1 + I23 <= I22]
2.32/2.33	
2.32/2.33	The dependency graph for this problem is:
2.32/2.33	  3 -> 7
2.32/2.33	  4 -> 5, 6
2.32/2.33	  5 -> 
2.32/2.33	  6 -> 
2.32/2.33	  7 -> 
2.32/2.33	Where:
2.32/2.33	  3) f5#(I6, I7, I8) -> f1#(I6, I7, I8) [1 <= I6]
2.32/2.33	  4) f5#(I9, I10, I11) -> f4#(I9, I10, I11) [I9 <= 0]
2.32/2.33	  5) f4#(I12, I13, I14) -> f2#(1, I14, I14) 
2.32/2.33	  6) f4#(I15, I16, I17) -> f2#(I15, I16, I17) [I15 <= 0]
2.32/2.33	  7) f1#(I21, I22, I23) -> f2#(I21, I22, I23) [1 + I23 <= I22]
2.32/2.33	
2.32/2.33	We have the following SCCs.
2.32/2.33	  
2.32/2.34	EOF