1.75/1.80	YES
1.75/1.80	
1.75/1.80	DP problem for innermost termination.
1.75/1.80	P =
1.75/1.80	  f8#(x1, x2, x3) -> f7#(x1, x2, x3) 
1.75/1.80	  f7#(I0, I1, I2) -> f4#(I0, 0, 10) 
1.75/1.80	  f2#(I3, I4, I5) -> f6#(I3, I4, I5) [2 * I5 <= I4]
1.75/1.80	  f2#(I6, I7, I8) -> f6#(I6, I7, I8) [1 + I7 <= 2 * I8]
1.75/1.80	  f3#(I12, I13, I14) -> f1#(I12, I13, I14) 
1.75/1.80	  f4#(I18, I19, I20) -> f3#(0, I19, I20) [2 <= I20]
1.75/1.80	  f1#(I21, I22, I23) -> f3#(1 + I21, 2 + I22, I23) [1 + I21 <= I23]
1.75/1.80	  f1#(I24, I25, I26) -> f2#(I24, I25, I26) [I26 <= I24]
1.75/1.80	R = 
1.75/1.80	  f8(x1, x2, x3) -> f7(x1, x2, x3) 
1.75/1.80	  f7(I0, I1, I2) -> f4(I0, 0, 10) 
1.75/1.80	  f2(I3, I4, I5) -> f6(I3, I4, I5) [2 * I5 <= I4]
1.75/1.80	  f2(I6, I7, I8) -> f6(I6, I7, I8) [1 + I7 <= 2 * I8]
1.75/1.80	  f6(I9, I10, I11) -> f5(I9, I10, I11) 
1.75/1.80	  f3(I12, I13, I14) -> f1(I12, I13, I14) 
1.75/1.80	  f4(I15, I16, I17) -> f5(I15, I16, I17) [I17 <= 1]
1.75/1.80	  f4(I18, I19, I20) -> f3(0, I19, I20) [2 <= I20]
1.75/1.80	  f1(I21, I22, I23) -> f3(1 + I21, 2 + I22, I23) [1 + I21 <= I23]
1.75/1.80	  f1(I24, I25, I26) -> f2(I24, I25, I26) [I26 <= I24]
1.75/1.80	
1.75/1.80	The dependency graph for this problem is:
1.75/1.80	  0 -> 1
1.75/1.80	  1 -> 5
1.75/1.80	  2 -> 
1.75/1.80	  3 -> 
1.75/1.80	  4 -> 6, 7
1.75/1.80	  5 -> 4
1.75/1.80	  6 -> 4
1.75/1.80	  7 -> 2, 3
1.75/1.80	Where:
1.75/1.80	  0) f8#(x1, x2, x3) -> f7#(x1, x2, x3) 
1.75/1.80	  1) f7#(I0, I1, I2) -> f4#(I0, 0, 10) 
1.75/1.80	  2) f2#(I3, I4, I5) -> f6#(I3, I4, I5) [2 * I5 <= I4]
1.75/1.80	  3) f2#(I6, I7, I8) -> f6#(I6, I7, I8) [1 + I7 <= 2 * I8]
1.75/1.80	  4) f3#(I12, I13, I14) -> f1#(I12, I13, I14) 
1.75/1.80	  5) f4#(I18, I19, I20) -> f3#(0, I19, I20) [2 <= I20]
1.75/1.80	  6) f1#(I21, I22, I23) -> f3#(1 + I21, 2 + I22, I23) [1 + I21 <= I23]
1.75/1.80	  7) f1#(I24, I25, I26) -> f2#(I24, I25, I26) [I26 <= I24]
1.75/1.80	
1.75/1.80	We have the following SCCs.
1.75/1.80	  { 4, 6 }
1.75/1.80	
1.75/1.80	DP problem for innermost termination.
1.75/1.80	P =
1.75/1.80	  f3#(I12, I13, I14) -> f1#(I12, I13, I14) 
1.75/1.80	  f1#(I21, I22, I23) -> f3#(1 + I21, 2 + I22, I23) [1 + I21 <= I23]
1.75/1.80	R = 
1.75/1.80	  f8(x1, x2, x3) -> f7(x1, x2, x3) 
1.75/1.80	  f7(I0, I1, I2) -> f4(I0, 0, 10) 
1.75/1.80	  f2(I3, I4, I5) -> f6(I3, I4, I5) [2 * I5 <= I4]
1.75/1.80	  f2(I6, I7, I8) -> f6(I6, I7, I8) [1 + I7 <= 2 * I8]
1.75/1.80	  f6(I9, I10, I11) -> f5(I9, I10, I11) 
1.75/1.80	  f3(I12, I13, I14) -> f1(I12, I13, I14) 
1.75/1.80	  f4(I15, I16, I17) -> f5(I15, I16, I17) [I17 <= 1]
1.75/1.80	  f4(I18, I19, I20) -> f3(0, I19, I20) [2 <= I20]
1.75/1.80	  f1(I21, I22, I23) -> f3(1 + I21, 2 + I22, I23) [1 + I21 <= I23]
1.75/1.80	  f1(I24, I25, I26) -> f2(I24, I25, I26) [I26 <= I24]
1.75/1.80	
1.75/1.80	We use the reverse value criterion with the projection function NU:
1.75/1.80	  NU[f1#(z1,z2,z3)] = z3 + -1 * (1 + z1)
1.75/1.80	  NU[f3#(z1,z2,z3)] = z3 + -1 * (1 + z1)
1.75/1.80	
1.75/1.80	This gives the following inequalities:
1.75/1.80	    ==>  I14 + -1 * (1 + I12) >= I14 + -1 * (1 + I12)
1.75/1.80	  1 + I21 <= I23  ==>  I23 + -1 * (1 + I21) > I23 + -1 * (1 + (1 + I21))  with  I23 + -1 * (1 + I21) >= 0
1.75/1.80	
1.75/1.80	We remove all the strictly oriented dependency pairs.
1.75/1.80	
1.75/1.80	DP problem for innermost termination.
1.75/1.80	P =
1.75/1.80	  f3#(I12, I13, I14) -> f1#(I12, I13, I14) 
1.75/1.80	R = 
1.75/1.80	  f8(x1, x2, x3) -> f7(x1, x2, x3) 
1.75/1.80	  f7(I0, I1, I2) -> f4(I0, 0, 10) 
1.75/1.80	  f2(I3, I4, I5) -> f6(I3, I4, I5) [2 * I5 <= I4]
1.75/1.80	  f2(I6, I7, I8) -> f6(I6, I7, I8) [1 + I7 <= 2 * I8]
1.75/1.80	  f6(I9, I10, I11) -> f5(I9, I10, I11) 
1.75/1.80	  f3(I12, I13, I14) -> f1(I12, I13, I14) 
1.75/1.80	  f4(I15, I16, I17) -> f5(I15, I16, I17) [I17 <= 1]
1.75/1.80	  f4(I18, I19, I20) -> f3(0, I19, I20) [2 <= I20]
1.75/1.80	  f1(I21, I22, I23) -> f3(1 + I21, 2 + I22, I23) [1 + I21 <= I23]
1.75/1.80	  f1(I24, I25, I26) -> f2(I24, I25, I26) [I26 <= I24]
1.75/1.80	
1.75/1.80	The dependency graph for this problem is:
1.75/1.80	  4 -> 
1.75/1.80	Where:
1.75/1.80	  4) f3#(I12, I13, I14) -> f1#(I12, I13, I14) 
1.75/1.80	
1.75/1.80	We have the following SCCs.
1.75/1.80	  
1.75/4.78	EOF