3.11/3.18	YES
3.11/3.18	
3.11/3.18	DP problem for innermost termination.
3.11/3.18	P =
3.11/3.18	  f8#(x1, x2, x3, x4) -> f7#(x1, x2, x3, x4) 
3.11/3.18	  f7#(I0, I1, I2, I3) -> f4#(I0, I1, 0, I0) 
3.11/3.18	  f2#(I4, I5, I6, I7) -> f6#(I4, I5, I6, I7) [2 * I7 <= I6]
3.11/3.18	  f2#(I8, I9, I10, I11) -> f6#(I8, I9, I10, I11) [1 + I10 <= 2 * I11]
3.11/3.18	  f3#(I16, I17, I18, I19) -> f1#(I16, I17, I18, I19) 
3.11/3.18	  f4#(I24, I25, I26, I27) -> f3#(I24, 0, I26, I27) [2 <= I27]
3.11/3.18	  f1#(I28, I29, I30, I31) -> f3#(I28, 1 + I29, 2 + I30, I31) [1 + I29 <= I31]
3.11/3.18	  f1#(I32, I33, I34, I35) -> f2#(I32, I33, I34, I35) [I35 <= I33]
3.11/3.18	R = 
3.11/3.18	  f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) 
3.11/3.18	  f7(I0, I1, I2, I3) -> f4(I0, I1, 0, I0) 
3.11/3.18	  f2(I4, I5, I6, I7) -> f6(I4, I5, I6, I7) [2 * I7 <= I6]
3.11/3.18	  f2(I8, I9, I10, I11) -> f6(I8, I9, I10, I11) [1 + I10 <= 2 * I11]
3.11/3.18	  f6(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) 
3.11/3.18	  f3(I16, I17, I18, I19) -> f1(I16, I17, I18, I19) 
3.11/3.18	  f4(I20, I21, I22, I23) -> f5(I20, I21, I22, I23) [I23 <= 1]
3.11/3.18	  f4(I24, I25, I26, I27) -> f3(I24, 0, I26, I27) [2 <= I27]
3.11/3.18	  f1(I28, I29, I30, I31) -> f3(I28, 1 + I29, 2 + I30, I31) [1 + I29 <= I31]
3.11/3.18	  f1(I32, I33, I34, I35) -> f2(I32, I33, I34, I35) [I35 <= I33]
3.11/3.18	
3.11/3.18	The dependency graph for this problem is:
3.11/3.18	  0 -> 1
3.11/3.18	  1 -> 5
3.11/3.18	  2 -> 
3.11/3.18	  3 -> 
3.11/3.18	  4 -> 6, 7
3.11/3.18	  5 -> 4
3.11/3.18	  6 -> 4
3.11/3.18	  7 -> 2, 3
3.11/3.18	Where:
3.11/3.18	  0) f8#(x1, x2, x3, x4) -> f7#(x1, x2, x3, x4) 
3.11/3.18	  1) f7#(I0, I1, I2, I3) -> f4#(I0, I1, 0, I0) 
3.11/3.18	  2) f2#(I4, I5, I6, I7) -> f6#(I4, I5, I6, I7) [2 * I7 <= I6]
3.11/3.18	  3) f2#(I8, I9, I10, I11) -> f6#(I8, I9, I10, I11) [1 + I10 <= 2 * I11]
3.11/3.18	  4) f3#(I16, I17, I18, I19) -> f1#(I16, I17, I18, I19) 
3.11/3.18	  5) f4#(I24, I25, I26, I27) -> f3#(I24, 0, I26, I27) [2 <= I27]
3.11/3.18	  6) f1#(I28, I29, I30, I31) -> f3#(I28, 1 + I29, 2 + I30, I31) [1 + I29 <= I31]
3.11/3.18	  7) f1#(I32, I33, I34, I35) -> f2#(I32, I33, I34, I35) [I35 <= I33]
3.11/3.18	
3.11/3.18	We have the following SCCs.
3.11/3.18	  { 4, 6 }
3.11/3.18	
3.11/3.18	DP problem for innermost termination.
3.11/3.18	P =
3.11/3.18	  f3#(I16, I17, I18, I19) -> f1#(I16, I17, I18, I19) 
3.11/3.18	  f1#(I28, I29, I30, I31) -> f3#(I28, 1 + I29, 2 + I30, I31) [1 + I29 <= I31]
3.11/3.18	R = 
3.11/3.18	  f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) 
3.11/3.18	  f7(I0, I1, I2, I3) -> f4(I0, I1, 0, I0) 
3.11/3.18	  f2(I4, I5, I6, I7) -> f6(I4, I5, I6, I7) [2 * I7 <= I6]
3.11/3.18	  f2(I8, I9, I10, I11) -> f6(I8, I9, I10, I11) [1 + I10 <= 2 * I11]
3.11/3.18	  f6(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) 
3.11/3.18	  f3(I16, I17, I18, I19) -> f1(I16, I17, I18, I19) 
3.11/3.18	  f4(I20, I21, I22, I23) -> f5(I20, I21, I22, I23) [I23 <= 1]
3.11/3.18	  f4(I24, I25, I26, I27) -> f3(I24, 0, I26, I27) [2 <= I27]
3.11/3.18	  f1(I28, I29, I30, I31) -> f3(I28, 1 + I29, 2 + I30, I31) [1 + I29 <= I31]
3.11/3.18	  f1(I32, I33, I34, I35) -> f2(I32, I33, I34, I35) [I35 <= I33]
3.11/3.18	
3.11/3.18	We use the reverse value criterion with the projection function NU:
3.11/3.18	  NU[f1#(z1,z2,z3,z4)] = z4 + -1 * (1 + z2)
3.11/3.18	  NU[f3#(z1,z2,z3,z4)] = z4 + -1 * (1 + z2)
3.11/3.18	
3.11/3.18	This gives the following inequalities:
3.11/3.18	    ==>  I19 + -1 * (1 + I17) >= I19 + -1 * (1 + I17)
3.11/3.18	  1 + I29 <= I31  ==>  I31 + -1 * (1 + I29) > I31 + -1 * (1 + (1 + I29))  with  I31 + -1 * (1 + I29) >= 0
3.11/3.18	
3.11/3.18	We remove all the strictly oriented dependency pairs.
3.11/3.18	
3.11/3.18	DP problem for innermost termination.
3.11/3.18	P =
3.11/3.18	  f3#(I16, I17, I18, I19) -> f1#(I16, I17, I18, I19) 
3.11/3.18	R = 
3.11/3.18	  f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) 
3.11/3.18	  f7(I0, I1, I2, I3) -> f4(I0, I1, 0, I0) 
3.11/3.18	  f2(I4, I5, I6, I7) -> f6(I4, I5, I6, I7) [2 * I7 <= I6]
3.11/3.18	  f2(I8, I9, I10, I11) -> f6(I8, I9, I10, I11) [1 + I10 <= 2 * I11]
3.11/3.18	  f6(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) 
3.11/3.18	  f3(I16, I17, I18, I19) -> f1(I16, I17, I18, I19) 
3.11/3.18	  f4(I20, I21, I22, I23) -> f5(I20, I21, I22, I23) [I23 <= 1]
3.11/3.18	  f4(I24, I25, I26, I27) -> f3(I24, 0, I26, I27) [2 <= I27]
3.11/3.18	  f1(I28, I29, I30, I31) -> f3(I28, 1 + I29, 2 + I30, I31) [1 + I29 <= I31]
3.11/3.18	  f1(I32, I33, I34, I35) -> f2(I32, I33, I34, I35) [I35 <= I33]
3.11/3.18	
3.11/3.18	The dependency graph for this problem is:
3.11/3.18	  4 -> 
3.11/3.18	Where:
3.11/3.18	  4) f3#(I16, I17, I18, I19) -> f1#(I16, I17, I18, I19) 
3.11/3.18	
3.11/3.18	We have the following SCCs.
3.11/3.18	  
3.11/6.16	EOF