6.64/6.59	YES
6.64/6.59	
6.64/6.59	DP problem for innermost termination.
6.64/6.59	P =
6.64/6.59	  f6#(x1, x2, x3, x4, x5, x6, x7) -> f1#(x1, x2, x3, x4, x5, x6, x7) 
6.64/6.59	  f2#(I0, I1, I2, I3, I4, I5, I6) -> f3#(I0, 1 + I1, I2, I3, I4, I5, I6) [1 + I1 <= I0]
6.64/6.59	  f5#(I14, I15, I16, I17, I18, I19, I20) -> f3#(I14, I15, I16, I17, I18, I19, I20) 
6.64/6.59	  f3#(I21, I22, I23, I24, I25, I26, I27) -> f5#(I21, 1 + I22, I23, I24, I25, I26, I27) [1 + I22 <= I21]
6.64/6.59	  f1#(I35, I36, I37, I38, I39, I40, I41) -> f2#(I35, rnd2, I39, I38, I39, I40, I41) [y1 = I40 /\ rnd2 = I39]
6.64/6.59	R = 
6.64/6.59	  f6(x1, x2, x3, x4, x5, x6, x7) -> f1(x1, x2, x3, x4, x5, x6, x7) 
6.64/6.59	  f2(I0, I1, I2, I3, I4, I5, I6) -> f3(I0, 1 + I1, I2, I3, I4, I5, I6) [1 + I1 <= I0]
6.64/6.59	  f2(I7, I8, I9, I10, I11, I12, I13) -> f4(I7, I8, I9, I13, I11, I12, I13) [I7 <= I8]
6.64/6.59	  f5(I14, I15, I16, I17, I18, I19, I20) -> f3(I14, I15, I16, I17, I18, I19, I20) 
6.64/6.59	  f3(I21, I22, I23, I24, I25, I26, I27) -> f5(I21, 1 + I22, I23, I24, I25, I26, I27) [1 + I22 <= I21]
6.64/6.59	  f3(I28, I29, I30, I31, I32, I33, I34) -> f4(I28, I29, I30, I34, I32, I33, I34) [I28 <= I29]
6.64/6.59	  f1(I35, I36, I37, I38, I39, I40, I41) -> f2(I35, rnd2, I39, I38, I39, I40, I41) [y1 = I40 /\ rnd2 = I39]
6.64/6.59	
6.64/6.59	The dependency graph for this problem is:
6.64/6.59	  0 -> 4
6.64/6.59	  1 -> 3
6.64/6.59	  2 -> 3
6.64/6.59	  3 -> 2
6.64/6.59	  4 -> 1
6.64/6.59	Where:
6.64/6.59	  0) f6#(x1, x2, x3, x4, x5, x6, x7) -> f1#(x1, x2, x3, x4, x5, x6, x7) 
6.64/6.59	  1) f2#(I0, I1, I2, I3, I4, I5, I6) -> f3#(I0, 1 + I1, I2, I3, I4, I5, I6) [1 + I1 <= I0]
6.64/6.59	  2) f5#(I14, I15, I16, I17, I18, I19, I20) -> f3#(I14, I15, I16, I17, I18, I19, I20) 
6.64/6.59	  3) f3#(I21, I22, I23, I24, I25, I26, I27) -> f5#(I21, 1 + I22, I23, I24, I25, I26, I27) [1 + I22 <= I21]
6.64/6.59	  4) f1#(I35, I36, I37, I38, I39, I40, I41) -> f2#(I35, rnd2, I39, I38, I39, I40, I41) [y1 = I40 /\ rnd2 = I39]
6.64/6.59	
6.64/6.59	We have the following SCCs.
6.64/6.59	  { 2, 3 }
6.64/6.59	
6.64/6.59	DP problem for innermost termination.
6.64/6.59	P =
6.64/6.59	  f5#(I14, I15, I16, I17, I18, I19, I20) -> f3#(I14, I15, I16, I17, I18, I19, I20) 
6.64/6.59	  f3#(I21, I22, I23, I24, I25, I26, I27) -> f5#(I21, 1 + I22, I23, I24, I25, I26, I27) [1 + I22 <= I21]
6.64/6.59	R = 
6.64/6.59	  f6(x1, x2, x3, x4, x5, x6, x7) -> f1(x1, x2, x3, x4, x5, x6, x7) 
6.64/6.59	  f2(I0, I1, I2, I3, I4, I5, I6) -> f3(I0, 1 + I1, I2, I3, I4, I5, I6) [1 + I1 <= I0]
6.64/6.59	  f2(I7, I8, I9, I10, I11, I12, I13) -> f4(I7, I8, I9, I13, I11, I12, I13) [I7 <= I8]
6.64/6.59	  f5(I14, I15, I16, I17, I18, I19, I20) -> f3(I14, I15, I16, I17, I18, I19, I20) 
6.64/6.59	  f3(I21, I22, I23, I24, I25, I26, I27) -> f5(I21, 1 + I22, I23, I24, I25, I26, I27) [1 + I22 <= I21]
6.64/6.59	  f3(I28, I29, I30, I31, I32, I33, I34) -> f4(I28, I29, I30, I34, I32, I33, I34) [I28 <= I29]
6.64/6.59	  f1(I35, I36, I37, I38, I39, I40, I41) -> f2(I35, rnd2, I39, I38, I39, I40, I41) [y1 = I40 /\ rnd2 = I39]
6.64/6.59	
6.64/6.59	We use the reverse value criterion with the projection function NU:
6.64/6.59	  NU[f3#(z1,z2,z3,z4,z5,z6,z7)] = z1 + -1 * (1 + z2)
6.64/6.59	  NU[f5#(z1,z2,z3,z4,z5,z6,z7)] = z1 + -1 * (1 + z2)
6.64/6.59	
6.64/6.59	This gives the following inequalities:
6.64/6.59	    ==>  I14 + -1 * (1 + I15) >= I14 + -1 * (1 + I15)
6.64/6.59	  1 + I22 <= I21  ==>  I21 + -1 * (1 + I22) > I21 + -1 * (1 + (1 + I22))  with  I21 + -1 * (1 + I22) >= 0
6.64/6.59	
6.64/6.59	We remove all the strictly oriented dependency pairs.
6.64/6.59	
6.64/6.59	DP problem for innermost termination.
6.64/6.59	P =
6.64/6.59	  f5#(I14, I15, I16, I17, I18, I19, I20) -> f3#(I14, I15, I16, I17, I18, I19, I20) 
6.64/6.59	R = 
6.64/6.59	  f6(x1, x2, x3, x4, x5, x6, x7) -> f1(x1, x2, x3, x4, x5, x6, x7) 
6.64/6.59	  f2(I0, I1, I2, I3, I4, I5, I6) -> f3(I0, 1 + I1, I2, I3, I4, I5, I6) [1 + I1 <= I0]
6.64/6.59	  f2(I7, I8, I9, I10, I11, I12, I13) -> f4(I7, I8, I9, I13, I11, I12, I13) [I7 <= I8]
6.64/6.59	  f5(I14, I15, I16, I17, I18, I19, I20) -> f3(I14, I15, I16, I17, I18, I19, I20) 
6.64/6.59	  f3(I21, I22, I23, I24, I25, I26, I27) -> f5(I21, 1 + I22, I23, I24, I25, I26, I27) [1 + I22 <= I21]
6.64/6.59	  f3(I28, I29, I30, I31, I32, I33, I34) -> f4(I28, I29, I30, I34, I32, I33, I34) [I28 <= I29]
6.64/6.59	  f1(I35, I36, I37, I38, I39, I40, I41) -> f2(I35, rnd2, I39, I38, I39, I40, I41) [y1 = I40 /\ rnd2 = I39]
6.64/6.59	
6.64/6.59	The dependency graph for this problem is:
6.64/6.59	  2 -> 
6.64/6.59	Where:
6.64/6.59	  2) f5#(I14, I15, I16, I17, I18, I19, I20) -> f3#(I14, I15, I16, I17, I18, I19, I20) 
6.64/6.59	
6.64/6.59	We have the following SCCs.
6.64/6.59	  
6.64/9.57	EOF