8.29/8.58	YES
8.29/8.58	
8.29/8.58	DP problem for innermost termination.
8.29/8.58	P =
8.29/8.58	  f7#(x1, x2, x3, x4, x5) -> f6#(x1, x2, x3, x4, x5) 
8.29/8.58	  f6#(I0, I1, I2, I3, I4) -> f3#(0, I1, I2, I3, I4) 
8.29/8.58	  f2#(I5, I6, I7, I8, I9) -> f4#(I5, I6, I7, I8, I9) 
8.29/8.58	  f4#(I10, I11, I12, I13, I14) -> f2#(I10, 1 + I11, I12, I13, I11) [1 + I11 <= 200]
8.29/8.58	  f3#(I20, I21, I22, I23, I24) -> f1#(I20, I21, I22, I23, I24) 
8.29/8.58	  f1#(I25, I26, I27, I28, I29) -> f3#(1 + I25, I26, I25, I25, I29) [1 + I25 <= 100]
8.29/8.58	  f1#(I30, I31, I32, I33, I34) -> f2#(I30, 100, I32, I33, I34) [100 <= I30]
8.29/8.58	R = 
8.29/8.58	  f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) 
8.29/8.58	  f6(I0, I1, I2, I3, I4) -> f3(0, I1, I2, I3, I4) 
8.29/8.58	  f2(I5, I6, I7, I8, I9) -> f4(I5, I6, I7, I8, I9) 
8.29/8.58	  f4(I10, I11, I12, I13, I14) -> f2(I10, 1 + I11, I12, I13, I11) [1 + I11 <= 200]
8.29/8.58	  f4(I15, I16, I17, I18, I19) -> f5(I15, I16, I17, I18, I19) [200 <= I16]
8.29/8.58	  f3(I20, I21, I22, I23, I24) -> f1(I20, I21, I22, I23, I24) 
8.29/8.58	  f1(I25, I26, I27, I28, I29) -> f3(1 + I25, I26, I25, I25, I29) [1 + I25 <= 100]
8.29/8.58	  f1(I30, I31, I32, I33, I34) -> f2(I30, 100, I32, I33, I34) [100 <= I30]
8.29/8.58	
8.29/8.58	The dependency graph for this problem is:
8.29/8.58	  0 -> 1
8.29/8.58	  1 -> 4
8.29/8.58	  2 -> 3
8.29/8.58	  3 -> 2
8.29/8.58	  4 -> 5, 6
8.29/8.58	  5 -> 4
8.29/8.58	  6 -> 2
8.29/8.58	Where:
8.29/8.58	  0) f7#(x1, x2, x3, x4, x5) -> f6#(x1, x2, x3, x4, x5) 
8.29/8.58	  1) f6#(I0, I1, I2, I3, I4) -> f3#(0, I1, I2, I3, I4) 
8.29/8.58	  2) f2#(I5, I6, I7, I8, I9) -> f4#(I5, I6, I7, I8, I9) 
8.29/8.58	  3) f4#(I10, I11, I12, I13, I14) -> f2#(I10, 1 + I11, I12, I13, I11) [1 + I11 <= 200]
8.29/8.58	  4) f3#(I20, I21, I22, I23, I24) -> f1#(I20, I21, I22, I23, I24) 
8.29/8.58	  5) f1#(I25, I26, I27, I28, I29) -> f3#(1 + I25, I26, I25, I25, I29) [1 + I25 <= 100]
8.29/8.58	  6) f1#(I30, I31, I32, I33, I34) -> f2#(I30, 100, I32, I33, I34) [100 <= I30]
8.29/8.58	
8.29/8.58	We have the following SCCs.
8.29/8.58	  { 4, 5 }
8.29/8.58	  { 2, 3 }
8.29/8.58	
8.29/8.58	DP problem for innermost termination.
8.29/8.58	P =
8.29/8.58	  f2#(I5, I6, I7, I8, I9) -> f4#(I5, I6, I7, I8, I9) 
8.29/8.58	  f4#(I10, I11, I12, I13, I14) -> f2#(I10, 1 + I11, I12, I13, I11) [1 + I11 <= 200]
8.29/8.58	R = 
8.29/8.58	  f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) 
8.29/8.58	  f6(I0, I1, I2, I3, I4) -> f3(0, I1, I2, I3, I4) 
8.29/8.58	  f2(I5, I6, I7, I8, I9) -> f4(I5, I6, I7, I8, I9) 
8.29/8.58	  f4(I10, I11, I12, I13, I14) -> f2(I10, 1 + I11, I12, I13, I11) [1 + I11 <= 200]
8.29/8.58	  f4(I15, I16, I17, I18, I19) -> f5(I15, I16, I17, I18, I19) [200 <= I16]
8.29/8.58	  f3(I20, I21, I22, I23, I24) -> f1(I20, I21, I22, I23, I24) 
8.29/8.58	  f1(I25, I26, I27, I28, I29) -> f3(1 + I25, I26, I25, I25, I29) [1 + I25 <= 100]
8.29/8.58	  f1(I30, I31, I32, I33, I34) -> f2(I30, 100, I32, I33, I34) [100 <= I30]
8.29/8.58	
8.29/8.58	We use the reverse value criterion with the projection function NU:
8.29/8.58	  NU[f4#(z1,z2,z3,z4,z5)] = 200 + -1 * (1 + z2)
8.29/8.58	  NU[f2#(z1,z2,z3,z4,z5)] = 200 + -1 * (1 + z2)
8.29/8.58	
8.29/8.58	This gives the following inequalities:
8.29/8.58	    ==>  200 + -1 * (1 + I6) >= 200 + -1 * (1 + I6)
8.29/8.58	  1 + I11 <= 200  ==>  200 + -1 * (1 + I11) > 200 + -1 * (1 + (1 + I11))  with  200 + -1 * (1 + I11) >= 0
8.29/8.58	
8.29/8.58	We remove all the strictly oriented dependency pairs.
8.29/8.58	
8.29/8.58	DP problem for innermost termination.
8.29/8.58	P =
8.29/8.58	  f2#(I5, I6, I7, I8, I9) -> f4#(I5, I6, I7, I8, I9) 
8.29/8.58	R = 
8.29/8.58	  f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) 
8.29/8.58	  f6(I0, I1, I2, I3, I4) -> f3(0, I1, I2, I3, I4) 
8.29/8.58	  f2(I5, I6, I7, I8, I9) -> f4(I5, I6, I7, I8, I9) 
8.29/8.58	  f4(I10, I11, I12, I13, I14) -> f2(I10, 1 + I11, I12, I13, I11) [1 + I11 <= 200]
8.29/8.58	  f4(I15, I16, I17, I18, I19) -> f5(I15, I16, I17, I18, I19) [200 <= I16]
8.29/8.58	  f3(I20, I21, I22, I23, I24) -> f1(I20, I21, I22, I23, I24) 
8.29/8.58	  f1(I25, I26, I27, I28, I29) -> f3(1 + I25, I26, I25, I25, I29) [1 + I25 <= 100]
8.29/8.58	  f1(I30, I31, I32, I33, I34) -> f2(I30, 100, I32, I33, I34) [100 <= I30]
8.29/8.58	
8.29/8.58	The dependency graph for this problem is:
8.29/8.58	  2 -> 
8.29/8.58	Where:
8.29/8.58	  2) f2#(I5, I6, I7, I8, I9) -> f4#(I5, I6, I7, I8, I9) 
8.29/8.58	
8.29/8.58	We have the following SCCs.
8.29/8.58	  
8.29/8.58	
8.29/8.58	DP problem for innermost termination.
8.29/8.58	P =
8.29/8.58	  f3#(I20, I21, I22, I23, I24) -> f1#(I20, I21, I22, I23, I24) 
8.29/8.58	  f1#(I25, I26, I27, I28, I29) -> f3#(1 + I25, I26, I25, I25, I29) [1 + I25 <= 100]
8.29/8.58	R = 
8.29/8.58	  f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) 
8.29/8.58	  f6(I0, I1, I2, I3, I4) -> f3(0, I1, I2, I3, I4) 
8.29/8.58	  f2(I5, I6, I7, I8, I9) -> f4(I5, I6, I7, I8, I9) 
8.29/8.58	  f4(I10, I11, I12, I13, I14) -> f2(I10, 1 + I11, I12, I13, I11) [1 + I11 <= 200]
8.29/8.58	  f4(I15, I16, I17, I18, I19) -> f5(I15, I16, I17, I18, I19) [200 <= I16]
8.29/8.58	  f3(I20, I21, I22, I23, I24) -> f1(I20, I21, I22, I23, I24) 
8.29/8.58	  f1(I25, I26, I27, I28, I29) -> f3(1 + I25, I26, I25, I25, I29) [1 + I25 <= 100]
8.29/8.58	  f1(I30, I31, I32, I33, I34) -> f2(I30, 100, I32, I33, I34) [100 <= I30]
8.29/8.58	
8.29/8.58	We use the reverse value criterion with the projection function NU:
8.29/8.58	  NU[f1#(z1,z2,z3,z4,z5)] = 100 + -1 * (1 + z1)
8.29/8.58	  NU[f3#(z1,z2,z3,z4,z5)] = 100 + -1 * (1 + z1)
8.29/8.58	
8.29/8.58	This gives the following inequalities:
8.29/8.58	    ==>  100 + -1 * (1 + I20) >= 100 + -1 * (1 + I20)
8.29/8.58	  1 + I25 <= 100  ==>  100 + -1 * (1 + I25) > 100 + -1 * (1 + (1 + I25))  with  100 + -1 * (1 + I25) >= 0
8.29/8.58	
8.29/8.58	We remove all the strictly oriented dependency pairs.
8.29/8.58	
8.29/8.58	DP problem for innermost termination.
8.29/8.58	P =
8.29/8.58	  f3#(I20, I21, I22, I23, I24) -> f1#(I20, I21, I22, I23, I24) 
8.29/8.58	R = 
8.29/8.58	  f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) 
8.29/8.58	  f6(I0, I1, I2, I3, I4) -> f3(0, I1, I2, I3, I4) 
8.29/8.58	  f2(I5, I6, I7, I8, I9) -> f4(I5, I6, I7, I8, I9) 
8.29/8.58	  f4(I10, I11, I12, I13, I14) -> f2(I10, 1 + I11, I12, I13, I11) [1 + I11 <= 200]
8.29/8.58	  f4(I15, I16, I17, I18, I19) -> f5(I15, I16, I17, I18, I19) [200 <= I16]
8.29/8.58	  f3(I20, I21, I22, I23, I24) -> f1(I20, I21, I22, I23, I24) 
8.29/8.58	  f1(I25, I26, I27, I28, I29) -> f3(1 + I25, I26, I25, I25, I29) [1 + I25 <= 100]
8.29/8.58	  f1(I30, I31, I32, I33, I34) -> f2(I30, 100, I32, I33, I34) [100 <= I30]
8.29/8.58	
8.29/8.58	The dependency graph for this problem is:
8.29/8.58	  4 -> 
8.29/8.58	Where:
8.29/8.58	  4) f3#(I20, I21, I22, I23, I24) -> f1#(I20, I21, I22, I23, I24) 
8.29/8.58	
8.29/8.58	We have the following SCCs.
8.29/8.58	  
8.29/11.56	EOF