1.70/1.77	YES
1.70/1.77	
1.70/1.77	DP problem for innermost termination.
1.70/1.77	P =
1.70/1.77	  f9#(x1, x2) -> f8#(x1, x2) 
1.70/1.77	  f8#(I0, I1) -> f3#(0, I1) 
1.70/1.77	  f4#(I2, I3) -> f2#(I2, 1 + I3) [1 + I3 <= 2]
1.70/1.77	  f4#(I4, I5) -> f7#(I4, I5) [2 <= I5]
1.70/1.77	  f7#(I6, I7) -> f5#(I6, I7) 
1.70/1.77	  f7#(I8, I9) -> f5#(I8, I9) 
1.70/1.77	  f2#(I12, I13) -> f4#(I12, I13) 
1.70/1.77	  f3#(I14, I15) -> f1#(I14, I15) 
1.70/1.77	  f1#(I16, I17) -> f3#(1 + I16, I17) [1 + I16 <= 2]
1.70/1.77	  f1#(I18, I19) -> f2#(I18, 0) [2 <= I18]
1.70/1.77	R = 
1.70/1.77	  f9(x1, x2) -> f8(x1, x2) 
1.70/1.77	  f8(I0, I1) -> f3(0, I1) 
1.70/1.77	  f4(I2, I3) -> f2(I2, 1 + I3) [1 + I3 <= 2]
1.70/1.77	  f4(I4, I5) -> f7(I4, I5) [2 <= I5]
1.70/1.77	  f7(I6, I7) -> f5(I6, I7) 
1.70/1.77	  f7(I8, I9) -> f5(I8, I9) 
1.70/1.77	  f5(I10, I11) -> f6(I10, I11) 
1.70/1.77	  f2(I12, I13) -> f4(I12, I13) 
1.70/1.77	  f3(I14, I15) -> f1(I14, I15) 
1.70/1.77	  f1(I16, I17) -> f3(1 + I16, I17) [1 + I16 <= 2]
1.70/1.77	  f1(I18, I19) -> f2(I18, 0) [2 <= I18]
1.70/1.77	
1.70/1.77	The dependency graph for this problem is:
1.70/1.77	  0 -> 1
1.70/1.77	  1 -> 7
1.70/1.77	  2 -> 6
1.70/1.77	  3 -> 4, 5
1.70/1.77	  4 -> 
1.70/1.77	  5 -> 
1.70/1.77	  6 -> 2, 3
1.70/1.77	  7 -> 8, 9
1.70/1.77	  8 -> 7
1.70/1.77	  9 -> 6
1.70/1.77	Where:
1.70/1.77	  0) f9#(x1, x2) -> f8#(x1, x2) 
1.70/1.77	  1) f8#(I0, I1) -> f3#(0, I1) 
1.70/1.77	  2) f4#(I2, I3) -> f2#(I2, 1 + I3) [1 + I3 <= 2]
1.70/1.77	  3) f4#(I4, I5) -> f7#(I4, I5) [2 <= I5]
1.70/1.77	  4) f7#(I6, I7) -> f5#(I6, I7) 
1.70/1.77	  5) f7#(I8, I9) -> f5#(I8, I9) 
1.70/1.77	  6) f2#(I12, I13) -> f4#(I12, I13) 
1.70/1.77	  7) f3#(I14, I15) -> f1#(I14, I15) 
1.70/1.77	  8) f1#(I16, I17) -> f3#(1 + I16, I17) [1 + I16 <= 2]
1.70/1.77	  9) f1#(I18, I19) -> f2#(I18, 0) [2 <= I18]
1.70/1.77	
1.70/1.77	We have the following SCCs.
1.70/1.77	  { 7, 8 }
1.70/1.77	  { 2, 6 }
1.70/1.77	
1.70/1.77	DP problem for innermost termination.
1.70/1.77	P =
1.70/1.77	  f4#(I2, I3) -> f2#(I2, 1 + I3) [1 + I3 <= 2]
1.70/1.77	  f2#(I12, I13) -> f4#(I12, I13) 
1.70/1.77	R = 
1.70/1.77	  f9(x1, x2) -> f8(x1, x2) 
1.70/1.77	  f8(I0, I1) -> f3(0, I1) 
1.70/1.77	  f4(I2, I3) -> f2(I2, 1 + I3) [1 + I3 <= 2]
1.70/1.77	  f4(I4, I5) -> f7(I4, I5) [2 <= I5]
1.70/1.77	  f7(I6, I7) -> f5(I6, I7) 
1.70/1.77	  f7(I8, I9) -> f5(I8, I9) 
1.70/1.77	  f5(I10, I11) -> f6(I10, I11) 
1.70/1.77	  f2(I12, I13) -> f4(I12, I13) 
1.70/1.77	  f3(I14, I15) -> f1(I14, I15) 
1.70/1.77	  f1(I16, I17) -> f3(1 + I16, I17) [1 + I16 <= 2]
1.70/1.77	  f1(I18, I19) -> f2(I18, 0) [2 <= I18]
1.70/1.77	
1.70/1.77	We use the reverse value criterion with the projection function NU:
1.70/1.77	  NU[f2#(z1,z2)] = 2 + -1 * (1 + z2)
1.70/1.77	  NU[f4#(z1,z2)] = 2 + -1 * (1 + z2)
1.70/1.77	
1.70/1.77	This gives the following inequalities:
1.70/1.77	  1 + I3 <= 2  ==>  2 + -1 * (1 + I3) > 2 + -1 * (1 + (1 + I3))  with  2 + -1 * (1 + I3) >= 0
1.70/1.77	    ==>  2 + -1 * (1 + I13) >= 2 + -1 * (1 + I13)
1.70/1.77	
1.70/1.77	We remove all the strictly oriented dependency pairs.
1.70/1.77	
1.70/1.77	DP problem for innermost termination.
1.70/1.77	P =
1.70/1.77	  f2#(I12, I13) -> f4#(I12, I13) 
1.70/1.77	R = 
1.70/1.77	  f9(x1, x2) -> f8(x1, x2) 
1.70/1.77	  f8(I0, I1) -> f3(0, I1) 
1.70/1.77	  f4(I2, I3) -> f2(I2, 1 + I3) [1 + I3 <= 2]
1.70/1.77	  f4(I4, I5) -> f7(I4, I5) [2 <= I5]
1.70/1.77	  f7(I6, I7) -> f5(I6, I7) 
1.70/1.77	  f7(I8, I9) -> f5(I8, I9) 
1.70/1.77	  f5(I10, I11) -> f6(I10, I11) 
1.70/1.77	  f2(I12, I13) -> f4(I12, I13) 
1.70/1.77	  f3(I14, I15) -> f1(I14, I15) 
1.70/1.77	  f1(I16, I17) -> f3(1 + I16, I17) [1 + I16 <= 2]
1.70/1.77	  f1(I18, I19) -> f2(I18, 0) [2 <= I18]
1.70/1.77	
1.70/1.77	The dependency graph for this problem is:
1.70/1.77	  6 -> 
1.70/1.77	Where:
1.70/1.77	  6) f2#(I12, I13) -> f4#(I12, I13) 
1.70/1.77	
1.70/1.77	We have the following SCCs.
1.70/1.77	  
1.70/1.77	
1.70/1.77	DP problem for innermost termination.
1.70/1.77	P =
1.70/1.77	  f3#(I14, I15) -> f1#(I14, I15) 
1.70/1.77	  f1#(I16, I17) -> f3#(1 + I16, I17) [1 + I16 <= 2]
1.70/1.77	R = 
1.70/1.77	  f9(x1, x2) -> f8(x1, x2) 
1.70/1.77	  f8(I0, I1) -> f3(0, I1) 
1.70/1.77	  f4(I2, I3) -> f2(I2, 1 + I3) [1 + I3 <= 2]
1.70/1.77	  f4(I4, I5) -> f7(I4, I5) [2 <= I5]
1.70/1.77	  f7(I6, I7) -> f5(I6, I7) 
1.70/1.77	  f7(I8, I9) -> f5(I8, I9) 
1.70/1.77	  f5(I10, I11) -> f6(I10, I11) 
1.70/1.77	  f2(I12, I13) -> f4(I12, I13) 
1.70/1.77	  f3(I14, I15) -> f1(I14, I15) 
1.70/1.77	  f1(I16, I17) -> f3(1 + I16, I17) [1 + I16 <= 2]
1.70/1.77	  f1(I18, I19) -> f2(I18, 0) [2 <= I18]
1.70/1.77	
1.70/1.77	We use the reverse value criterion with the projection function NU:
1.70/1.77	  NU[f1#(z1,z2)] = 2 + -1 * (1 + z1)
1.70/1.77	  NU[f3#(z1,z2)] = 2 + -1 * (1 + z1)
1.70/1.77	
1.70/1.77	This gives the following inequalities:
1.70/1.77	    ==>  2 + -1 * (1 + I14) >= 2 + -1 * (1 + I14)
1.70/1.77	  1 + I16 <= 2  ==>  2 + -1 * (1 + I16) > 2 + -1 * (1 + (1 + I16))  with  2 + -1 * (1 + I16) >= 0
1.70/1.77	
1.70/1.77	We remove all the strictly oriented dependency pairs.
1.70/1.77	
1.70/1.77	DP problem for innermost termination.
1.70/1.77	P =
1.70/1.77	  f3#(I14, I15) -> f1#(I14, I15) 
1.70/1.77	R = 
1.70/1.77	  f9(x1, x2) -> f8(x1, x2) 
1.70/1.77	  f8(I0, I1) -> f3(0, I1) 
1.70/1.77	  f4(I2, I3) -> f2(I2, 1 + I3) [1 + I3 <= 2]
1.70/1.77	  f4(I4, I5) -> f7(I4, I5) [2 <= I5]
1.70/1.77	  f7(I6, I7) -> f5(I6, I7) 
1.70/1.77	  f7(I8, I9) -> f5(I8, I9) 
1.70/1.77	  f5(I10, I11) -> f6(I10, I11) 
1.70/1.77	  f2(I12, I13) -> f4(I12, I13) 
1.70/1.77	  f3(I14, I15) -> f1(I14, I15) 
1.70/1.77	  f1(I16, I17) -> f3(1 + I16, I17) [1 + I16 <= 2]
1.70/1.77	  f1(I18, I19) -> f2(I18, 0) [2 <= I18]
1.70/1.77	
1.70/1.77	The dependency graph for this problem is:
1.70/1.77	  7 -> 
1.70/1.77	Where:
1.70/1.77	  7) f3#(I14, I15) -> f1#(I14, I15) 
1.70/1.77	
1.70/1.77	We have the following SCCs.
1.70/1.77	  
1.70/4.75	EOF