5.06/5.08	YES
5.06/5.08	
5.06/5.08	DP problem for innermost termination.
5.06/5.08	P =
5.06/5.08	  f10#(x1, x2, x3) -> f9#(x1, x2, x3) 
5.06/5.08	  f9#(I0, I1, I2) -> f7#(0, I1, I2) 
5.06/5.08	  f8#(I3, I4, I5) -> f7#(I3, I4, I5) 
5.06/5.08	  f7#(I6, I7, I8) -> f8#(I6, -1 + I7, I8) [1 + I8 <= I7]
5.06/5.08	  f7#(I9, I10, I11) -> f5#(I9, I10, I11) [I10 <= I11]
5.06/5.08	  f6#(I12, I13, I14) -> f5#(I12, I13, I14) 
5.06/5.08	  f5#(I15, I16, I17) -> f6#(1 + I15, I16, I17) [1 + I15 <= I17]
5.06/5.08	  f5#(I18, I19, I20) -> f3#(I18, I19, I20) [I20 <= I18]
5.06/5.08	  f4#(I21, I22, I23) -> f3#(I21, I22, I23) 
5.06/5.08	  f3#(I24, I25, I26) -> f4#(I24, -1 + I25, I26) [1 <= I25]
5.06/5.08	  f3#(I27, I28, I29) -> f1#(I27, I28, I29) [I28 <= 0]
5.06/5.08	  f2#(I30, I31, I32) -> f1#(I30, I31, I32) 
5.06/5.08	  f1#(I33, I34, I35) -> f2#(-1 + I33, I34, I35) [1 <= I33]
5.06/5.08	R = 
5.06/5.08	  f10(x1, x2, x3) -> f9(x1, x2, x3) 
5.06/5.08	  f9(I0, I1, I2) -> f7(0, I1, I2) 
5.06/5.08	  f8(I3, I4, I5) -> f7(I3, I4, I5) 
5.06/5.08	  f7(I6, I7, I8) -> f8(I6, -1 + I7, I8) [1 + I8 <= I7]
5.06/5.08	  f7(I9, I10, I11) -> f5(I9, I10, I11) [I10 <= I11]
5.06/5.08	  f6(I12, I13, I14) -> f5(I12, I13, I14) 
5.06/5.08	  f5(I15, I16, I17) -> f6(1 + I15, I16, I17) [1 + I15 <= I17]
5.06/5.08	  f5(I18, I19, I20) -> f3(I18, I19, I20) [I20 <= I18]
5.06/5.08	  f4(I21, I22, I23) -> f3(I21, I22, I23) 
5.06/5.08	  f3(I24, I25, I26) -> f4(I24, -1 + I25, I26) [1 <= I25]
5.06/5.08	  f3(I27, I28, I29) -> f1(I27, I28, I29) [I28 <= 0]
5.06/5.08	  f2(I30, I31, I32) -> f1(I30, I31, I32) 
5.06/5.08	  f1(I33, I34, I35) -> f2(-1 + I33, I34, I35) [1 <= I33]
5.06/5.08	
5.06/5.08	The dependency graph for this problem is:
5.06/5.08	  0 -> 1
5.06/5.08	  1 -> 3, 4
5.06/5.08	  2 -> 3, 4
5.06/5.08	  3 -> 2
5.06/5.08	  4 -> 6, 7
5.06/5.08	  5 -> 6, 7
5.06/5.08	  6 -> 5
5.06/5.08	  7 -> 9, 10
5.06/5.08	  8 -> 9, 10
5.06/5.08	  9 -> 8
5.06/5.08	  10 -> 12
5.06/5.08	  11 -> 12
5.06/5.08	  12 -> 11
5.06/5.08	Where:
5.06/5.08	  0) f10#(x1, x2, x3) -> f9#(x1, x2, x3) 
5.06/5.08	  1) f9#(I0, I1, I2) -> f7#(0, I1, I2) 
5.06/5.08	  2) f8#(I3, I4, I5) -> f7#(I3, I4, I5) 
5.06/5.08	  3) f7#(I6, I7, I8) -> f8#(I6, -1 + I7, I8) [1 + I8 <= I7]
5.06/5.08	  4) f7#(I9, I10, I11) -> f5#(I9, I10, I11) [I10 <= I11]
5.06/5.08	  5) f6#(I12, I13, I14) -> f5#(I12, I13, I14) 
5.06/5.08	  6) f5#(I15, I16, I17) -> f6#(1 + I15, I16, I17) [1 + I15 <= I17]
5.06/5.08	  7) f5#(I18, I19, I20) -> f3#(I18, I19, I20) [I20 <= I18]
5.06/5.08	  8) f4#(I21, I22, I23) -> f3#(I21, I22, I23) 
5.06/5.08	  9) f3#(I24, I25, I26) -> f4#(I24, -1 + I25, I26) [1 <= I25]
5.06/5.08	  10) f3#(I27, I28, I29) -> f1#(I27, I28, I29) [I28 <= 0]
5.06/5.08	  11) f2#(I30, I31, I32) -> f1#(I30, I31, I32) 
5.06/5.08	  12) f1#(I33, I34, I35) -> f2#(-1 + I33, I34, I35) [1 <= I33]
5.06/5.08	
5.06/5.08	We have the following SCCs.
5.06/5.08	  { 2, 3 }
5.06/5.08	  { 5, 6 }
5.06/5.08	  { 8, 9 }
5.06/5.08	  { 11, 12 }
5.06/5.08	
5.06/5.08	DP problem for innermost termination.
5.06/5.08	P =
5.06/5.08	  f2#(I30, I31, I32) -> f1#(I30, I31, I32) 
5.06/5.08	  f1#(I33, I34, I35) -> f2#(-1 + I33, I34, I35) [1 <= I33]
5.06/5.08	R = 
5.06/5.08	  f10(x1, x2, x3) -> f9(x1, x2, x3) 
5.06/5.08	  f9(I0, I1, I2) -> f7(0, I1, I2) 
5.06/5.08	  f8(I3, I4, I5) -> f7(I3, I4, I5) 
5.06/5.08	  f7(I6, I7, I8) -> f8(I6, -1 + I7, I8) [1 + I8 <= I7]
5.06/5.08	  f7(I9, I10, I11) -> f5(I9, I10, I11) [I10 <= I11]
5.06/5.08	  f6(I12, I13, I14) -> f5(I12, I13, I14) 
5.06/5.08	  f5(I15, I16, I17) -> f6(1 + I15, I16, I17) [1 + I15 <= I17]
5.06/5.08	  f5(I18, I19, I20) -> f3(I18, I19, I20) [I20 <= I18]
5.06/5.08	  f4(I21, I22, I23) -> f3(I21, I22, I23) 
5.06/5.08	  f3(I24, I25, I26) -> f4(I24, -1 + I25, I26) [1 <= I25]
5.06/5.08	  f3(I27, I28, I29) -> f1(I27, I28, I29) [I28 <= 0]
5.06/5.08	  f2(I30, I31, I32) -> f1(I30, I31, I32) 
5.06/5.08	  f1(I33, I34, I35) -> f2(-1 + I33, I34, I35) [1 <= I33]
5.06/5.08	
5.06/5.08	We use the basic value criterion with the projection function NU:
5.06/5.08	  NU[f1#(z1,z2,z3)] = z1
5.06/5.08	  NU[f2#(z1,z2,z3)] = z1
5.06/5.08	
5.06/5.08	This gives the following inequalities:
5.06/5.08	    ==>  I30 (>! \union =) I30
5.06/5.08	  1 <= I33  ==>  I33 >! -1 + I33
5.06/5.08	
5.06/5.08	We remove all the strictly oriented dependency pairs.
5.06/5.08	
5.06/5.08	DP problem for innermost termination.
5.06/5.08	P =
5.06/5.08	  f2#(I30, I31, I32) -> f1#(I30, I31, I32) 
5.06/5.08	R = 
5.06/5.08	  f10(x1, x2, x3) -> f9(x1, x2, x3) 
5.06/5.08	  f9(I0, I1, I2) -> f7(0, I1, I2) 
5.06/5.08	  f8(I3, I4, I5) -> f7(I3, I4, I5) 
5.06/5.08	  f7(I6, I7, I8) -> f8(I6, -1 + I7, I8) [1 + I8 <= I7]
5.06/5.08	  f7(I9, I10, I11) -> f5(I9, I10, I11) [I10 <= I11]
5.06/5.08	  f6(I12, I13, I14) -> f5(I12, I13, I14) 
5.06/5.08	  f5(I15, I16, I17) -> f6(1 + I15, I16, I17) [1 + I15 <= I17]
5.06/5.08	  f5(I18, I19, I20) -> f3(I18, I19, I20) [I20 <= I18]
5.06/5.08	  f4(I21, I22, I23) -> f3(I21, I22, I23) 
5.06/5.08	  f3(I24, I25, I26) -> f4(I24, -1 + I25, I26) [1 <= I25]
5.06/5.08	  f3(I27, I28, I29) -> f1(I27, I28, I29) [I28 <= 0]
5.06/5.08	  f2(I30, I31, I32) -> f1(I30, I31, I32) 
5.06/5.08	  f1(I33, I34, I35) -> f2(-1 + I33, I34, I35) [1 <= I33]
5.06/5.08	
5.06/5.08	The dependency graph for this problem is:
5.06/5.08	  11 -> 
5.06/5.08	Where:
5.06/5.08	  11) f2#(I30, I31, I32) -> f1#(I30, I31, I32) 
5.06/5.08	
5.06/5.08	We have the following SCCs.
5.06/5.08	  
5.06/5.08	
5.06/5.08	DP problem for innermost termination.
5.06/5.08	P =
5.06/5.08	  f4#(I21, I22, I23) -> f3#(I21, I22, I23) 
5.06/5.08	  f3#(I24, I25, I26) -> f4#(I24, -1 + I25, I26) [1 <= I25]
5.06/5.08	R = 
5.06/5.08	  f10(x1, x2, x3) -> f9(x1, x2, x3) 
5.06/5.08	  f9(I0, I1, I2) -> f7(0, I1, I2) 
5.06/5.08	  f8(I3, I4, I5) -> f7(I3, I4, I5) 
5.06/5.08	  f7(I6, I7, I8) -> f8(I6, -1 + I7, I8) [1 + I8 <= I7]
5.06/5.08	  f7(I9, I10, I11) -> f5(I9, I10, I11) [I10 <= I11]
5.06/5.08	  f6(I12, I13, I14) -> f5(I12, I13, I14) 
5.06/5.08	  f5(I15, I16, I17) -> f6(1 + I15, I16, I17) [1 + I15 <= I17]
5.06/5.08	  f5(I18, I19, I20) -> f3(I18, I19, I20) [I20 <= I18]
5.06/5.08	  f4(I21, I22, I23) -> f3(I21, I22, I23) 
5.06/5.08	  f3(I24, I25, I26) -> f4(I24, -1 + I25, I26) [1 <= I25]
5.06/5.08	  f3(I27, I28, I29) -> f1(I27, I28, I29) [I28 <= 0]
5.06/5.08	  f2(I30, I31, I32) -> f1(I30, I31, I32) 
5.06/5.08	  f1(I33, I34, I35) -> f2(-1 + I33, I34, I35) [1 <= I33]
5.06/5.08	
5.06/5.08	We use the basic value criterion with the projection function NU:
5.06/5.08	  NU[f3#(z1,z2,z3)] = z2
5.06/5.08	  NU[f4#(z1,z2,z3)] = z2
5.06/5.08	
5.06/5.08	This gives the following inequalities:
5.06/5.08	    ==>  I22 (>! \union =) I22
5.06/5.08	  1 <= I25  ==>  I25 >! -1 + I25
5.06/5.08	
5.06/5.08	We remove all the strictly oriented dependency pairs.
5.06/5.08	
5.06/5.08	DP problem for innermost termination.
5.06/5.08	P =
5.06/5.08	  f4#(I21, I22, I23) -> f3#(I21, I22, I23) 
5.06/5.08	R = 
5.06/5.08	  f10(x1, x2, x3) -> f9(x1, x2, x3) 
5.06/5.08	  f9(I0, I1, I2) -> f7(0, I1, I2) 
5.06/5.08	  f8(I3, I4, I5) -> f7(I3, I4, I5) 
5.06/5.08	  f7(I6, I7, I8) -> f8(I6, -1 + I7, I8) [1 + I8 <= I7]
5.06/5.08	  f7(I9, I10, I11) -> f5(I9, I10, I11) [I10 <= I11]
5.06/5.08	  f6(I12, I13, I14) -> f5(I12, I13, I14) 
5.06/5.08	  f5(I15, I16, I17) -> f6(1 + I15, I16, I17) [1 + I15 <= I17]
5.06/5.08	  f5(I18, I19, I20) -> f3(I18, I19, I20) [I20 <= I18]
5.06/5.08	  f4(I21, I22, I23) -> f3(I21, I22, I23) 
5.06/5.08	  f3(I24, I25, I26) -> f4(I24, -1 + I25, I26) [1 <= I25]
5.06/5.08	  f3(I27, I28, I29) -> f1(I27, I28, I29) [I28 <= 0]
5.06/5.08	  f2(I30, I31, I32) -> f1(I30, I31, I32) 
5.06/5.08	  f1(I33, I34, I35) -> f2(-1 + I33, I34, I35) [1 <= I33]
5.06/5.08	
5.06/5.08	The dependency graph for this problem is:
5.06/5.08	  8 -> 
5.06/5.08	Where:
5.06/5.08	  8) f4#(I21, I22, I23) -> f3#(I21, I22, I23) 
5.06/5.08	
5.06/5.08	We have the following SCCs.
5.06/5.08	  
5.06/5.08	
5.06/5.08	DP problem for innermost termination.
5.06/5.08	P =
5.06/5.08	  f6#(I12, I13, I14) -> f5#(I12, I13, I14) 
5.06/5.08	  f5#(I15, I16, I17) -> f6#(1 + I15, I16, I17) [1 + I15 <= I17]
5.06/5.08	R = 
5.06/5.08	  f10(x1, x2, x3) -> f9(x1, x2, x3) 
5.06/5.08	  f9(I0, I1, I2) -> f7(0, I1, I2) 
5.06/5.08	  f8(I3, I4, I5) -> f7(I3, I4, I5) 
5.06/5.08	  f7(I6, I7, I8) -> f8(I6, -1 + I7, I8) [1 + I8 <= I7]
5.06/5.08	  f7(I9, I10, I11) -> f5(I9, I10, I11) [I10 <= I11]
5.06/5.08	  f6(I12, I13, I14) -> f5(I12, I13, I14) 
5.06/5.08	  f5(I15, I16, I17) -> f6(1 + I15, I16, I17) [1 + I15 <= I17]
5.06/5.08	  f5(I18, I19, I20) -> f3(I18, I19, I20) [I20 <= I18]
5.06/5.08	  f4(I21, I22, I23) -> f3(I21, I22, I23) 
5.06/5.08	  f3(I24, I25, I26) -> f4(I24, -1 + I25, I26) [1 <= I25]
5.06/5.08	  f3(I27, I28, I29) -> f1(I27, I28, I29) [I28 <= 0]
5.06/5.08	  f2(I30, I31, I32) -> f1(I30, I31, I32) 
5.06/5.08	  f1(I33, I34, I35) -> f2(-1 + I33, I34, I35) [1 <= I33]
5.06/5.08	
5.06/5.08	We use the reverse value criterion with the projection function NU:
5.06/5.08	  NU[f5#(z1,z2,z3)] = z3 + -1 * (1 + z1)
5.06/5.08	  NU[f6#(z1,z2,z3)] = z3 + -1 * (1 + z1)
5.06/5.08	
5.06/5.08	This gives the following inequalities:
5.06/5.08	    ==>  I14 + -1 * (1 + I12) >= I14 + -1 * (1 + I12)
5.06/5.08	  1 + I15 <= I17  ==>  I17 + -1 * (1 + I15) > I17 + -1 * (1 + (1 + I15))  with  I17 + -1 * (1 + I15) >= 0
5.06/5.08	
5.06/5.08	We remove all the strictly oriented dependency pairs.
5.06/5.08	
5.06/5.08	DP problem for innermost termination.
5.06/5.08	P =
5.06/5.08	  f6#(I12, I13, I14) -> f5#(I12, I13, I14) 
5.06/5.08	R = 
5.06/5.08	  f10(x1, x2, x3) -> f9(x1, x2, x3) 
5.06/5.08	  f9(I0, I1, I2) -> f7(0, I1, I2) 
5.06/5.08	  f8(I3, I4, I5) -> f7(I3, I4, I5) 
5.06/5.08	  f7(I6, I7, I8) -> f8(I6, -1 + I7, I8) [1 + I8 <= I7]
5.06/5.08	  f7(I9, I10, I11) -> f5(I9, I10, I11) [I10 <= I11]
5.06/5.08	  f6(I12, I13, I14) -> f5(I12, I13, I14) 
5.06/5.08	  f5(I15, I16, I17) -> f6(1 + I15, I16, I17) [1 + I15 <= I17]
5.06/5.08	  f5(I18, I19, I20) -> f3(I18, I19, I20) [I20 <= I18]
5.06/5.08	  f4(I21, I22, I23) -> f3(I21, I22, I23) 
5.06/5.08	  f3(I24, I25, I26) -> f4(I24, -1 + I25, I26) [1 <= I25]
5.06/5.08	  f3(I27, I28, I29) -> f1(I27, I28, I29) [I28 <= 0]
5.06/5.08	  f2(I30, I31, I32) -> f1(I30, I31, I32) 
5.06/5.08	  f1(I33, I34, I35) -> f2(-1 + I33, I34, I35) [1 <= I33]
5.06/5.08	
5.06/5.08	The dependency graph for this problem is:
5.06/5.08	  5 -> 
5.06/5.08	Where:
5.06/5.08	  5) f6#(I12, I13, I14) -> f5#(I12, I13, I14) 
5.06/5.08	
5.06/5.08	We have the following SCCs.
5.06/5.08	  
5.06/5.08	
5.06/5.08	DP problem for innermost termination.
5.06/5.08	P =
5.06/5.08	  f8#(I3, I4, I5) -> f7#(I3, I4, I5) 
5.06/5.08	  f7#(I6, I7, I8) -> f8#(I6, -1 + I7, I8) [1 + I8 <= I7]
5.06/5.08	R = 
5.06/5.08	  f10(x1, x2, x3) -> f9(x1, x2, x3) 
5.06/5.08	  f9(I0, I1, I2) -> f7(0, I1, I2) 
5.06/5.08	  f8(I3, I4, I5) -> f7(I3, I4, I5) 
5.06/5.08	  f7(I6, I7, I8) -> f8(I6, -1 + I7, I8) [1 + I8 <= I7]
5.06/5.08	  f7(I9, I10, I11) -> f5(I9, I10, I11) [I10 <= I11]
5.06/5.08	  f6(I12, I13, I14) -> f5(I12, I13, I14) 
5.06/5.08	  f5(I15, I16, I17) -> f6(1 + I15, I16, I17) [1 + I15 <= I17]
5.06/5.08	  f5(I18, I19, I20) -> f3(I18, I19, I20) [I20 <= I18]
5.06/5.08	  f4(I21, I22, I23) -> f3(I21, I22, I23) 
5.06/5.08	  f3(I24, I25, I26) -> f4(I24, -1 + I25, I26) [1 <= I25]
5.06/5.08	  f3(I27, I28, I29) -> f1(I27, I28, I29) [I28 <= 0]
5.06/5.08	  f2(I30, I31, I32) -> f1(I30, I31, I32) 
5.06/5.08	  f1(I33, I34, I35) -> f2(-1 + I33, I34, I35) [1 <= I33]
5.06/5.08	
5.06/5.08	We use the reverse value criterion with the projection function NU:
5.06/5.08	  NU[f7#(z1,z2,z3)] = z2 + -1 * (1 + z3)
5.06/5.08	  NU[f8#(z1,z2,z3)] = z2 + -1 * (1 + z3)
5.06/5.08	
5.06/5.08	This gives the following inequalities:
5.06/5.08	    ==>  I4 + -1 * (1 + I5) >= I4 + -1 * (1 + I5)
5.06/5.08	  1 + I8 <= I7  ==>  I7 + -1 * (1 + I8) > -1 + I7 + -1 * (1 + I8)  with  I7 + -1 * (1 + I8) >= 0
5.06/5.08	
5.06/5.08	We remove all the strictly oriented dependency pairs.
5.06/5.08	
5.06/5.08	DP problem for innermost termination.
5.06/5.08	P =
5.06/5.08	  f8#(I3, I4, I5) -> f7#(I3, I4, I5) 
5.06/5.08	R = 
5.06/5.08	  f10(x1, x2, x3) -> f9(x1, x2, x3) 
5.06/5.08	  f9(I0, I1, I2) -> f7(0, I1, I2) 
5.06/5.08	  f8(I3, I4, I5) -> f7(I3, I4, I5) 
5.06/5.08	  f7(I6, I7, I8) -> f8(I6, -1 + I7, I8) [1 + I8 <= I7]
5.06/5.08	  f7(I9, I10, I11) -> f5(I9, I10, I11) [I10 <= I11]
5.06/5.08	  f6(I12, I13, I14) -> f5(I12, I13, I14) 
5.06/5.08	  f5(I15, I16, I17) -> f6(1 + I15, I16, I17) [1 + I15 <= I17]
5.06/5.08	  f5(I18, I19, I20) -> f3(I18, I19, I20) [I20 <= I18]
5.06/5.08	  f4(I21, I22, I23) -> f3(I21, I22, I23) 
5.06/5.08	  f3(I24, I25, I26) -> f4(I24, -1 + I25, I26) [1 <= I25]
5.06/5.08	  f3(I27, I28, I29) -> f1(I27, I28, I29) [I28 <= 0]
5.06/5.08	  f2(I30, I31, I32) -> f1(I30, I31, I32) 
5.06/5.08	  f1(I33, I34, I35) -> f2(-1 + I33, I34, I35) [1 <= I33]
5.06/5.08	
5.06/5.08	The dependency graph for this problem is:
5.06/5.08	  2 -> 
5.06/5.08	Where:
5.06/5.08	  2) f8#(I3, I4, I5) -> f7#(I3, I4, I5) 
5.06/5.08	
5.06/5.08	We have the following SCCs.
5.06/5.08	  
5.06/8.06	EOF