15.59/15.36	YES
15.59/15.36	
15.59/15.36	DP problem for innermost termination.
15.59/15.36	P =
15.59/15.36	  f7#(x1, x2, x3, x4, x5, x6, x7) -> f6#(x1, x2, x3, x4, x5, x6, x7) 
15.59/15.36	  f6#(I0, I1, I2, I3, I4, I5, I6) -> f3#(I0, I1, 0, I3, I4, I5, I6) 
15.59/15.36	  f2#(I7, I8, I9, I10, I11, I12, I13) -> f4#(I7, I8, I9, I10, I11, I12, I13) 
15.59/15.36	  f4#(I14, I15, I16, I17, I18, I19, I20) -> f2#(I14, I15, I16, 1 + I17, I18, I19, I17) [1 + I17 <= I15]
15.59/15.36	  f3#(I28, I29, I30, I31, I32, I33, I34) -> f1#(I28, I29, I30, I31, I32, I33, I34) 
15.59/15.36	  f1#(I35, I36, I37, I38, I39, I40, I41) -> f3#(I35, I36, 1 + I37, I38, I37, I37, I41) [1 + I37 <= I35]
15.59/15.36	  f1#(I42, I43, I44, I45, I46, I47, I48) -> f2#(I42, I43, I44, I42, I46, I47, I48) [I42 <= I44]
15.59/15.36	R = 
15.59/15.36	  f7(x1, x2, x3, x4, x5, x6, x7) -> f6(x1, x2, x3, x4, x5, x6, x7) 
15.59/15.36	  f6(I0, I1, I2, I3, I4, I5, I6) -> f3(I0, I1, 0, I3, I4, I5, I6) 
15.59/15.36	  f2(I7, I8, I9, I10, I11, I12, I13) -> f4(I7, I8, I9, I10, I11, I12, I13) 
15.59/15.36	  f4(I14, I15, I16, I17, I18, I19, I20) -> f2(I14, I15, I16, 1 + I17, I18, I19, I17) [1 + I17 <= I15]
15.59/15.36	  f4(I21, I22, I23, I24, I25, I26, I27) -> f5(I21, I22, I23, I24, I25, I26, I27) [I22 <= I24]
15.59/15.36	  f3(I28, I29, I30, I31, I32, I33, I34) -> f1(I28, I29, I30, I31, I32, I33, I34) 
15.59/15.36	  f1(I35, I36, I37, I38, I39, I40, I41) -> f3(I35, I36, 1 + I37, I38, I37, I37, I41) [1 + I37 <= I35]
15.59/15.36	  f1(I42, I43, I44, I45, I46, I47, I48) -> f2(I42, I43, I44, I42, I46, I47, I48) [I42 <= I44]
15.59/15.36	
15.59/15.36	The dependency graph for this problem is:
15.59/15.36	  0 -> 1
15.59/15.36	  1 -> 4
15.59/15.36	  2 -> 3
15.59/15.36	  3 -> 2
15.59/15.36	  4 -> 5, 6
15.59/15.36	  5 -> 4
15.59/15.36	  6 -> 2
15.59/15.36	Where:
15.59/15.36	  0) f7#(x1, x2, x3, x4, x5, x6, x7) -> f6#(x1, x2, x3, x4, x5, x6, x7) 
15.59/15.36	  1) f6#(I0, I1, I2, I3, I4, I5, I6) -> f3#(I0, I1, 0, I3, I4, I5, I6) 
15.59/15.36	  2) f2#(I7, I8, I9, I10, I11, I12, I13) -> f4#(I7, I8, I9, I10, I11, I12, I13) 
15.59/15.36	  3) f4#(I14, I15, I16, I17, I18, I19, I20) -> f2#(I14, I15, I16, 1 + I17, I18, I19, I17) [1 + I17 <= I15]
15.59/15.36	  4) f3#(I28, I29, I30, I31, I32, I33, I34) -> f1#(I28, I29, I30, I31, I32, I33, I34) 
15.59/15.36	  5) f1#(I35, I36, I37, I38, I39, I40, I41) -> f3#(I35, I36, 1 + I37, I38, I37, I37, I41) [1 + I37 <= I35]
15.59/15.36	  6) f1#(I42, I43, I44, I45, I46, I47, I48) -> f2#(I42, I43, I44, I42, I46, I47, I48) [I42 <= I44]
15.59/15.36	
15.59/15.36	We have the following SCCs.
15.59/15.36	  { 4, 5 }
15.59/15.36	  { 2, 3 }
15.59/15.36	
15.59/15.36	DP problem for innermost termination.
15.59/15.36	P =
15.59/15.36	  f2#(I7, I8, I9, I10, I11, I12, I13) -> f4#(I7, I8, I9, I10, I11, I12, I13) 
15.59/15.36	  f4#(I14, I15, I16, I17, I18, I19, I20) -> f2#(I14, I15, I16, 1 + I17, I18, I19, I17) [1 + I17 <= I15]
15.59/15.36	R = 
15.59/15.36	  f7(x1, x2, x3, x4, x5, x6, x7) -> f6(x1, x2, x3, x4, x5, x6, x7) 
15.59/15.36	  f6(I0, I1, I2, I3, I4, I5, I6) -> f3(I0, I1, 0, I3, I4, I5, I6) 
15.59/15.36	  f2(I7, I8, I9, I10, I11, I12, I13) -> f4(I7, I8, I9, I10, I11, I12, I13) 
15.59/15.36	  f4(I14, I15, I16, I17, I18, I19, I20) -> f2(I14, I15, I16, 1 + I17, I18, I19, I17) [1 + I17 <= I15]
15.59/15.36	  f4(I21, I22, I23, I24, I25, I26, I27) -> f5(I21, I22, I23, I24, I25, I26, I27) [I22 <= I24]
15.59/15.36	  f3(I28, I29, I30, I31, I32, I33, I34) -> f1(I28, I29, I30, I31, I32, I33, I34) 
15.59/15.36	  f1(I35, I36, I37, I38, I39, I40, I41) -> f3(I35, I36, 1 + I37, I38, I37, I37, I41) [1 + I37 <= I35]
15.59/15.36	  f1(I42, I43, I44, I45, I46, I47, I48) -> f2(I42, I43, I44, I42, I46, I47, I48) [I42 <= I44]
15.59/15.36	
15.59/15.36	We use the reverse value criterion with the projection function NU:
15.59/15.36	  NU[f4#(z1,z2,z3,z4,z5,z6,z7)] = z2 + -1 * (1 + z4)
15.59/15.36	  NU[f2#(z1,z2,z3,z4,z5,z6,z7)] = z2 + -1 * (1 + z4)
15.59/15.36	
15.59/15.36	This gives the following inequalities:
15.59/15.36	    ==>  I8 + -1 * (1 + I10) >= I8 + -1 * (1 + I10)
15.59/15.36	  1 + I17 <= I15  ==>  I15 + -1 * (1 + I17) > I15 + -1 * (1 + (1 + I17))  with  I15 + -1 * (1 + I17) >= 0
15.59/15.36	
15.59/15.36	We remove all the strictly oriented dependency pairs.
15.59/15.36	
15.59/15.36	DP problem for innermost termination.
15.59/15.36	P =
15.59/15.36	  f2#(I7, I8, I9, I10, I11, I12, I13) -> f4#(I7, I8, I9, I10, I11, I12, I13) 
15.59/15.36	R = 
15.59/15.36	  f7(x1, x2, x3, x4, x5, x6, x7) -> f6(x1, x2, x3, x4, x5, x6, x7) 
15.59/15.36	  f6(I0, I1, I2, I3, I4, I5, I6) -> f3(I0, I1, 0, I3, I4, I5, I6) 
15.59/15.36	  f2(I7, I8, I9, I10, I11, I12, I13) -> f4(I7, I8, I9, I10, I11, I12, I13) 
15.59/15.36	  f4(I14, I15, I16, I17, I18, I19, I20) -> f2(I14, I15, I16, 1 + I17, I18, I19, I17) [1 + I17 <= I15]
15.59/15.36	  f4(I21, I22, I23, I24, I25, I26, I27) -> f5(I21, I22, I23, I24, I25, I26, I27) [I22 <= I24]
15.59/15.36	  f3(I28, I29, I30, I31, I32, I33, I34) -> f1(I28, I29, I30, I31, I32, I33, I34) 
15.59/15.36	  f1(I35, I36, I37, I38, I39, I40, I41) -> f3(I35, I36, 1 + I37, I38, I37, I37, I41) [1 + I37 <= I35]
15.59/15.36	  f1(I42, I43, I44, I45, I46, I47, I48) -> f2(I42, I43, I44, I42, I46, I47, I48) [I42 <= I44]
15.59/15.36	
15.59/15.36	The dependency graph for this problem is:
15.59/15.36	  2 -> 
15.59/15.36	Where:
15.59/15.36	  2) f2#(I7, I8, I9, I10, I11, I12, I13) -> f4#(I7, I8, I9, I10, I11, I12, I13) 
15.59/15.36	
15.59/15.36	We have the following SCCs.
15.59/15.36	  
15.59/15.36	
15.59/15.36	DP problem for innermost termination.
15.59/15.36	P =
15.59/15.36	  f3#(I28, I29, I30, I31, I32, I33, I34) -> f1#(I28, I29, I30, I31, I32, I33, I34) 
15.59/15.36	  f1#(I35, I36, I37, I38, I39, I40, I41) -> f3#(I35, I36, 1 + I37, I38, I37, I37, I41) [1 + I37 <= I35]
15.59/15.36	R = 
15.59/15.36	  f7(x1, x2, x3, x4, x5, x6, x7) -> f6(x1, x2, x3, x4, x5, x6, x7) 
15.59/15.36	  f6(I0, I1, I2, I3, I4, I5, I6) -> f3(I0, I1, 0, I3, I4, I5, I6) 
15.59/15.36	  f2(I7, I8, I9, I10, I11, I12, I13) -> f4(I7, I8, I9, I10, I11, I12, I13) 
15.59/15.36	  f4(I14, I15, I16, I17, I18, I19, I20) -> f2(I14, I15, I16, 1 + I17, I18, I19, I17) [1 + I17 <= I15]
15.59/15.36	  f4(I21, I22, I23, I24, I25, I26, I27) -> f5(I21, I22, I23, I24, I25, I26, I27) [I22 <= I24]
15.59/15.36	  f3(I28, I29, I30, I31, I32, I33, I34) -> f1(I28, I29, I30, I31, I32, I33, I34) 
15.59/15.36	  f1(I35, I36, I37, I38, I39, I40, I41) -> f3(I35, I36, 1 + I37, I38, I37, I37, I41) [1 + I37 <= I35]
15.59/15.36	  f1(I42, I43, I44, I45, I46, I47, I48) -> f2(I42, I43, I44, I42, I46, I47, I48) [I42 <= I44]
15.59/15.36	
15.59/15.36	We use the reverse value criterion with the projection function NU:
15.59/15.36	  NU[f1#(z1,z2,z3,z4,z5,z6,z7)] = z1 + -1 * (1 + z3)
15.59/15.36	  NU[f3#(z1,z2,z3,z4,z5,z6,z7)] = z1 + -1 * (1 + z3)
15.59/15.36	
15.59/15.36	This gives the following inequalities:
15.59/15.36	    ==>  I28 + -1 * (1 + I30) >= I28 + -1 * (1 + I30)
15.59/15.36	  1 + I37 <= I35  ==>  I35 + -1 * (1 + I37) > I35 + -1 * (1 + (1 + I37))  with  I35 + -1 * (1 + I37) >= 0
15.59/15.36	
15.59/15.36	We remove all the strictly oriented dependency pairs.
15.59/15.36	
15.59/15.36	DP problem for innermost termination.
15.59/15.36	P =
15.59/15.36	  f3#(I28, I29, I30, I31, I32, I33, I34) -> f1#(I28, I29, I30, I31, I32, I33, I34) 
15.59/15.36	R = 
15.59/15.36	  f7(x1, x2, x3, x4, x5, x6, x7) -> f6(x1, x2, x3, x4, x5, x6, x7) 
15.59/15.36	  f6(I0, I1, I2, I3, I4, I5, I6) -> f3(I0, I1, 0, I3, I4, I5, I6) 
15.59/15.36	  f2(I7, I8, I9, I10, I11, I12, I13) -> f4(I7, I8, I9, I10, I11, I12, I13) 
15.59/15.36	  f4(I14, I15, I16, I17, I18, I19, I20) -> f2(I14, I15, I16, 1 + I17, I18, I19, I17) [1 + I17 <= I15]
15.59/15.36	  f4(I21, I22, I23, I24, I25, I26, I27) -> f5(I21, I22, I23, I24, I25, I26, I27) [I22 <= I24]
15.59/15.36	  f3(I28, I29, I30, I31, I32, I33, I34) -> f1(I28, I29, I30, I31, I32, I33, I34) 
15.59/15.36	  f1(I35, I36, I37, I38, I39, I40, I41) -> f3(I35, I36, 1 + I37, I38, I37, I37, I41) [1 + I37 <= I35]
15.59/15.36	  f1(I42, I43, I44, I45, I46, I47, I48) -> f2(I42, I43, I44, I42, I46, I47, I48) [I42 <= I44]
15.59/15.36	
15.59/15.36	The dependency graph for this problem is:
15.59/15.36	  4 -> 
15.59/15.36	Where:
15.59/15.36	  4) f3#(I28, I29, I30, I31, I32, I33, I34) -> f1#(I28, I29, I30, I31, I32, I33, I34) 
15.59/15.36	
15.59/15.36	We have the following SCCs.
15.59/15.36	  
15.59/18.34	EOF