2.37/2.36	YES
2.37/2.36	
2.37/2.36	DP problem for innermost termination.
2.37/2.36	P =
2.37/2.36	  b15#(sv14_14, sv23_37, sv24_38) -> b10#(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) 
2.37/2.36	  b14#(I0, I1, I2) -> Cond_b14#(I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1, I0, I1, I2) 
2.37/2.36	  Cond_b14#(true, I3, I4, I5) -> b15#(I3, I4, I5) 
2.37/2.36	  b10#(I6, I7, I8) -> b14#(I6, I7, I8) 
2.37/2.36	R = 
2.37/2.36	  b15(sv14_14, sv23_37, sv24_38) -> b10(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) 
2.37/2.36	  b14(I0, I1, I2) -> Cond_b14(I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1, I0, I1, I2) 
2.37/2.36	  Cond_b14(true, I3, I4, I5) -> b15(I3, I4, I5) 
2.37/2.36	  b10(I6, I7, I8) -> b14(I6, I7, I8) 
2.37/2.36	
2.37/2.36	This problem is converted using chaining, where edges between chained DPs are removed.
2.37/2.36	
2.37/2.36	DP problem for innermost termination.
2.37/2.36	P =
2.37/2.36	  b15#(sv14_14, sv23_37, sv24_38) -> b10#(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) 
2.37/2.36	  b14#(I0, I1, I2) -> Cond_b14#(I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1, I0, I1, I2) 
2.37/2.36	  Cond_b14#(true, I3, I4, I5) -> b15#(I3, I4, I5) 
2.37/2.36	  b10#(I6, I7, I8) -> b14#(I6, I7, I8) 
2.37/2.36	  b14#(I0, I1, I2) -> b15#(I0, I1, I2) [I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1]
2.37/2.37	R = 
2.37/2.37	  b15(sv14_14, sv23_37, sv24_38) -> b10(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) 
2.37/2.37	  b14(I0, I1, I2) -> Cond_b14(I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1, I0, I1, I2) 
2.37/2.37	  Cond_b14(true, I3, I4, I5) -> b15(I3, I4, I5) 
2.37/2.37	  b10(I6, I7, I8) -> b14(I6, I7, I8) 
2.37/2.37	
2.37/2.37	The dependency graph for this problem is:
2.37/2.37	  0 -> 3
2.37/2.37	  1 -> 
2.37/2.37	  2 -> 0
2.37/2.37	  3 -> 4, 1
2.37/2.37	  4 -> 0
2.37/2.37	Where:
2.37/2.37	  0) b15#(sv14_14, sv23_37, sv24_38) -> b10#(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) 
2.37/2.37	  1) b14#(I0, I1, I2) -> Cond_b14#(I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1, I0, I1, I2) 
2.37/2.37	  2) Cond_b14#(true, I3, I4, I5) -> b15#(I3, I4, I5) 
2.37/2.37	  3) b10#(I6, I7, I8) -> b14#(I6, I7, I8) 
2.37/2.37	  4) b14#(I0, I1, I2) -> b15#(I0, I1, I2) [I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1]
2.37/2.37	
2.37/2.37	We have the following SCCs.
2.37/2.37	  { 0, 3, 4 }
2.37/2.37	
2.37/2.37	DP problem for innermost termination.
2.37/2.37	P =
2.37/2.37	  b15#(sv14_14, sv23_37, sv24_38) -> b10#(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) 
2.37/2.37	  b10#(I6, I7, I8) -> b14#(I6, I7, I8) 
2.37/2.37	  b14#(I0, I1, I2) -> b15#(I0, I1, I2) [I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1]
2.37/2.37	R = 
2.37/2.37	  b15(sv14_14, sv23_37, sv24_38) -> b10(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) 
2.37/2.37	  b14(I0, I1, I2) -> Cond_b14(I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1, I0, I1, I2) 
2.37/2.37	  Cond_b14(true, I3, I4, I5) -> b15(I3, I4, I5) 
2.37/2.37	  b10(I6, I7, I8) -> b14(I6, I7, I8) 
2.37/2.37	
2.37/2.37	We use the extended value criterion with the projection function NU:
2.37/2.37	  NU[b14#(x0,x1,x2)] = x1 - 2
2.37/2.37	  NU[b10#(x0,x1,x2)] = x1 - 2
2.37/2.37	  NU[b15#(x0,x1,x2)] = -x0 + x1 - 2
2.37/2.37	
2.37/2.37	This gives the following inequalities:
2.37/2.37	    ==>  -sv14_14 + sv23_37 - 2 >= (sv23_37 - sv14_14) - 2
2.37/2.37	    ==>  I7 - 2 >= I7 - 2
2.37/2.37	  I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1  ==>  I1 - 2 > -I0 + I1 - 2  with  I1 - 2 >= 0
2.37/2.37	
2.37/2.37	We remove all the strictly oriented dependency pairs.
2.37/2.37	
2.37/2.37	DP problem for innermost termination.
2.37/2.37	P =
2.37/2.37	  b15#(sv14_14, sv23_37, sv24_38) -> b10#(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) 
2.37/2.37	  b10#(I6, I7, I8) -> b14#(I6, I7, I8) 
2.37/2.37	R = 
2.37/2.37	  b15(sv14_14, sv23_37, sv24_38) -> b10(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) 
2.37/2.37	  b14(I0, I1, I2) -> Cond_b14(I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1, I0, I1, I2) 
2.37/2.37	  Cond_b14(true, I3, I4, I5) -> b15(I3, I4, I5) 
2.37/2.37	  b10(I6, I7, I8) -> b14(I6, I7, I8) 
2.37/2.37	
2.37/2.37	The dependency graph for this problem is:
2.37/2.37	  0 -> 3
2.37/2.37	  3 -> 
2.37/2.37	Where:
2.37/2.37	  0) b15#(sv14_14, sv23_37, sv24_38) -> b10#(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) 
2.37/2.37	  3) b10#(I6, I7, I8) -> b14#(I6, I7, I8) 
2.37/2.37	
2.37/2.37	We have the following SCCs.
2.37/2.37	  
2.37/5.34	EOF