2.25/2.24	YES
2.25/2.24	
2.25/2.24	DP problem for innermost termination.
2.25/2.24	P =
2.25/2.24	  b14#(sv14_14, sv23_37, sv24_38) -> Cond_b14#(sv23_37 >= sv14_14 && (1 < sv14_14 && (true && (0 < sv24_38 && 1 < sv23_37))), sv14_14, sv23_37, sv24_38) 
2.25/2.24	  b15#(I0, I1, I2) -> b10#(I0, I1 - I0, I2 + 1) 
2.25/2.24	  Cond_b14#(true, I3, I4, I5) -> b15#(I3, I4, I5) 
2.25/2.24	  b10#(I6, I7, I8) -> b14#(I6, I7, I8) 
2.25/2.24	R = 
2.25/2.24	  b14(sv14_14, sv23_37, sv24_38) -> Cond_b14(sv23_37 >= sv14_14 && (1 < sv14_14 && (true && (0 < sv24_38 && 1 < sv23_37))), sv14_14, sv23_37, sv24_38) 
2.25/2.24	  b15(I0, I1, I2) -> b10(I0, I1 - I0, I2 + 1) 
2.25/2.24	  Cond_b14(true, I3, I4, I5) -> b15(I3, I4, I5) 
2.25/2.24	  b10(I6, I7, I8) -> b14(I6, I7, I8) 
2.25/2.24	
2.25/2.24	This problem is converted using chaining, where edges between chained DPs are removed.
2.25/2.24	
2.25/2.24	DP problem for innermost termination.
2.25/2.24	P =
2.25/2.24	  b14#(sv14_14, sv23_37, sv24_38) -> Cond_b14#(sv23_37 >= sv14_14 && (1 < sv14_14 && (true && (0 < sv24_38 && 1 < sv23_37))), sv14_14, sv23_37, sv24_38) 
2.25/2.24	  b15#(I0, I1, I2) -> b10#(I0, I1 - I0, I2 + 1) 
2.25/2.24	  Cond_b14#(true, I3, I4, I5) -> b15#(I3, I4, I5) 
2.25/2.24	  b10#(I6, I7, I8) -> b14#(I6, I7, I8) 
2.25/2.24	  b14#(sv14_14, sv23_37, sv24_38) -> b15#(sv14_14, sv23_37, sv24_38) [sv23_37 >= sv14_14 && (1 < sv14_14 && (true && (0 < sv24_38 && 1 < sv23_37)))]
2.25/2.24	R = 
2.25/2.24	  b14(sv14_14, sv23_37, sv24_38) -> Cond_b14(sv23_37 >= sv14_14 && (1 < sv14_14 && (true && (0 < sv24_38 && 1 < sv23_37))), sv14_14, sv23_37, sv24_38) 
2.25/2.24	  b15(I0, I1, I2) -> b10(I0, I1 - I0, I2 + 1) 
2.25/2.24	  Cond_b14(true, I3, I4, I5) -> b15(I3, I4, I5) 
2.25/2.24	  b10(I6, I7, I8) -> b14(I6, I7, I8) 
2.25/2.24	
2.25/2.24	The dependency graph for this problem is:
2.25/2.24	  0 -> 
2.25/2.24	  1 -> 3
2.25/2.24	  2 -> 1
2.25/2.24	  3 -> 4, 0
2.25/2.24	  4 -> 1
2.25/2.24	Where:
2.25/2.24	  0) b14#(sv14_14, sv23_37, sv24_38) -> Cond_b14#(sv23_37 >= sv14_14 && (1 < sv14_14 && (true && (0 < sv24_38 && 1 < sv23_37))), sv14_14, sv23_37, sv24_38) 
2.25/2.24	  1) b15#(I0, I1, I2) -> b10#(I0, I1 - I0, I2 + 1) 
2.25/2.24	  2) Cond_b14#(true, I3, I4, I5) -> b15#(I3, I4, I5) 
2.25/2.24	  3) b10#(I6, I7, I8) -> b14#(I6, I7, I8) 
2.25/2.24	  4) b14#(sv14_14, sv23_37, sv24_38) -> b15#(sv14_14, sv23_37, sv24_38) [sv23_37 >= sv14_14 && (1 < sv14_14 && (true && (0 < sv24_38 && 1 < sv23_37)))]
2.25/2.24	
2.25/2.24	We have the following SCCs.
2.25/2.24	  { 1, 3, 4 }
2.25/2.24	
2.25/2.24	DP problem for innermost termination.
2.25/2.24	P =
2.25/2.24	  b15#(I0, I1, I2) -> b10#(I0, I1 - I0, I2 + 1) 
2.25/2.24	  b10#(I6, I7, I8) -> b14#(I6, I7, I8) 
2.25/2.24	  b14#(sv14_14, sv23_37, sv24_38) -> b15#(sv14_14, sv23_37, sv24_38) [sv23_37 >= sv14_14 && (1 < sv14_14 && (true && (0 < sv24_38 && 1 < sv23_37)))]
2.25/2.24	R = 
2.25/2.24	  b14(sv14_14, sv23_37, sv24_38) -> Cond_b14(sv23_37 >= sv14_14 && (1 < sv14_14 && (true && (0 < sv24_38 && 1 < sv23_37))), sv14_14, sv23_37, sv24_38) 
2.25/2.24	  b15(I0, I1, I2) -> b10(I0, I1 - I0, I2 + 1) 
2.25/2.24	  Cond_b14(true, I3, I4, I5) -> b15(I3, I4, I5) 
2.25/2.24	  b10(I6, I7, I8) -> b14(I6, I7, I8) 
2.25/2.24	
2.25/2.24	We use the extended value criterion with the projection function NU:
2.25/2.24	  NU[b14#(x0,x1,x2)] = x1 - 2
2.25/2.24	  NU[b10#(x0,x1,x2)] = x1 - 2
2.25/2.24	  NU[b15#(x0,x1,x2)] = -x0 + x1 - 2
2.25/2.24	
2.25/2.24	This gives the following inequalities:
2.25/2.24	    ==>  -I0 + I1 - 2 >= (I1 - I0) - 2
2.25/2.24	    ==>  I7 - 2 >= I7 - 2
2.25/2.24	  sv23_37 >= sv14_14 && (1 < sv14_14 && (true && (0 < sv24_38 && 1 < sv23_37)))  ==>  sv23_37 - 2 > -sv14_14 + sv23_37 - 2  with  sv23_37 - 2 >= 0
2.25/2.24	
2.25/2.24	We remove all the strictly oriented dependency pairs.
2.25/2.24	
2.25/2.24	DP problem for innermost termination.
2.25/2.24	P =
2.25/2.24	  b15#(I0, I1, I2) -> b10#(I0, I1 - I0, I2 + 1) 
2.25/2.24	  b10#(I6, I7, I8) -> b14#(I6, I7, I8) 
2.25/2.24	R = 
2.25/2.24	  b14(sv14_14, sv23_37, sv24_38) -> Cond_b14(sv23_37 >= sv14_14 && (1 < sv14_14 && (true && (0 < sv24_38 && 1 < sv23_37))), sv14_14, sv23_37, sv24_38) 
2.25/2.24	  b15(I0, I1, I2) -> b10(I0, I1 - I0, I2 + 1) 
2.25/2.24	  Cond_b14(true, I3, I4, I5) -> b15(I3, I4, I5) 
2.25/2.24	  b10(I6, I7, I8) -> b14(I6, I7, I8) 
2.25/2.24	
2.25/2.24	The dependency graph for this problem is:
2.25/2.24	  1 -> 3
2.25/2.24	  3 -> 
2.25/2.24	Where:
2.25/2.24	  1) b15#(I0, I1, I2) -> b10#(I0, I1 - I0, I2 + 1) 
2.25/2.24	  3) b10#(I6, I7, I8) -> b14#(I6, I7, I8) 
2.25/2.24	
2.25/2.24	We have the following SCCs.
2.25/2.24	  
2.25/2.24	EOF