0.00/0.10	MAYBE
0.00/0.10	
0.00/0.10	DP problem for innermost termination.
0.00/0.10	P =
0.00/0.10	  if_3#(pair(m, zs), x, ys) -> msort#(zs) 
0.00/0.10	  msort#(ins(I0, B0)) -> if_3#(min(I0, B0), I0, B0) 
0.00/0.10	  msort#(ins(I0, B0)) -> min#(I0, B0) 
0.00/0.10	  min#(I3, ins(I4, B2)) -> if_2#(min(I4, B2), I3, I4, B2) 
0.00/0.10	  min#(I3, ins(I4, B2)) -> min#(I4, B2) 
0.00/0.10	  min#(I8, ins(I9, B5)) -> if_1#(min(I9, B5), I8, I9, B5) 
0.00/0.10	  min#(I8, ins(I9, B5)) -> min#(I9, B5) 
0.00/0.10	R = 
0.00/0.10	  if_3(pair(m, zs), x, ys) -> cons(m, msort(zs)) 
0.00/0.10	  msort(ins(I0, B0)) -> if_3(min(I0, B0), I0, B0) 
0.00/0.10	  msort(e) -> nil 
0.00/0.10	  if_2(pair(I1, zh), I2, y, B1) -> pair(I1, ins(I2, zh)) [I2 >= I1]
0.00/0.10	  min(I3, ins(I4, B2)) -> if_2(min(I4, B2), I3, I4, B2) 
0.00/0.10	  if_1(pair(I5, B3), I6, I7, B4) -> pair(I6, ins(I5, B3)) [I5 > I6]
0.00/0.10	  min(I8, ins(I9, B5)) -> if_1(min(I9, B5), I8, I9, B5) 
0.00/0.10	  min(I10, e) -> pair(I10, e) 
0.00/0.10	
0.00/0.10	The dependency graph for this problem is:
0.00/0.10	  0 -> 1, 2
0.00/0.10	  1 -> 0
0.00/0.10	  2 -> 3, 4, 5, 6
0.00/0.10	  3 -> 
0.00/0.10	  4 -> 3, 4, 5, 6
0.00/0.10	  5 -> 
0.00/0.10	  6 -> 3, 4, 5, 6
0.00/0.10	Where:
0.00/0.10	  0) if_3#(pair(m, zs), x, ys) -> msort#(zs) 
0.00/0.10	  1) msort#(ins(I0, B0)) -> if_3#(min(I0, B0), I0, B0) 
0.00/0.10	  2) msort#(ins(I0, B0)) -> min#(I0, B0) 
0.00/0.10	  3) min#(I3, ins(I4, B2)) -> if_2#(min(I4, B2), I3, I4, B2) 
0.00/0.10	  4) min#(I3, ins(I4, B2)) -> min#(I4, B2) 
0.00/0.10	  5) min#(I8, ins(I9, B5)) -> if_1#(min(I9, B5), I8, I9, B5) 
0.00/0.10	  6) min#(I8, ins(I9, B5)) -> min#(I9, B5) 
0.00/0.10	
0.00/0.10	We have the following SCCs.
0.00/0.10	  { 0, 1 }
0.00/0.10	  { 4, 6 }
0.00/0.10	
0.00/0.10	DP problem for innermost termination.
0.00/0.10	P =
0.00/0.10	  min#(I3, ins(I4, B2)) -> min#(I4, B2) 
0.00/0.10	  min#(I8, ins(I9, B5)) -> min#(I9, B5) 
0.00/0.10	R = 
0.00/0.10	  if_3(pair(m, zs), x, ys) -> cons(m, msort(zs)) 
0.00/0.10	  msort(ins(I0, B0)) -> if_3(min(I0, B0), I0, B0) 
0.00/0.10	  msort(e) -> nil 
0.00/0.10	  if_2(pair(I1, zh), I2, y, B1) -> pair(I1, ins(I2, zh)) [I2 >= I1]
0.00/0.10	  min(I3, ins(I4, B2)) -> if_2(min(I4, B2), I3, I4, B2) 
0.00/0.10	  if_1(pair(I5, B3), I6, I7, B4) -> pair(I6, ins(I5, B3)) [I5 > I6]
0.00/0.10	  min(I8, ins(I9, B5)) -> if_1(min(I9, B5), I8, I9, B5) 
0.00/0.10	  min(I10, e) -> pair(I10, e) 
0.00/0.10	
0.00/0.10	We use the subterm criterion with the projection function NU:
0.00/0.10	  NU[min#(x0,x1)] = x1
0.00/0.10	
0.00/0.10	This gives the following inequalities:
0.00/0.10	  ins(I4, B2) |> B2
0.00/0.10	  ins(I9, B5) |> B5
0.00/0.10	
0.00/0.10	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/0.10	
0.00/0.10	DP problem for innermost termination.
0.00/0.10	P =
0.00/0.10	  if_3#(pair(m, zs), x, ys) -> msort#(zs) 
0.00/0.10	  msort#(ins(I0, B0)) -> if_3#(min(I0, B0), I0, B0) 
0.00/0.10	R = 
0.00/0.10	  if_3(pair(m, zs), x, ys) -> cons(m, msort(zs)) 
0.00/0.10	  msort(ins(I0, B0)) -> if_3(min(I0, B0), I0, B0) 
0.00/0.10	  msort(e) -> nil 
0.00/0.10	  if_2(pair(I1, zh), I2, y, B1) -> pair(I1, ins(I2, zh)) [I2 >= I1]
0.00/0.10	  min(I3, ins(I4, B2)) -> if_2(min(I4, B2), I3, I4, B2) 
0.00/0.10	  if_1(pair(I5, B3), I6, I7, B4) -> pair(I6, ins(I5, B3)) [I5 > I6]
0.00/0.10	  min(I8, ins(I9, B5)) -> if_1(min(I9, B5), I8, I9, B5) 
0.00/0.10	  min(I10, e) -> pair(I10, e) 
0.00/0.10	
0.00/3.08	EOF