/export/starexec/sandbox2/solver/bin/starexec_run_Itrs /export/starexec/sandbox2/benchmark/theBenchmark.itrs /export/starexec/sandbox2/output/output_files


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MAYBE

DP problem for innermost termination.
P =
  if_3#(pair(m, zs), x, ys) -> msort#(zs) 
  msort#(ins(I0, B0)) -> if_3#(min(I0, B0), I0, B0) 
  msort#(ins(I0, B0)) -> min#(I0, B0) 
  min#(I3, ins(I4, B2)) -> if_2#(min(I4, B2), I3, I4, B2) 
  min#(I3, ins(I4, B2)) -> min#(I4, B2) 
  min#(I8, ins(I9, B5)) -> if_1#(min(I9, B5), I8, I9, B5) 
  min#(I8, ins(I9, B5)) -> min#(I9, B5) 
R = 
  if_3(pair(m, zs), x, ys) -> cons(m, msort(zs)) 
  msort(ins(I0, B0)) -> if_3(min(I0, B0), I0, B0) 
  msort(e) -> nil 
  if_2(pair(I1, zh), I2, y, B1) -> pair(I1, ins(I2, zh)) [I2 >= I1]
  min(I3, ins(I4, B2)) -> if_2(min(I4, B2), I3, I4, B2) 
  if_1(pair(I5, B3), I6, I7, B4) -> pair(I6, ins(I5, B3)) [I5 > I6]
  min(I8, ins(I9, B5)) -> if_1(min(I9, B5), I8, I9, B5) 
  min(I10, e) -> pair(I10, e) 

The dependency graph for this problem is:
  0 -> 1, 2
  1 -> 0
  2 -> 3, 4, 5, 6
  3 -> 
  4 -> 3, 4, 5, 6
  5 -> 
  6 -> 3, 4, 5, 6
Where:
  0) if_3#(pair(m, zs), x, ys) -> msort#(zs) 
  1) msort#(ins(I0, B0)) -> if_3#(min(I0, B0), I0, B0) 
  2) msort#(ins(I0, B0)) -> min#(I0, B0) 
  3) min#(I3, ins(I4, B2)) -> if_2#(min(I4, B2), I3, I4, B2) 
  4) min#(I3, ins(I4, B2)) -> min#(I4, B2) 
  5) min#(I8, ins(I9, B5)) -> if_1#(min(I9, B5), I8, I9, B5) 
  6) min#(I8, ins(I9, B5)) -> min#(I9, B5) 

We have the following SCCs.
  { 0, 1 }
  { 4, 6 }

DP problem for innermost termination.
P =
  min#(I3, ins(I4, B2)) -> min#(I4, B2) 
  min#(I8, ins(I9, B5)) -> min#(I9, B5) 
R = 
  if_3(pair(m, zs), x, ys) -> cons(m, msort(zs)) 
  msort(ins(I0, B0)) -> if_3(min(I0, B0), I0, B0) 
  msort(e) -> nil 
  if_2(pair(I1, zh), I2, y, B1) -> pair(I1, ins(I2, zh)) [I2 >= I1]
  min(I3, ins(I4, B2)) -> if_2(min(I4, B2), I3, I4, B2) 
  if_1(pair(I5, B3), I6, I7, B4) -> pair(I6, ins(I5, B3)) [I5 > I6]
  min(I8, ins(I9, B5)) -> if_1(min(I9, B5), I8, I9, B5) 
  min(I10, e) -> pair(I10, e) 

We use the subterm criterion with the projection function NU:
  NU[min#(x0,x1)] = x1

This gives the following inequalities:
  ins(I4, B2) |> B2
  ins(I9, B5) |> B5

All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.

DP problem for innermost termination.
P =
  if_3#(pair(m, zs), x, ys) -> msort#(zs) 
  msort#(ins(I0, B0)) -> if_3#(min(I0, B0), I0, B0) 
R = 
  if_3(pair(m, zs), x, ys) -> cons(m, msort(zs)) 
  msort(ins(I0, B0)) -> if_3(min(I0, B0), I0, B0) 
  msort(e) -> nil 
  if_2(pair(I1, zh), I2, y, B1) -> pair(I1, ins(I2, zh)) [I2 >= I1]
  min(I3, ins(I4, B2)) -> if_2(min(I4, B2), I3, I4, B2) 
  if_1(pair(I5, B3), I6, I7, B4) -> pair(I6, ins(I5, B3)) [I5 > I6]
  min(I8, ins(I9, B5)) -> if_1(min(I9, B5), I8, I9, B5) 
  min(I10, e) -> pair(I10, e)