Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following rules:

msort1(nil) -> nil
msort1(.2(x, y)) -> .2(min2(x, y), msort1(del2(min2(x, y), .2(x, y))))
min2(x, nil) -> x
min2(x, .2(y, z)) -> if3(<=2(x, y), min2(x, z), min2(y, z))
del2(x, nil) -> nil
del2(x, .2(y, z)) -> if3(=2(x, y), z, .2(y, del2(x, z)))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

Q restricted rewrite system:
The TRS R consists of the following rules:

msort1(nil) -> nil
msort1(.2(x, y)) -> .2(min2(x, y), msort1(del2(min2(x, y), .2(x, y))))
min2(x, nil) -> x
min2(x, .2(y, z)) -> if3(<=2(x, y), min2(x, z), min2(y, z))
del2(x, nil) -> nil
del2(x, .2(y, z)) -> if3(=2(x, y), z, .2(y, del2(x, z)))

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

msort1(nil) -> nil
msort1(.2(x, y)) -> .2(min2(x, y), msort1(del2(min2(x, y), .2(x, y))))
min2(x, nil) -> x
min2(x, .2(y, z)) -> if3(<=2(x, y), min2(x, z), min2(y, z))
del2(x, nil) -> nil
del2(x, .2(y, z)) -> if3(=2(x, y), z, .2(y, del2(x, z)))

The set Q consists of the following terms:

msort1(nil)
msort1(.2(x0, x1))
min2(x0, nil)
min2(x0, .2(x1, x2))
del2(x0, nil)
del2(x0, .2(x1, x2))


Q DP problem:
The TRS P consists of the following rules:

MIN2(x, .2(y, z)) -> MIN2(y, z)
DEL2(x, .2(y, z)) -> DEL2(x, z)
MSORT1(.2(x, y)) -> DEL2(min2(x, y), .2(x, y))
MSORT1(.2(x, y)) -> MIN2(x, y)
MSORT1(.2(x, y)) -> MSORT1(del2(min2(x, y), .2(x, y)))
MIN2(x, .2(y, z)) -> MIN2(x, z)

The TRS R consists of the following rules:

msort1(nil) -> nil
msort1(.2(x, y)) -> .2(min2(x, y), msort1(del2(min2(x, y), .2(x, y))))
min2(x, nil) -> x
min2(x, .2(y, z)) -> if3(<=2(x, y), min2(x, z), min2(y, z))
del2(x, nil) -> nil
del2(x, .2(y, z)) -> if3(=2(x, y), z, .2(y, del2(x, z)))

The set Q consists of the following terms:

msort1(nil)
msort1(.2(x0, x1))
min2(x0, nil)
min2(x0, .2(x1, x2))
del2(x0, nil)
del2(x0, .2(x1, x2))

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

MIN2(x, .2(y, z)) -> MIN2(y, z)
DEL2(x, .2(y, z)) -> DEL2(x, z)
MSORT1(.2(x, y)) -> DEL2(min2(x, y), .2(x, y))
MSORT1(.2(x, y)) -> MIN2(x, y)
MSORT1(.2(x, y)) -> MSORT1(del2(min2(x, y), .2(x, y)))
MIN2(x, .2(y, z)) -> MIN2(x, z)

The TRS R consists of the following rules:

msort1(nil) -> nil
msort1(.2(x, y)) -> .2(min2(x, y), msort1(del2(min2(x, y), .2(x, y))))
min2(x, nil) -> x
min2(x, .2(y, z)) -> if3(<=2(x, y), min2(x, z), min2(y, z))
del2(x, nil) -> nil
del2(x, .2(y, z)) -> if3(=2(x, y), z, .2(y, del2(x, z)))

The set Q consists of the following terms:

msort1(nil)
msort1(.2(x0, x1))
min2(x0, nil)
min2(x0, .2(x1, x2))
del2(x0, nil)
del2(x0, .2(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs with 2 less nodes.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
              ↳ QDP
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

DEL2(x, .2(y, z)) -> DEL2(x, z)

The TRS R consists of the following rules:

msort1(nil) -> nil
msort1(.2(x, y)) -> .2(min2(x, y), msort1(del2(min2(x, y), .2(x, y))))
min2(x, nil) -> x
min2(x, .2(y, z)) -> if3(<=2(x, y), min2(x, z), min2(y, z))
del2(x, nil) -> nil
del2(x, .2(y, z)) -> if3(=2(x, y), z, .2(y, del2(x, z)))

The set Q consists of the following terms:

msort1(nil)
msort1(.2(x0, x1))
min2(x0, nil)
min2(x0, .2(x1, x2))
del2(x0, nil)
del2(x0, .2(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

DEL2(x, .2(y, z)) -> DEL2(x, z)
Used argument filtering: DEL2(x1, x2)  =  x2
.2(x1, x2)  =  .1(x2)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ PisEmptyProof
              ↳ QDP
              ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

msort1(nil) -> nil
msort1(.2(x, y)) -> .2(min2(x, y), msort1(del2(min2(x, y), .2(x, y))))
min2(x, nil) -> x
min2(x, .2(y, z)) -> if3(<=2(x, y), min2(x, z), min2(y, z))
del2(x, nil) -> nil
del2(x, .2(y, z)) -> if3(=2(x, y), z, .2(y, del2(x, z)))

The set Q consists of the following terms:

msort1(nil)
msort1(.2(x0, x1))
min2(x0, nil)
min2(x0, .2(x1, x2))
del2(x0, nil)
del2(x0, .2(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

MIN2(x, .2(y, z)) -> MIN2(y, z)
MIN2(x, .2(y, z)) -> MIN2(x, z)

The TRS R consists of the following rules:

msort1(nil) -> nil
msort1(.2(x, y)) -> .2(min2(x, y), msort1(del2(min2(x, y), .2(x, y))))
min2(x, nil) -> x
min2(x, .2(y, z)) -> if3(<=2(x, y), min2(x, z), min2(y, z))
del2(x, nil) -> nil
del2(x, .2(y, z)) -> if3(=2(x, y), z, .2(y, del2(x, z)))

The set Q consists of the following terms:

msort1(nil)
msort1(.2(x0, x1))
min2(x0, nil)
min2(x0, .2(x1, x2))
del2(x0, nil)
del2(x0, .2(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

MIN2(x, .2(y, z)) -> MIN2(y, z)
MIN2(x, .2(y, z)) -> MIN2(x, z)
Used argument filtering: MIN2(x1, x2)  =  x2
.2(x1, x2)  =  .1(x2)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ PisEmptyProof
              ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

msort1(nil) -> nil
msort1(.2(x, y)) -> .2(min2(x, y), msort1(del2(min2(x, y), .2(x, y))))
min2(x, nil) -> x
min2(x, .2(y, z)) -> if3(<=2(x, y), min2(x, z), min2(y, z))
del2(x, nil) -> nil
del2(x, .2(y, z)) -> if3(=2(x, y), z, .2(y, del2(x, z)))

The set Q consists of the following terms:

msort1(nil)
msort1(.2(x0, x1))
min2(x0, nil)
min2(x0, .2(x1, x2))
del2(x0, nil)
del2(x0, .2(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof

Q DP problem:
The TRS P consists of the following rules:

MSORT1(.2(x, y)) -> MSORT1(del2(min2(x, y), .2(x, y)))

The TRS R consists of the following rules:

msort1(nil) -> nil
msort1(.2(x, y)) -> .2(min2(x, y), msort1(del2(min2(x, y), .2(x, y))))
min2(x, nil) -> x
min2(x, .2(y, z)) -> if3(<=2(x, y), min2(x, z), min2(y, z))
del2(x, nil) -> nil
del2(x, .2(y, z)) -> if3(=2(x, y), z, .2(y, del2(x, z)))

The set Q consists of the following terms:

msort1(nil)
msort1(.2(x0, x1))
min2(x0, nil)
min2(x0, .2(x1, x2))
del2(x0, nil)
del2(x0, .2(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

MSORT1(.2(x, y)) -> MSORT1(del2(min2(x, y), .2(x, y)))
Used argument filtering: MSORT1(x1)  =  x1
.2(x1, x2)  =  .
del2(x1, x2)  =  del
if3(x1, x2, x3)  =  if
Used ordering: Quasi Precedence: . > del > if 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

msort1(nil) -> nil
msort1(.2(x, y)) -> .2(min2(x, y), msort1(del2(min2(x, y), .2(x, y))))
min2(x, nil) -> x
min2(x, .2(y, z)) -> if3(<=2(x, y), min2(x, z), min2(y, z))
del2(x, nil) -> nil
del2(x, .2(y, z)) -> if3(=2(x, y), z, .2(y, del2(x, z)))

The set Q consists of the following terms:

msort1(nil)
msort1(.2(x0, x1))
min2(x0, nil)
min2(x0, .2(x1, x2))
del2(x0, nil)
del2(x0, .2(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.