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))


Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 3 SCCs with 2 less nodes.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
              ↳ 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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


DEL2(x, .2(y, z)) -> DEL2(x, z)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
DEL2(x1, x2)  =  DEL1(x2)
.2(x1, x2)  =  .1(x2)

Lexicographic Path Order [19].
Precedence:
[DEL1, .1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ 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
                ↳ QDPOrderProof
              ↳ 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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


MIN2(x, .2(y, z)) -> MIN2(y, z)
MIN2(x, .2(y, z)) -> MIN2(x, z)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MIN2(x1, x2)  =  MIN1(x2)
.2(x1, x2)  =  .2(x1, x2)

Lexicographic Path Order [19].
Precedence:
.2 > MIN1


The following usable rules [14] were oriented: none



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

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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


MSORT1(.2(x, y)) -> MSORT1(del2(min2(x, y), .2(x, y)))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MSORT1(x1)  =  MSORT1(x1)
.2(x1, x2)  =  .2(x1, x2)
del2(x1, x2)  =  x1
min2(x1, x2)  =  min1(x1)
nil  =  nil
if3(x1, x2, x3)  =  x1
<=2(x1, x2)  =  <=1(x1)
=2(x1, x2)  =  x1

Lexicographic Path Order [19].
Precedence:
[.2, min1, <=1] > MSORT1 > nil


The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ 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.