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
  ↳ 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)))

Q is empty.

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)))

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

↳ 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)))

Q is empty.
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
  ↳ 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)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


DEL2(x, .2(y, z)) -> DEL2(x, z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(.2(x1, x2)) = 1 + x2   
POL(DEL2(x1, x2)) = x2   

The following usable rules [14] were oriented: none



↳ 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)))

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

↳ 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)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly 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 be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(.2(x1, x2)) = 1 + x1 + 2·x2   
POL(MIN2(x1, x2)) = 2·x1 + 3·x2   

The following usable rules [14] were oriented: none



↳ 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)))

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

↳ 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)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


MSORT1(.2(x, y)) -> MSORT1(del2(min2(x, y), .2(x, y)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(.2(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(<=2(x1, x2)) = x2   
POL(=2(x1, x2)) = 0   
POL(MSORT1(x1)) = x1   
POL(del2(x1, x2)) = 1 + x1   
POL(if3(x1, x2, x3)) = 1 + x1   
POL(min2(x1, x2)) = x1 + x2   
POL(nil) = 0   

The following usable rules [14] were oriented:

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



↳ 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)))

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