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:

msort(nil) → nil
msort(.(x, y)) → .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) → x
min(x, .(y, z)) → if(<=(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, .(y, z)) → if(=(x, y), z, .(y, del(x, z)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

msort(nil) → nil
msort(.(x, y)) → .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) → x
min(x, .(y, z)) → if(<=(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, .(y, z)) → if(=(x, y), z, .(y, del(x, z)))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

MSORT(.(x, y)) → MIN(x, y)
MIN(x, .(y, z)) → MIN(x, z)
MSORT(.(x, y)) → MSORT(del(min(x, y), .(x, y)))
DEL(x, .(y, z)) → DEL(x, z)
MSORT(.(x, y)) → DEL(min(x, y), .(x, y))
MIN(x, .(y, z)) → MIN(y, z)

The TRS R consists of the following rules:

msort(nil) → nil
msort(.(x, y)) → .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) → x
min(x, .(y, z)) → if(<=(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, .(y, z)) → if(=(x, y), z, .(y, del(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:

MSORT(.(x, y)) → MIN(x, y)
MIN(x, .(y, z)) → MIN(x, z)
MSORT(.(x, y)) → MSORT(del(min(x, y), .(x, y)))
DEL(x, .(y, z)) → DEL(x, z)
MSORT(.(x, y)) → DEL(min(x, y), .(x, y))
MIN(x, .(y, z)) → MIN(y, z)

The TRS R consists of the following rules:

msort(nil) → nil
msort(.(x, y)) → .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) → x
min(x, .(y, z)) → if(<=(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, .(y, z)) → if(=(x, y), z, .(y, del(x, z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] 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:

DEL(x, .(y, z)) → DEL(x, z)

The TRS R consists of the following rules:

msort(nil) → nil
msort(.(x, y)) → .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) → x
min(x, .(y, z)) → if(<=(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, .(y, z)) → if(=(x, y), z, .(y, del(x, z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


DEL(x, .(y, z)) → DEL(x, z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(DEL(x1, x2)) = (2)x_2   
POL(.(x1, x2)) = 1/4 + (7/2)x_2   
The value of delta used in the strict ordering is 1/2.
The following usable rules [17] 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:

msort(nil) → nil
msort(.(x, y)) → .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) → x
min(x, .(y, z)) → if(<=(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, .(y, z)) → if(=(x, y), z, .(y, del(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:

MIN(x, .(y, z)) → MIN(x, z)
MIN(x, .(y, z)) → MIN(y, z)

The TRS R consists of the following rules:

msort(nil) → nil
msort(.(x, y)) → .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) → x
min(x, .(y, z)) → if(<=(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, .(y, z)) → if(=(x, y), z, .(y, del(x, z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


MIN(x, .(y, z)) → MIN(x, z)
MIN(x, .(y, z)) → MIN(y, z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(MIN(x1, x2)) = (15/4)x_1 + (2)x_2   
POL(.(x1, x2)) = 1/4 + (4)x_1 + (5/2)x_2   
The value of delta used in the strict ordering is 1/2.
The following usable rules [17] 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:

msort(nil) → nil
msort(.(x, y)) → .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) → x
min(x, .(y, z)) → if(<=(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, .(y, z)) → if(=(x, y), z, .(y, del(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:

MSORT(.(x, y)) → MSORT(del(min(x, y), .(x, y)))

The TRS R consists of the following rules:

msort(nil) → nil
msort(.(x, y)) → .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) → x
min(x, .(y, z)) → if(<=(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, .(y, z)) → if(=(x, y), z, .(y, del(x, z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


MSORT(.(x, y)) → MSORT(del(min(x, y), .(x, y)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(if(x1, x2, x3)) = 0   
POL(=(x1, x2)) = 4 + (4)x_1 + (2)x_2   
POL(MSORT(x1)) = (4)x_1   
POL(.(x1, x2)) = 1/4 + (1/4)x_1   
POL(min(x1, x2)) = 15/4 + (5/2)x_2   
POL(del(x1, x2)) = (1/4)x_2   
POL(nil) = 7/4   
POL(<=(x1, x2)) = 1/4 + (2)x_1 + (7/2)x_2   
The value of delta used in the strict ordering is 3/4.
The following usable rules [17] were oriented:

del(x, .(y, z)) → if(=(x, y), z, .(y, del(x, 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:

msort(nil) → nil
msort(.(x, y)) → .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) → x
min(x, .(y, z)) → if(<=(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, .(y, z)) → if(=(x, y), z, .(y, del(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.