Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
    ↳5 QDP
        ↳6 QDPSizeChangeProof (⇔)
        ↳7 TRUE
    ↳8 QDP
        ↳9 QDPSizeChangeProof (⇔)
        ↳10 TRUE
    ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔)
        ↳13 TRUE

(0) Obligation:

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.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 2 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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.

(6) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
.(x1, x2)  =  .(x2)

From the DPs we obtained the following set of size-change graphs:

  • DEL(x, .(y, z)) → DEL(x, z) (allowed arguments on rhs = {1, 2})
    The graph contains the following edges 1 >= 1, 2 > 2

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(7) TRUE

(8) Obligation:

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

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

(9) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
.(x1, x2)  =  .(x2)

From the DPs we obtained the following set of size-change graphs:

  • MIN(x, .(y, z)) → MIN(y, z) (allowed arguments on rhs = {1, 2})
    The graph contains the following edges 2 > 2

  • MIN(x, .(y, z)) → MIN(x, z) (allowed arguments on rhs = {1, 2})
    The graph contains the following edges 1 >= 1, 2 > 2

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(10) TRUE

(11) Obligation:

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.

(12) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Polynomial interpretation [POLO]:


POL(.(x1, x2)) = 1 + x1   
POL(<=(x1, x2)) = x1   
POL(=(x1, x2)) = 0   
POL(del(x1, x2)) = x1   
POL(if(x1, x2, x3)) = x1   
POL(min(x1, x2)) = x1   
POL(nil) = 0   

From the DPs we obtained the following set of size-change graphs:

  • MSORT(.(x, y)) → MSORT(del(min(x, y), .(x, y))) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 > 1

We oriented the following set of usable rules [AAECC05,FROCOS05].


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

(13) TRUE