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

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 PisEmptyProof (⇔)
    ↳11 TRUE
  ↳12 QDP
    ↳13 QDPOrderProof (⇔)
    ↳14 QDP
    ↳15 PisEmptyProof (⇔)
    ↳16 TRUE
  ↳17 QDP
    ↳18 QDPOrderProof (⇔)
    ↳19 QDP
    ↳20 PisEmptyProof (⇔)
    ↳21 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) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) 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)))

The set Q consists of the following terms:

msort(nil)
msort(.(x0, x1))
min(x0, nil)
min(x0, .(x1, x2))
del(x0, nil)
del(x0, .(x1, x2))

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) 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)))

The set Q consists of the following terms:

msort(nil)
msort(.(x0, x1))
min(x0, nil)
min(x0, .(x1, x2))
del(x0, nil)
del(x0, .(x1, x2))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Complex Obligation (AND)

(7) 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)))

The set Q consists of the following terms:

msort(nil)
msort(.(x0, x1))
min(x0, nil)
min(x0, .(x1, x2))
del(x0, nil)
del(x0, .(x1, x2))

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


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.
Used ordering: Combined order from the following AFS and order.
DEL(x1, x2)  =  x2
.(x1, x2)  =  .(x2)
msort(x1)  =  msort(x1)
nil  =  nil
min(x1, x2)  =  min(x1)
del(x1, x2)  =  del
if(x1, x2, x3)  =  if
<=(x1, x2)  =  <=(x1, x2)
=(x1, x2)  =  =

Lexicographic path order with status [LPO].
Precedence:
msort1 > .1 > del > nil > <=2
msort1 > .1 > del > if > <=2
msort1 > min1 > if > <=2
= > <=2

Status:
.1: [1]
msort1: [1]
=: []
if: []
min1: [1]
<=2: [1,2]
nil: []
del: []

The following usable rules [FROCOS05] were oriented:

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)))

(9) Obligation:

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)))

The set Q consists of the following terms:

msort(nil)
msort(.(x0, x1))
min(x0, nil)
min(x0, .(x1, x2))
del(x0, nil)
del(x0, .(x1, x2))

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

(10) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(11) TRUE

(12) 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)))

The set Q consists of the following terms:

msort(nil)
msort(.(x0, x1))
min(x0, nil)
min(x0, .(x1, x2))
del(x0, nil)
del(x0, .(x1, x2))

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MIN(x, .(y, z)) → MIN(y, z)
MIN(x, .(y, z)) → MIN(x, z)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MIN(x1, x2)  =  x2
.(x1, x2)  =  .(x2)
msort(x1)  =  msort(x1)
nil  =  nil
min(x1, x2)  =  min(x1, x2)
del(x1, x2)  =  del
if(x1, x2, x3)  =  if
<=(x1, x2)  =  <=(x1, x2)
=(x1, x2)  =  =

Lexicographic path order with status [LPO].
Precedence:
msort1 > .1 > del > nil > if
msort1 > .1 > <=2 > if
min2 > <=2 > if
= > if

Status:
.1: [1]
msort1: [1]
=: []
if: []
min2: [2,1]
<=2: [1,2]
nil: []
del: []

The following usable rules [FROCOS05] were oriented:

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)))

(14) Obligation:

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)))

The set Q consists of the following terms:

msort(nil)
msort(.(x0, x1))
min(x0, nil)
min(x0, .(x1, x2))
del(x0, nil)
del(x0, .(x1, x2))

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

(15) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(16) TRUE

(17) 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)))

The set Q consists of the following terms:

msort(nil)
msort(.(x0, x1))
min(x0, nil)
min(x0, .(x1, x2))
del(x0, nil)
del(x0, .(x1, x2))

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

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


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.
Used ordering: Combined order from the following AFS and order.
MSORT(x1)  =  x1
.(x1, x2)  =  .(x1)
del(x1, x2)  =  x1
min(x1, x2)  =  min(x1)
msort(x1)  =  msort(x1)
nil  =  nil
if(x1, x2, x3)  =  x1
<=(x1, x2)  =  <=(x1)
=(x1, x2)  =  x1

Lexicographic path order with status [LPO].
Precedence:
msort1 > .1 > min1 > <=1 > nil

Status:
.1: [1]
<=1: [1]
msort1: [1]
min1: [1]
nil: []

The following usable rules [FROCOS05] were oriented:

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)))

(19) Obligation:

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)))

The set Q consists of the following terms:

msort(nil)
msort(.(x0, x1))
min(x0, nil)
min(x0, .(x1, x2))
del(x0, nil)
del(x0, .(x1, x2))

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

(20) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(21) TRUE