Termination w.r.t. Q of the given QTRS could not be shown:

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

(0) Obligation:

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

app(app(app(sort, f), g), nil) → nil
app(app(app(sort, f), g), app(app(cons, x), y)) → app(app(app(app(insert, f), g), app(app(app(sort, f), g), y)), x)
app(app(app(app(insert, f), g), nil), y) → app(app(cons, y), nil)
app(app(app(app(insert, f), g), app(app(cons, x), z)), y) → app(app(cons, app(app(f, x), y)), app(app(app(app(insert, f), g), z), app(app(g, x), y)))
app(app(max, 0), y) → y
app(app(max, x), 0) → x
app(app(max, app(s, x)), app(s, y)) → app(app(max, x), y)
app(app(min, 0), y) → 0
app(app(min, x), 0) → 0
app(app(min, app(s, x)), app(s, y)) → app(app(min, x), y)
app(asort, z) → app(app(app(sort, min), max), z)
app(dsort, z) → app(app(app(sort, max), min), 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:

app(app(app(sort, f), g), nil) → nil
app(app(app(sort, f), g), app(app(cons, x), y)) → app(app(app(app(insert, f), g), app(app(app(sort, f), g), y)), x)
app(app(app(app(insert, f), g), nil), y) → app(app(cons, y), nil)
app(app(app(app(insert, f), g), app(app(cons, x), z)), y) → app(app(cons, app(app(f, x), y)), app(app(app(app(insert, f), g), z), app(app(g, x), y)))
app(app(max, 0), y) → y
app(app(max, x), 0) → x
app(app(max, app(s, x)), app(s, y)) → app(app(max, x), y)
app(app(min, 0), y) → 0
app(app(min, x), 0) → 0
app(app(min, app(s, x)), app(s, y)) → app(app(min, x), y)
app(asort, z) → app(app(app(sort, min), max), z)
app(dsort, z) → app(app(app(sort, max), min), z)

The set Q consists of the following terms:

app(app(app(sort, x0), x1), nil)
app(app(app(sort, x0), x1), app(app(cons, x2), x3))
app(app(app(app(insert, x0), x1), nil), x2)
app(app(app(app(insert, x0), x1), app(app(cons, x2), x3)), x4)
app(app(max, 0), x0)
app(app(max, x0), 0)
app(app(max, app(s, x0)), app(s, x1))
app(app(min, 0), x0)
app(app(min, x0), 0)
app(app(min, app(s, x0)), app(s, x1))
app(asort, x0)
app(dsort, x0)

(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:

APP(app(app(sort, f), g), app(app(cons, x), y)) → APP(app(app(app(insert, f), g), app(app(app(sort, f), g), y)), x)
APP(app(app(sort, f), g), app(app(cons, x), y)) → APP(app(app(insert, f), g), app(app(app(sort, f), g), y))
APP(app(app(sort, f), g), app(app(cons, x), y)) → APP(app(insert, f), g)
APP(app(app(sort, f), g), app(app(cons, x), y)) → APP(insert, f)
APP(app(app(sort, f), g), app(app(cons, x), y)) → APP(app(app(sort, f), g), y)
APP(app(app(app(insert, f), g), nil), y) → APP(app(cons, y), nil)
APP(app(app(app(insert, f), g), nil), y) → APP(cons, y)
APP(app(app(app(insert, f), g), app(app(cons, x), z)), y) → APP(app(cons, app(app(f, x), y)), app(app(app(app(insert, f), g), z), app(app(g, x), y)))
APP(app(app(app(insert, f), g), app(app(cons, x), z)), y) → APP(cons, app(app(f, x), y))
APP(app(app(app(insert, f), g), app(app(cons, x), z)), y) → APP(app(f, x), y)
APP(app(app(app(insert, f), g), app(app(cons, x), z)), y) → APP(f, x)
APP(app(app(app(insert, f), g), app(app(cons, x), z)), y) → APP(app(app(app(insert, f), g), z), app(app(g, x), y))
APP(app(app(app(insert, f), g), app(app(cons, x), z)), y) → APP(app(app(insert, f), g), z)
APP(app(app(app(insert, f), g), app(app(cons, x), z)), y) → APP(app(g, x), y)
APP(app(app(app(insert, f), g), app(app(cons, x), z)), y) → APP(g, x)
APP(app(max, app(s, x)), app(s, y)) → APP(app(max, x), y)
APP(app(max, app(s, x)), app(s, y)) → APP(max, x)
APP(app(min, app(s, x)), app(s, y)) → APP(app(min, x), y)
APP(app(min, app(s, x)), app(s, y)) → APP(min, x)
APP(asort, z) → APP(app(app(sort, min), max), z)
APP(asort, z) → APP(app(sort, min), max)
APP(asort, z) → APP(sort, min)
APP(dsort, z) → APP(app(app(sort, max), min), z)
APP(dsort, z) → APP(app(sort, max), min)
APP(dsort, z) → APP(sort, max)

The TRS R consists of the following rules:

app(app(app(sort, f), g), nil) → nil
app(app(app(sort, f), g), app(app(cons, x), y)) → app(app(app(app(insert, f), g), app(app(app(sort, f), g), y)), x)
app(app(app(app(insert, f), g), nil), y) → app(app(cons, y), nil)
app(app(app(app(insert, f), g), app(app(cons, x), z)), y) → app(app(cons, app(app(f, x), y)), app(app(app(app(insert, f), g), z), app(app(g, x), y)))
app(app(max, 0), y) → y
app(app(max, x), 0) → x
app(app(max, app(s, x)), app(s, y)) → app(app(max, x), y)
app(app(min, 0), y) → 0
app(app(min, x), 0) → 0
app(app(min, app(s, x)), app(s, y)) → app(app(min, x), y)
app(asort, z) → app(app(app(sort, min), max), z)
app(dsort, z) → app(app(app(sort, max), min), z)

The set Q consists of the following terms:

app(app(app(sort, x0), x1), nil)
app(app(app(sort, x0), x1), app(app(cons, x2), x3))
app(app(app(app(insert, x0), x1), nil), x2)
app(app(app(app(insert, x0), x1), app(app(cons, x2), x3)), x4)
app(app(max, 0), x0)
app(app(max, x0), 0)
app(app(max, app(s, x0)), app(s, x1))
app(app(min, 0), x0)
app(app(min, x0), 0)
app(app(min, app(s, x0)), app(s, x1))
app(asort, x0)
app(dsort, x0)

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 14 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

APP(app(min, app(s, x)), app(s, y)) → APP(app(min, x), y)

The TRS R consists of the following rules:

app(app(app(sort, f), g), nil) → nil
app(app(app(sort, f), g), app(app(cons, x), y)) → app(app(app(app(insert, f), g), app(app(app(sort, f), g), y)), x)
app(app(app(app(insert, f), g), nil), y) → app(app(cons, y), nil)
app(app(app(app(insert, f), g), app(app(cons, x), z)), y) → app(app(cons, app(app(f, x), y)), app(app(app(app(insert, f), g), z), app(app(g, x), y)))
app(app(max, 0), y) → y
app(app(max, x), 0) → x
app(app(max, app(s, x)), app(s, y)) → app(app(max, x), y)
app(app(min, 0), y) → 0
app(app(min, x), 0) → 0
app(app(min, app(s, x)), app(s, y)) → app(app(min, x), y)
app(asort, z) → app(app(app(sort, min), max), z)
app(dsort, z) → app(app(app(sort, max), min), z)

The set Q consists of the following terms:

app(app(app(sort, x0), x1), nil)
app(app(app(sort, x0), x1), app(app(cons, x2), x3))
app(app(app(app(insert, x0), x1), nil), x2)
app(app(app(app(insert, x0), x1), app(app(cons, x2), x3)), x4)
app(app(max, 0), x0)
app(app(max, x0), 0)
app(app(max, app(s, x0)), app(s, x1))
app(app(min, 0), x0)
app(app(min, x0), 0)
app(app(min, app(s, x0)), app(s, x1))
app(asort, x0)
app(dsort, x0)

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04]. Here, we combined the reduction pair processor with the A-transformation [FROCOS05] which results in the following intermediate Q-DP Problem.
The a-transformed P is

min1(s(x), s(y)) → min1(x, y)

The a-transformed usable rules are
none


The following pairs can be oriented strictly and are deleted.


APP(app(min, app(s, x)), app(s, y)) → APP(app(min, x), y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
min1(x1, x2)  =  x1
s(x1)  =  s(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:
trivial

Status:
s1: [1]


The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

app(app(app(sort, f), g), nil) → nil
app(app(app(sort, f), g), app(app(cons, x), y)) → app(app(app(app(insert, f), g), app(app(app(sort, f), g), y)), x)
app(app(app(app(insert, f), g), nil), y) → app(app(cons, y), nil)
app(app(app(app(insert, f), g), app(app(cons, x), z)), y) → app(app(cons, app(app(f, x), y)), app(app(app(app(insert, f), g), z), app(app(g, x), y)))
app(app(max, 0), y) → y
app(app(max, x), 0) → x
app(app(max, app(s, x)), app(s, y)) → app(app(max, x), y)
app(app(min, 0), y) → 0
app(app(min, x), 0) → 0
app(app(min, app(s, x)), app(s, y)) → app(app(min, x), y)
app(asort, z) → app(app(app(sort, min), max), z)
app(dsort, z) → app(app(app(sort, max), min), z)

The set Q consists of the following terms:

app(app(app(sort, x0), x1), nil)
app(app(app(sort, x0), x1), app(app(cons, x2), x3))
app(app(app(app(insert, x0), x1), nil), x2)
app(app(app(app(insert, x0), x1), app(app(cons, x2), x3)), x4)
app(app(max, 0), x0)
app(app(max, x0), 0)
app(app(max, app(s, x0)), app(s, x1))
app(app(min, 0), x0)
app(app(min, x0), 0)
app(app(min, app(s, x0)), app(s, x1))
app(asort, x0)
app(dsort, x0)

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:

APP(app(max, app(s, x)), app(s, y)) → APP(app(max, x), y)

The TRS R consists of the following rules:

app(app(app(sort, f), g), nil) → nil
app(app(app(sort, f), g), app(app(cons, x), y)) → app(app(app(app(insert, f), g), app(app(app(sort, f), g), y)), x)
app(app(app(app(insert, f), g), nil), y) → app(app(cons, y), nil)
app(app(app(app(insert, f), g), app(app(cons, x), z)), y) → app(app(cons, app(app(f, x), y)), app(app(app(app(insert, f), g), z), app(app(g, x), y)))
app(app(max, 0), y) → y
app(app(max, x), 0) → x
app(app(max, app(s, x)), app(s, y)) → app(app(max, x), y)
app(app(min, 0), y) → 0
app(app(min, x), 0) → 0
app(app(min, app(s, x)), app(s, y)) → app(app(min, x), y)
app(asort, z) → app(app(app(sort, min), max), z)
app(dsort, z) → app(app(app(sort, max), min), z)

The set Q consists of the following terms:

app(app(app(sort, x0), x1), nil)
app(app(app(sort, x0), x1), app(app(cons, x2), x3))
app(app(app(app(insert, x0), x1), nil), x2)
app(app(app(app(insert, x0), x1), app(app(cons, x2), x3)), x4)
app(app(max, 0), x0)
app(app(max, x0), 0)
app(app(max, app(s, x0)), app(s, x1))
app(app(min, 0), x0)
app(app(min, x0), 0)
app(app(min, app(s, x0)), app(s, x1))
app(asort, x0)
app(dsort, x0)

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04]. Here, we combined the reduction pair processor with the A-transformation [FROCOS05] which results in the following intermediate Q-DP Problem.
The a-transformed P is

max1(s(x), s(y)) → max1(x, y)

The a-transformed usable rules are
none


The following pairs can be oriented strictly and are deleted.


APP(app(max, app(s, x)), app(s, y)) → APP(app(max, x), y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
max1(x1, x2)  =  x1
s(x1)  =  s(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:
trivial

Status:
s1: [1]


The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

app(app(app(sort, f), g), nil) → nil
app(app(app(sort, f), g), app(app(cons, x), y)) → app(app(app(app(insert, f), g), app(app(app(sort, f), g), y)), x)
app(app(app(app(insert, f), g), nil), y) → app(app(cons, y), nil)
app(app(app(app(insert, f), g), app(app(cons, x), z)), y) → app(app(cons, app(app(f, x), y)), app(app(app(app(insert, f), g), z), app(app(g, x), y)))
app(app(max, 0), y) → y
app(app(max, x), 0) → x
app(app(max, app(s, x)), app(s, y)) → app(app(max, x), y)
app(app(min, 0), y) → 0
app(app(min, x), 0) → 0
app(app(min, app(s, x)), app(s, y)) → app(app(min, x), y)
app(asort, z) → app(app(app(sort, min), max), z)
app(dsort, z) → app(app(app(sort, max), min), z)

The set Q consists of the following terms:

app(app(app(sort, x0), x1), nil)
app(app(app(sort, x0), x1), app(app(cons, x2), x3))
app(app(app(app(insert, x0), x1), nil), x2)
app(app(app(app(insert, x0), x1), app(app(cons, x2), x3)), x4)
app(app(max, 0), x0)
app(app(max, x0), 0)
app(app(max, app(s, x0)), app(s, x1))
app(app(min, 0), x0)
app(app(min, x0), 0)
app(app(min, app(s, x0)), app(s, x1))
app(asort, x0)
app(dsort, x0)

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:

APP(app(app(app(insert, f), g), app(app(cons, x), z)), y) → APP(app(f, x), y)
APP(app(app(sort, f), g), app(app(cons, x), y)) → APP(app(app(app(insert, f), g), app(app(app(sort, f), g), y)), x)
APP(app(app(app(insert, f), g), app(app(cons, x), z)), y) → APP(f, x)
APP(app(app(sort, f), g), app(app(cons, x), y)) → APP(app(app(sort, f), g), y)
APP(app(app(app(insert, f), g), app(app(cons, x), z)), y) → APP(app(app(app(insert, f), g), z), app(app(g, x), y))
APP(app(app(app(insert, f), g), app(app(cons, x), z)), y) → APP(app(g, x), y)
APP(app(app(app(insert, f), g), app(app(cons, x), z)), y) → APP(g, x)
APP(asort, z) → APP(app(app(sort, min), max), z)
APP(dsort, z) → APP(app(app(sort, max), min), z)

The TRS R consists of the following rules:

app(app(app(sort, f), g), nil) → nil
app(app(app(sort, f), g), app(app(cons, x), y)) → app(app(app(app(insert, f), g), app(app(app(sort, f), g), y)), x)
app(app(app(app(insert, f), g), nil), y) → app(app(cons, y), nil)
app(app(app(app(insert, f), g), app(app(cons, x), z)), y) → app(app(cons, app(app(f, x), y)), app(app(app(app(insert, f), g), z), app(app(g, x), y)))
app(app(max, 0), y) → y
app(app(max, x), 0) → x
app(app(max, app(s, x)), app(s, y)) → app(app(max, x), y)
app(app(min, 0), y) → 0
app(app(min, x), 0) → 0
app(app(min, app(s, x)), app(s, y)) → app(app(min, x), y)
app(asort, z) → app(app(app(sort, min), max), z)
app(dsort, z) → app(app(app(sort, max), min), z)

The set Q consists of the following terms:

app(app(app(sort, x0), x1), nil)
app(app(app(sort, x0), x1), app(app(cons, x2), x3))
app(app(app(app(insert, x0), x1), nil), x2)
app(app(app(app(insert, x0), x1), app(app(cons, x2), x3)), x4)
app(app(max, 0), x0)
app(app(max, x0), 0)
app(app(max, app(s, x0)), app(s, x1))
app(app(min, 0), x0)
app(app(min, x0), 0)
app(app(min, app(s, x0)), app(s, x1))
app(asort, x0)
app(dsort, x0)

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