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 QDPOrderProof (⇔)
    ↳11 QDP
    ↳12 QDPOrderProof (⇔)
    ↳13 QDP
    ↳14 PisEmptyProof (⇔)
    ↳15 TRUE
  ↳16 QDP
    ↳17 QDPOrderProof (⇔)
    ↳18 QDP
    ↳19 PisEmptyProof (⇔)
    ↳20 TRUE
  ↳21 QDP
    ↳22 QDPOrderProof (⇔)
    ↳23 QDP
    ↳24 PisEmptyProof (⇔)
    ↳25 TRUE

(0) Obligation:

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

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(a(x, y, h), l))
a(h, h, x) → s(x)
a(x, s(y), h) → a(x, y, s(h))
a(x, s(y), s(z)) → a(x, y, a(x, s(y), z))
a(s(x), h, z) → a(x, z, 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(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(a(x, y, h), l))
a(h, h, x) → s(x)
a(x, s(y), h) → a(x, y, s(h))
a(x, s(y), s(z)) → a(x, y, a(x, s(y), z))
a(s(x), h, z) → a(x, z, z)

The set Q consists of the following terms:

app(nil, x0)
app(x0, nil)
app(cons(x0, x1), x2)
sum(cons(x0, nil))
sum(cons(x0, cons(x1, x2)))
a(h, h, x0)
a(x0, s(x1), h)
a(x0, s(x1), s(x2))
a(s(x0), h, x1)

(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(cons(x, l), k) → APP(l, k)
SUM(cons(x, cons(y, l))) → SUM(cons(a(x, y, h), l))
SUM(cons(x, cons(y, l))) → A(x, y, h)
A(x, s(y), h) → A(x, y, s(h))
A(x, s(y), s(z)) → A(x, y, a(x, s(y), z))
A(x, s(y), s(z)) → A(x, s(y), z)
A(s(x), h, z) → A(x, z, z)

The TRS R consists of the following rules:

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(a(x, y, h), l))
a(h, h, x) → s(x)
a(x, s(y), h) → a(x, y, s(h))
a(x, s(y), s(z)) → a(x, y, a(x, s(y), z))
a(s(x), h, z) → a(x, z, z)

The set Q consists of the following terms:

app(nil, x0)
app(x0, nil)
app(cons(x0, x1), x2)
sum(cons(x0, nil))
sum(cons(x0, cons(x1, x2)))
a(h, h, x0)
a(x0, s(x1), h)
a(x0, s(x1), s(x2))
a(s(x0), h, x1)

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 1 less node.

(6) Complex Obligation (AND)

(7) Obligation:

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

A(x, s(y), s(z)) → A(x, y, a(x, s(y), z))
A(x, s(y), s(z)) → A(x, s(y), z)
A(x, s(y), h) → A(x, y, s(h))
A(s(x), h, z) → A(x, z, z)

The TRS R consists of the following rules:

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(a(x, y, h), l))
a(h, h, x) → s(x)
a(x, s(y), h) → a(x, y, s(h))
a(x, s(y), s(z)) → a(x, y, a(x, s(y), z))
a(s(x), h, z) → a(x, z, z)

The set Q consists of the following terms:

app(nil, x0)
app(x0, nil)
app(cons(x0, x1), x2)
sum(cons(x0, nil))
sum(cons(x0, cons(x1, x2)))
a(h, h, x0)
a(x0, s(x1), h)
a(x0, s(x1), s(x2))
a(s(x0), h, x1)

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.


A(s(x), h, z) → A(x, z, z)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
A(x1, x2, x3)  =  A(x1)
s(x1)  =  s(x1)
a(x1, x2, x3)  =  a(x1, x2, x3)
h  =  h
app(x1, x2)  =  app(x1, x2)
nil  =  nil
cons(x1, x2)  =  x2
sum(x1)  =  sum(x1)

Lexicographic path order with status [LPO].
Precedence:
a3 > s1 > h > A1
app2 > A1
nil > A1
sum1 > h > A1

Status:
A1: [1]
s1: [1]
a3: [1,2,3]
h: []
app2: [1,2]
nil: []
sum1: [1]

The following usable rules [FROCOS05] were oriented:

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(a(x, y, h), l))
a(h, h, x) → s(x)
a(x, s(y), h) → a(x, y, s(h))
a(x, s(y), s(z)) → a(x, y, a(x, s(y), z))
a(s(x), h, z) → a(x, z, z)

(9) Obligation:

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

A(x, s(y), s(z)) → A(x, y, a(x, s(y), z))
A(x, s(y), s(z)) → A(x, s(y), z)
A(x, s(y), h) → A(x, y, s(h))

The TRS R consists of the following rules:

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(a(x, y, h), l))
a(h, h, x) → s(x)
a(x, s(y), h) → a(x, y, s(h))
a(x, s(y), s(z)) → a(x, y, a(x, s(y), z))
a(s(x), h, z) → a(x, z, z)

The set Q consists of the following terms:

app(nil, x0)
app(x0, nil)
app(cons(x0, x1), x2)
sum(cons(x0, nil))
sum(cons(x0, cons(x1, x2)))
a(h, h, x0)
a(x0, s(x1), h)
a(x0, s(x1), s(x2))
a(s(x0), h, x1)

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

(10) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


A(x, s(y), s(z)) → A(x, y, a(x, s(y), z))
A(x, s(y), h) → A(x, y, s(h))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
A(x1, x2, x3)  =  A(x1, x2)
s(x1)  =  s(x1)
a(x1, x2, x3)  =  a(x1, x2, x3)
h  =  h
app(x1, x2)  =  app(x1, x2)
nil  =  nil
cons(x1, x2)  =  x2
sum(x1)  =  sum(x1)

Lexicographic path order with status [LPO].
Precedence:
a3 > s1 > h > A2
app2 > A2
nil > A2
sum1 > h > A2

Status:
A2: [2,1]
s1: [1]
a3: [1,2,3]
h: []
app2: [1,2]
nil: []
sum1: [1]

The following usable rules [FROCOS05] were oriented:

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(a(x, y, h), l))
a(h, h, x) → s(x)
a(x, s(y), h) → a(x, y, s(h))
a(x, s(y), s(z)) → a(x, y, a(x, s(y), z))
a(s(x), h, z) → a(x, z, z)

(11) Obligation:

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

A(x, s(y), s(z)) → A(x, s(y), z)

The TRS R consists of the following rules:

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(a(x, y, h), l))
a(h, h, x) → s(x)
a(x, s(y), h) → a(x, y, s(h))
a(x, s(y), s(z)) → a(x, y, a(x, s(y), z))
a(s(x), h, z) → a(x, z, z)

The set Q consists of the following terms:

app(nil, x0)
app(x0, nil)
app(cons(x0, x1), x2)
sum(cons(x0, nil))
sum(cons(x0, cons(x1, x2)))
a(h, h, x0)
a(x0, s(x1), h)
a(x0, s(x1), s(x2))
a(s(x0), h, x1)

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

(12) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


A(x, s(y), s(z)) → A(x, s(y), z)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
A(x1, x2, x3)  =  A(x1, x2, x3)
s(x1)  =  s(x1)
app(x1, x2)  =  app(x1, x2)
nil  =  nil
cons(x1, x2)  =  x2
sum(x1)  =  sum(x1)
a(x1, x2, x3)  =  a(x1, x2, x3)
h  =  h

Lexicographic path order with status [LPO].
Precedence:
app2 > A3
nil > A3
sum1 > h > A3
a3 > s1 > h > A3

Status:
A3: [2,3,1]
s1: [1]
app2: [1,2]
nil: []
sum1: [1]
a3: [1,2,3]
h: []

The following usable rules [FROCOS05] were oriented:

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(a(x, y, h), l))
a(h, h, x) → s(x)
a(x, s(y), h) → a(x, y, s(h))
a(x, s(y), s(z)) → a(x, y, a(x, s(y), z))
a(s(x), h, z) → a(x, z, z)

(13) Obligation:

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

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(a(x, y, h), l))
a(h, h, x) → s(x)
a(x, s(y), h) → a(x, y, s(h))
a(x, s(y), s(z)) → a(x, y, a(x, s(y), z))
a(s(x), h, z) → a(x, z, z)

The set Q consists of the following terms:

app(nil, x0)
app(x0, nil)
app(cons(x0, x1), x2)
sum(cons(x0, nil))
sum(cons(x0, cons(x1, x2)))
a(h, h, x0)
a(x0, s(x1), h)
a(x0, s(x1), s(x2))
a(s(x0), h, x1)

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

(14) PisEmptyProof (EQUIVALENT transformation)

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

(15) TRUE

(16) Obligation:

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

SUM(cons(x, cons(y, l))) → SUM(cons(a(x, y, h), l))

The TRS R consists of the following rules:

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(a(x, y, h), l))
a(h, h, x) → s(x)
a(x, s(y), h) → a(x, y, s(h))
a(x, s(y), s(z)) → a(x, y, a(x, s(y), z))
a(s(x), h, z) → a(x, z, z)

The set Q consists of the following terms:

app(nil, x0)
app(x0, nil)
app(cons(x0, x1), x2)
sum(cons(x0, nil))
sum(cons(x0, cons(x1, x2)))
a(h, h, x0)
a(x0, s(x1), h)
a(x0, s(x1), s(x2))
a(s(x0), h, x1)

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

(17) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


SUM(cons(x, cons(y, l))) → SUM(cons(a(x, y, h), l))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SUM(x1)  =  SUM(x1)
cons(x1, x2)  =  cons(x2)
a(x1, x2, x3)  =  x3
h  =  h
app(x1, x2)  =  app(x1, x2)
nil  =  nil
sum(x1)  =  sum(x1)
s(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:
SUM1 > cons1
SUM1 > h
app2 > cons1
sum1 > cons1
sum1 > h
sum1 > nil

Status:
SUM1: [1]
cons1: [1]
h: []
app2: [1,2]
nil: []
sum1: [1]

The following usable rules [FROCOS05] were oriented:

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(a(x, y, h), l))
a(h, h, x) → s(x)
a(x, s(y), h) → a(x, y, s(h))
a(x, s(y), s(z)) → a(x, y, a(x, s(y), z))
a(s(x), h, z) → a(x, z, z)

(18) Obligation:

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

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(a(x, y, h), l))
a(h, h, x) → s(x)
a(x, s(y), h) → a(x, y, s(h))
a(x, s(y), s(z)) → a(x, y, a(x, s(y), z))
a(s(x), h, z) → a(x, z, z)

The set Q consists of the following terms:

app(nil, x0)
app(x0, nil)
app(cons(x0, x1), x2)
sum(cons(x0, nil))
sum(cons(x0, cons(x1, x2)))
a(h, h, x0)
a(x0, s(x1), h)
a(x0, s(x1), s(x2))
a(s(x0), h, x1)

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

(19) PisEmptyProof (EQUIVALENT transformation)

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

(20) TRUE

(21) Obligation:

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

APP(cons(x, l), k) → APP(l, k)

The TRS R consists of the following rules:

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(a(x, y, h), l))
a(h, h, x) → s(x)
a(x, s(y), h) → a(x, y, s(h))
a(x, s(y), s(z)) → a(x, y, a(x, s(y), z))
a(s(x), h, z) → a(x, z, z)

The set Q consists of the following terms:

app(nil, x0)
app(x0, nil)
app(cons(x0, x1), x2)
sum(cons(x0, nil))
sum(cons(x0, cons(x1, x2)))
a(h, h, x0)
a(x0, s(x1), h)
a(x0, s(x1), s(x2))
a(s(x0), h, x1)

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

(22) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


APP(cons(x, l), k) → APP(l, k)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  APP(x1, x2)
cons(x1, x2)  =  cons(x2)
app(x1, x2)  =  app(x1, x2)
nil  =  nil
sum(x1)  =  sum(x1)
a(x1, x2, x3)  =  x3
h  =  h
s(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:
app2 > cons1 > APP2
nil > APP2
sum1 > cons1 > APP2
sum1 > h > APP2

Status:
APP2: [1,2]
cons1: [1]
app2: [1,2]
nil: []
sum1: [1]
h: []

The following usable rules [FROCOS05] were oriented:

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(a(x, y, h), l))
a(h, h, x) → s(x)
a(x, s(y), h) → a(x, y, s(h))
a(x, s(y), s(z)) → a(x, y, a(x, s(y), z))
a(s(x), h, z) → a(x, z, z)

(23) Obligation:

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

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(a(x, y, h), l))
a(h, h, x) → s(x)
a(x, s(y), h) → a(x, y, s(h))
a(x, s(y), s(z)) → a(x, y, a(x, s(y), z))
a(s(x), h, z) → a(x, z, z)

The set Q consists of the following terms:

app(nil, x0)
app(x0, nil)
app(cons(x0, x1), x2)
sum(cons(x0, nil))
sum(cons(x0, cons(x1, x2)))
a(h, h, x0)
a(x0, s(x1), h)
a(x0, s(x1), s(x2))
a(s(x0), h, x1)

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

(24) PisEmptyProof (EQUIVALENT transformation)

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

(25) TRUE