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

(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)
h  =  h

Recursive path order with status [RPO].
Quasi-Precedence:

s₁ > a₂ > [A₁, h]

Status:

a₂: [2,1]
A₁: multiset
s₁: multiset
h: multiset

The following usable rules [FROCOS05] were oriented: none

(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), s(z)) → 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: Recursive path order with status [RPO].
Quasi-Precedence:

h > [A₃, a₃] > s₁

Status:

a₃: [1,2,3]
A₃: [1,2,3]
s₁: multiset
h: multiset

The following usable rules [FROCOS05] were oriented:

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

(11) 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.

(12) PisEmptyProof (EQUIVALENT transformation)

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

(13) TRUE

(14) 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.

(15) 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(x1, x2)
a(x1, x2, x3)  =  a
h  =  h
s(x1)  =  s

Recursive path order with status [RPO].
Quasi-Precedence:

[SUM₁, cons₂, a, s] > h

Status:

cons₂: multiset
a: multiset
SUM₁: multiset
h: multiset
s: multiset

The following usable rules [FROCOS05] were oriented:

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

(16) 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.

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) TRUE

(19) 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.

(20) 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: Recursive path order with status [RPO].
Quasi-Precedence:

[APP₂, cons₂]

Status:

cons₂: multiset
APP₂: multiset

The following usable rules [FROCOS05] were oriented: none

(21) 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.

(22) PisEmptyProof (EQUIVALENT transformation)

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

(23) TRUE