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, x3)
h  =  h
app(x1, x2)  =  app(x1, x2)
nil  =  nil
cons(x1, x2)  =  cons
sum(x1)  =  sum(x1)

Recursive path order with status [RPO].
Precedence:

app₂ > h
nil > h
cons > a₃ > s₁ > A₁ > h
cons > sum₁ > h

Status:

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

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)  =  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)  =  cons
sum(x1)  =  sum(x1)

Recursive path order with status [RPO].
Precedence:

app₂ > cons > a₃ > s₁ > h
app₂ > cons > nil > h
app₂ > cons > sum₁ > h

Status:

s₁: multiset
a₃: [1,2,3]
h: multiset
app₂: multiset
nil: multiset
cons: []
sum₁: multiset

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)  =  cons
sum(x1)  =  sum(x1)
a(x1, x2, x3)  =  a(x1, x2, x3)
h  =  h

Recursive path order with status [RPO].
Precedence:

A₃ > s₁ > app₂
cons > sum₁ > nil > app₂
a₃ > s₁ > app₂
h > s₁ > app₂

Status:

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

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(x1, x2)
a(x1, x2, x3)  =  x3
h  =  h
app(x1, x2)  =  app(x1, x2)
nil  =  nil
sum(x1)  =  sum(x1)
s(x1)  =  x1

Recursive path order with status [RPO].
Precedence:

app₂ > cons₂ > SUM₁ > h
sum₁ > cons₂ > SUM₁ > h
sum₁ > nil > h

Status:

SUM₁: multiset
cons₂: multiset
h: multiset
app₂: [2,1]
nil: multiset
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(x1, x2)
app(x1, x2)  =  app(x1, x2)
nil  =  nil
sum(x1)  =  sum(x1)
a(x1, x2, x3)  =  x3
h  =  h
s(x1)  =  x1

Recursive path order with status [RPO].
Precedence:

app₂ > cons₂ > h
sum₁ > h

Status:

APP₂: [1,2]
cons₂: [2,1]
app₂: [2,1]
nil: multiset
sum₁: multiset
h: multiset

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