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 QDP
    ↳9 QDPOrderProof (⇔)
    ↳10 QDP
    ↳11 PisEmptyProof (⇔)
    ↳12 TRUE
  ↳13 QDP
    ↳14 QDPOrderProof (⇔)
    ↳15 QDP
    ↳16 PisEmptyProof (⇔)
    ↳17 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) 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.

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

Recursive Path Order [RPO].
Precedence:
app2 > cons1
sum > cons1
sum > nil
a > s

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)

(10) 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.

(11) PisEmptyProof (EQUIVALENT transformation)

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

(12) TRUE

(13) 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.

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

Recursive Path Order [RPO].
Precedence:
app2 > cons1
sum > cons1
sum > nil
a > s

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)

(15) 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.

(16) PisEmptyProof (EQUIVALENT transformation)

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

(17) TRUE