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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 QDPOrderProof (⇔)
    ↳7 QDP
    ↳8 PisEmptyProof (⇔)
    ↳9 TRUE
  ↳10 QDP
    ↳11 QDPOrderProof (⇔)
    ↳12 QDP
    ↳13 PisEmptyProof (⇔)
    ↳14 TRUE
  ↳15 QDP
    ↳16 QDPOrderProof (⇔)
    ↳17 QDP
    ↳18 PisEmptyProof (⇔)
    ↳19 TRUE
  ↳20 QDP
    ↳21 QDPOrderProof (⇔)
    ↳22 QDP
    ↳23 PisEmptyProof (⇔)
    ↳24 TRUE

(0) Obligation:

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

a(h, h, h, x) → s(x)
a(l, x, s(y), h) → a(l, x, y, s(h))
a(l, x, s(y), s(z)) → a(l, x, y, a(l, x, s(y), z))
a(l, s(x), h, z) → a(l, x, z, z)
a(s(l), h, h, z) → a(l, z, h, z)
+(x, h) → x
+(h, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
s(h) → 1
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, h), l))

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

A(h, h, h, x) → S(x)
A(l, x, s(y), h) → A(l, x, y, s(h))
A(l, x, s(y), h) → S(h)
A(l, x, s(y), s(z)) → A(l, x, y, a(l, x, s(y), z))
A(l, x, s(y), s(z)) → A(l, x, s(y), z)
A(l, s(x), h, z) → A(l, x, z, z)
A(s(l), h, h, z) → A(l, z, h, z)
+1(s(x), s(y)) → S(s(+(x, y)))
+1(s(x), s(y)) → S(+(x, y))
+1(s(x), s(y)) → +1(x, y)
+1(+(x, y), z) → +1(x, +(y, z))
+1(+(x, y), z) → +1(y, z)
APP(cons(x, l), k) → APP(l, k)
SUM(cons(x, cons(y, l))) → SUM(cons(a(x, y, h, h), l))
SUM(cons(x, cons(y, l))) → A(x, y, h, h)

The TRS R consists of the following rules:

a(h, h, h, x) → s(x)
a(l, x, s(y), h) → a(l, x, y, s(h))
a(l, x, s(y), s(z)) → a(l, x, y, a(l, x, s(y), z))
a(l, s(x), h, z) → a(l, x, z, z)
a(s(l), h, h, z) → a(l, z, h, z)
+(x, h) → x
+(h, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
s(h) → 1
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, h), l))

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 5 less nodes.

(4) Complex Obligation (AND)

(5) 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:

a(h, h, h, x) → s(x)
a(l, x, s(y), h) → a(l, x, y, s(h))
a(l, x, s(y), s(z)) → a(l, x, y, a(l, x, s(y), z))
a(l, s(x), h, z) → a(l, x, z, z)
a(s(l), h, h, z) → a(l, z, h, z)
+(x, h) → x
+(h, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
s(h) → 1
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, h), l))

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

(6) 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)
a(x1, x2, x3, x4)  =  a
h  =  h
s(x1)  =  s
+(x1, x2)  =  +(x1, x2)
1  =  1
app(x1, x2)  =  app(x1, x2)
nil  =  nil
sum(x1)  =  sum(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:
[a, nil, sum1] > [s, 1] > h > cons1
[a, nil, sum1] > [s, 1] > +2 > cons1
app2 > cons1

Status:
cons1: [1]
a: []
h: []
s: []
+2: [1,2]
1: []
app2: [1,2]
nil: []
sum1: [1]


The following usable rules [FROCOS05] were oriented:

a(h, h, h, x) → s(x)
a(l, x, s(y), h) → a(l, x, y, s(h))
a(l, x, s(y), s(z)) → a(l, x, y, a(l, x, s(y), z))
a(l, s(x), h, z) → a(l, x, z, z)
a(s(l), h, h, z) → a(l, z, h, z)
+(x, h) → x
+(h, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
s(h) → 1
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, h), l))

(7) Obligation:

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

a(h, h, h, x) → s(x)
a(l, x, s(y), h) → a(l, x, y, s(h))
a(l, x, s(y), s(z)) → a(l, x, y, a(l, x, s(y), z))
a(l, s(x), h, z) → a(l, x, z, z)
a(s(l), h, h, z) → a(l, z, h, z)
+(x, h) → x
+(h, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
s(h) → 1
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, h), l))

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

+1(+(x, y), z) → +1(x, +(y, z))
+1(s(x), s(y)) → +1(x, y)
+1(+(x, y), z) → +1(y, z)

The TRS R consists of the following rules:

a(h, h, h, x) → s(x)
a(l, x, s(y), h) → a(l, x, y, s(h))
a(l, x, s(y), s(z)) → a(l, x, y, a(l, x, s(y), z))
a(l, s(x), h, z) → a(l, x, z, z)
a(s(l), h, h, z) → a(l, z, h, z)
+(x, h) → x
+(h, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
s(h) → 1
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, h), l))

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


+1(+(x, y), z) → +1(x, +(y, z))
+1(s(x), s(y)) → +1(x, y)
+1(+(x, y), z) → +1(y, z)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
+1(x1, x2)  =  x1
+(x1, x2)  =  +(x1, x2)
s(x1)  =  s(x1)
a(x1, x2, x3, x4)  =  a(x1, x2, x3, x4)
h  =  h
1  =  1
app(x1, x2)  =  app(x1, x2)
nil  =  nil
cons(x1, x2)  =  cons
sum(x1)  =  sum(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:
+2 > s1 > cons
h > [a4, sum1] > s1 > cons
h > [a4, sum1] > nil > cons
h > 1 > cons
app2 > cons

Status:
+2: [1,2]
s1: [1]
a4: [1,2,3,4]
h: []
1: []
app2: [2,1]
nil: []
cons: []
sum1: [1]


The following usable rules [FROCOS05] were oriented:

a(h, h, h, x) → s(x)
a(l, x, s(y), h) → a(l, x, y, s(h))
a(l, x, s(y), s(z)) → a(l, x, y, a(l, x, s(y), z))
a(l, s(x), h, z) → a(l, x, z, z)
a(s(l), h, h, z) → a(l, z, h, z)
+(x, h) → x
+(h, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
s(h) → 1
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, h), l))

(12) Obligation:

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

a(h, h, h, x) → s(x)
a(l, x, s(y), h) → a(l, x, y, s(h))
a(l, x, s(y), s(z)) → a(l, x, y, a(l, x, s(y), z))
a(l, s(x), h, z) → a(l, x, z, z)
a(s(l), h, h, z) → a(l, z, h, z)
+(x, h) → x
+(h, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
s(h) → 1
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, h), l))

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) Obligation:

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

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

The TRS R consists of the following rules:

a(h, h, h, x) → s(x)
a(l, x, s(y), h) → a(l, x, y, s(h))
a(l, x, s(y), s(z)) → a(l, x, y, a(l, x, s(y), z))
a(l, s(x), h, z) → a(l, x, z, z)
a(s(l), h, h, z) → a(l, z, h, z)
+(x, h) → x
+(h, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
s(h) → 1
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, h), l))

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

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


A(l, x, s(y), s(z)) → A(l, x, y, a(l, x, s(y), z))
A(l, x, s(y), h) → A(l, x, y, s(h))
A(l, x, s(y), s(z)) → A(l, x, s(y), z)
A(l, s(x), h, z) → A(l, x, z, z)
A(s(l), h, h, z) → A(l, z, h, z)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
A(x1, x2, x3, x4)  =  A(x1, x2, x3, x4)
s(x1)  =  s(x1)
a(x1, x2, x3, x4)  =  a(x1, x2, x3, x4)
h  =  h
+(x1, x2)  =  +(x1, x2)
1  =  1
app(x1, x2)  =  app(x1, x2)
nil  =  nil
cons(x1, x2)  =  cons
sum(x1)  =  sum(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:
+2 > s1
sum1 > [A4, a4] > h > s1
sum1 > [A4, a4] > h > 1
sum1 > cons > app2
sum1 > cons > nil

Status:
A4: [1,2,3,4]
s1: [1]
a4: [1,2,3,4]
h: []
+2: [1,2]
1: []
app2: [2,1]
nil: []
cons: []
sum1: [1]


The following usable rules [FROCOS05] were oriented:

a(h, h, h, x) → s(x)
a(l, x, s(y), h) → a(l, x, y, s(h))
a(l, x, s(y), s(z)) → a(l, x, y, a(l, x, s(y), z))
a(l, s(x), h, z) → a(l, x, z, z)
a(s(l), h, h, z) → a(l, z, h, z)
+(x, h) → x
+(h, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
s(h) → 1
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, h), l))

(17) Obligation:

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

a(h, h, h, x) → s(x)
a(l, x, s(y), h) → a(l, x, y, s(h))
a(l, x, s(y), s(z)) → a(l, x, y, a(l, x, s(y), z))
a(l, s(x), h, z) → a(l, x, z, z)
a(s(l), h, h, z) → a(l, z, h, z)
+(x, h) → x
+(h, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
s(h) → 1
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, h), l))

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

(18) PisEmptyProof (EQUIVALENT transformation)

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

(19) TRUE

(20) Obligation:

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

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

The TRS R consists of the following rules:

a(h, h, h, x) → s(x)
a(l, x, s(y), h) → a(l, x, y, s(h))
a(l, x, s(y), s(z)) → a(l, x, y, a(l, x, s(y), z))
a(l, s(x), h, z) → a(l, x, z, z)
a(s(l), h, h, z) → a(l, z, h, z)
+(x, h) → x
+(h, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
s(h) → 1
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, h), l))

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

(21) 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, 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(x1, x2)
a(x1, x2, x3, x4)  =  a
h  =  h
s(x1)  =  s
+(x1, x2)  =  +(x1, x2)
1  =  1
app(x1, x2)  =  app(x1, x2)
nil  =  nil
sum(x1)  =  sum(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:
[cons2, app2] > nil > [a, h, s, 1]
[cons2, app2] > sum1 > [a, h, s, 1]
+2 > [a, h, s, 1]

Status:
cons2: [1,2]
a: []
h: []
s: []
+2: [1,2]
1: []
app2: [1,2]
nil: []
sum1: [1]


The following usable rules [FROCOS05] were oriented:

a(h, h, h, x) → s(x)
a(l, x, s(y), h) → a(l, x, y, s(h))
a(l, x, s(y), s(z)) → a(l, x, y, a(l, x, s(y), z))
a(l, s(x), h, z) → a(l, x, z, z)
a(s(l), h, h, z) → a(l, z, h, z)
+(x, h) → x
+(h, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
s(h) → 1
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, h), l))

(22) Obligation:

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

a(h, h, h, x) → s(x)
a(l, x, s(y), h) → a(l, x, y, s(h))
a(l, x, s(y), s(z)) → a(l, x, y, a(l, x, s(y), z))
a(l, s(x), h, z) → a(l, x, z, z)
a(s(l), h, h, z) → a(l, z, h, z)
+(x, h) → x
+(h, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
s(h) → 1
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, h), l))

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

(23) PisEmptyProof (EQUIVALENT transformation)

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

(24) TRUE