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 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 PisEmptyProof (⇔)
    ↳11 TRUE
  ↳12 QDP
    ↳13 QDPOrderProof (⇔)
    ↳14 QDP
    ↳15 PisEmptyProof (⇔)
    ↳16 TRUE
  ↳17 QDP
    ↳18 QDPOrderProof (⇔)
    ↳19 QDP
    ↳20 QDPOrderProof (⇔)
    ↳21 QDP
    ↳22 QDPOrderProof (⇔)
    ↳23 QDP
    ↳24 QDPOrderProof (⇔)
    ↳25 QDP
    ↳26 PisEmptyProof (⇔)
    ↳27 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
*(h, x) → h
*(x, h) → h
*(s(x), s(y)) → s(+(+(*(x, y), x), y))

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)
*1(s(x), s(y)) → S(+(+(*(x, y), x), y))
*1(s(x), s(y)) → +1(+(*(x, y), x), y)
*1(s(x), s(y)) → +1(*(x, y), x)
*1(s(x), s(y)) → *1(x, y)

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
*(h, x) → h
*(x, h) → h
*(s(x), s(y)) → s(+(+(*(x, y), x), y))

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 3 SCCs with 7 less nodes.

(4) Complex Obligation (AND)

(5) 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
*(h, x) → h
*(x, h) → h
*(s(x), s(y)) → s(+(+(*(x, y), x), y))

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.


+1(+(x, y), z) → +1(x, +(y, z))
+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)  =  +1(x1)
+(x1, x2)  =  +(x1, x2)
s(x1)  =  x1
h  =  h
1  =  1

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
h: []
1: []
+2: [2,1]
+^11: [1]

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

+1(s(x), s(y)) → +1(x, y)

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
*(h, x) → h
*(x, h) → h
*(s(x), s(y)) → s(+(+(*(x, y), x), y))

Q is empty.
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.


+1(s(x), s(y)) → +1(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
+1(x1, x2)  =  x1
s(x1)  =  s(x1)

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
s1: [1]

The following usable rules [FROCOS05] were oriented: none

(9) 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
*(h, x) → h
*(x, h) → h
*(s(x), s(y)) → s(+(+(*(x, y), x), y))

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

*1(s(x), s(y)) → *1(x, y)

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
*(h, x) → h
*(x, h) → h
*(s(x), s(y)) → s(+(+(*(x, y), x), y))

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


*1(s(x), s(y)) → *1(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
*1(x1, x2)  =  x1
s(x1)  =  s(x1)

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
s1: [1]

The following usable rules [FROCOS05] were oriented: none

(14) 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
*(h, x) → h
*(x, h) → h
*(s(x), s(y)) → s(+(+(*(x, y), x), y))

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

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE

(17) 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
*(h, x) → h
*(x, h) → h
*(s(x), s(y)) → s(+(+(*(x, y), x), y))

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

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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)
s(x1)  =  s(x1)
a(x1, x2, x3, x4)  =  a(x1, x2, x3)
h  =  h
1  =  1

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

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

The following usable rules [FROCOS05] were oriented: none

(19) 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)

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
*(h, x) → h
*(x, h) → h
*(s(x), s(y)) → s(+(+(*(x, y), x), y))

Q is empty.
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.


A(l, s(x), h, z) → A(l, 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, x4)  =  A(x1, x2)
s(x1)  =  s(x1)
a(x1, x2, x3, x4)  =  a(x1, x2, x3, x4)
h  =  h
1  =  1

Lexicographic path order with status [LPO].
Precedence:
s1 > A2 > a4
s1 > 1 > a4
h > A2 > a4
h > 1 > a4

Status:
a4: [2,1,4,3]
A2: [1,2]
s1: [1]
h: []
1: []

The following usable rules [FROCOS05] were oriented: none

(21) 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)

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
*(h, x) → h
*(x, h) → h
*(s(x), s(y)) → s(+(+(*(x, y), x), y))

Q is empty.
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.


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))
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(x3)
s(x1)  =  s(x1)
a(x1, x2, x3, x4)  =  a(x1, x2, x3, x4)
h  =  h
1  =  1

Lexicographic path order with status [LPO].
Precedence:
h > A1 > s1
h > A1 > a4
h > 1

Status:
a4: [2,1,4,3]
A1: [1]
s1: [1]
h: []
1: []

The following usable rules [FROCOS05] were oriented:

s(h) → 1

(23) Obligation:

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

A(l, x, s(y), s(z)) → A(l, x, s(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
*(h, x) → h
*(x, h) → h
*(s(x), s(y)) → s(+(+(*(x, y), x), y))

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

(24) 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, 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, x4)  =  A(x2, x3, x4)
s(x1)  =  s(x1)
h  =  h
1  =  1

Lexicographic path order with status [LPO].
Precedence:
s1 > 1
h > 1

Status:
A3: [1,2,3]
s1: [1]
h: []
1: []

The following usable rules [FROCOS05] were oriented:

s(h) → 1

(25) 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
*(h, x) → h
*(x, h) → h
*(s(x), s(y)) → s(+(+(*(x, y), x), y))

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

(26) PisEmptyProof (EQUIVALENT transformation)

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

(27) TRUE