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:

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(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: Lexicographic path order with status [LPO].
Quasi-Precedence:

*₂ > [+^1₂, +₂] > s₁ > 1
*₂ > [a₄, h] > s₁ > 1

Status:

*₂: [2,1]
a₄: [1,2,3,4]
s₁: [1]
h: []
1: []
+^1₂: [1,2]
+₂: [1,2]

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

(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
*(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) 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(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.

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

*^1₁ > [s₁, 1]
*₂ > [a₄, h] > [s₁, 1]
*₂ > +₂ > [s₁, 1]

Status:

*^1₁: [1]
*₂: [2,1]
a₄: [1,2,3,4]
s₁: [1]
h: []
1: []
+₂: [1,2]

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

(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
*(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) 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
*(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.

(16) 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, x4)
h  =  h
+(x1, x2)  =  +(x1, x2)
1  =  1
*(x1, x2)  =  *(x1, x2)

Lexicographic path order with status [LPO].
Quasi-Precedence:

[h, 1] > A₁ > a₄ > s₁
*₂ > +₂ > s₁

Status:

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

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

(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)

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(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 remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Quasi-Precedence:

A₄ > a₄ > s₁ > 1 > h
*₂ > +₂ > s₁ > 1 > h

Status:

*₂: [2,1]
a₄: [1,2,3,4]
A₄: [1,2,3,4]
s₁: [1]
h: []
1: []
+₂: [1,2]

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

(19) 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.

(20) PisEmptyProof (EQUIVALENT transformation)

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

(21) TRUE