(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) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
s(x1)  =  s(x1)
+(x1, x2)  =  +(x1, x2)

From the DPs we obtained the following set of size-change graphs:

  • +1(+(x, y), z) → +1(x, +(y, z)) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 > 1

  • +1(s(x), s(y)) → +1(x, y) (allowed arguments on rhs = {1, 2})
    The graph contains the following edges 1 > 1, 2 > 2

  • +1(+(x, y), z) → +1(y, z) (allowed arguments on rhs = {1, 2})
    The graph contains the following edges 1 > 1, 2 >= 2

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(7) TRUE

(8) 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.

(9) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
s(x1)  =  s(x1)

From the DPs we obtained the following set of size-change graphs:

  • *1(s(x), s(y)) → *1(x, y) (allowed arguments on rhs = {1, 2})
    The graph contains the following edges 1 > 1, 2 > 2

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(10) TRUE

(11) 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.

(12) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Recursive path order with status [RPO].
Quasi-Precedence:

s1 > [h, 1]

Status:
s1: multiset
h: multiset
1: multiset

AFS:
s(x1)  =  s(x1)
h  =  h
1  =  1

From the DPs we obtained the following set of size-change graphs:

  • A(s(l), h, h, z) → A(l, z, h, z) (allowed arguments on rhs = {1, 2, 3, 4})
    The graph contains the following edges 1 > 1, 4 >= 2, 2 >= 3, 3 >= 3, 4 >= 4

  • A(l, s(x), h, z) → A(l, x, z, z) (allowed arguments on rhs = {1, 2, 3, 4})
    The graph contains the following edges 1 >= 1, 2 > 2, 4 >= 3, 4 >= 4

  • A(l, x, s(y), s(z)) → A(l, x, y, a(l, x, s(y), z)) (allowed arguments on rhs = {1, 2, 3})
    The graph contains the following edges 1 >= 1, 2 >= 2, 3 > 3

  • A(l, x, s(y), h) → A(l, x, y, s(h)) (allowed arguments on rhs = {1, 2, 3})
    The graph contains the following edges 1 >= 1, 2 >= 2, 3 > 3

  • A(l, x, s(y), s(z)) → A(l, x, s(y), z) (allowed arguments on rhs = {1, 2, 3, 4})
    The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3, 4 > 4

We oriented the following set of usable rules [AAECC05,FROCOS05].


s(h) → 1

(13) TRUE