Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
f(s(x)) → f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))

Q is empty.


QTRS
  ↳ AAECC Innermost

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
f(s(x)) → f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))

Q is empty.

We have applied [15,7] to switch to innermost. The TRS R 1 is

+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)

The TRS R 2 is

f(s(x)) → f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))

The signature Sigma is {f}

↳ QTRS
  ↳ AAECC Innermost
QTRS
      ↳ DependencyPairsProof

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
f(s(x)) → f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
f(s(x0))


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

*1(x, s(y)) → +1(x, *(x, y))
*1(x, s(y)) → *1(x, y)
-1(s(x), s(y)) → -1(x, y)
F(s(x)) → +1(*(s(x), s(s(x))), s(s(0)))
+1(s(x), y) → +1(x, y)
F(s(x)) → *1(s(x), s(s(x)))
F(s(x)) → -1(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0))))
F(s(x)) → F(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))
TWICE(s(x)) → TWICE(x)
F(s(x)) → *1(s(s(x)), s(s(x)))

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
f(s(x)) → f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
f(s(x0))

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

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
QDP
          ↳ EdgeDeletionProof

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

*1(x, s(y)) → +1(x, *(x, y))
*1(x, s(y)) → *1(x, y)
-1(s(x), s(y)) → -1(x, y)
F(s(x)) → +1(*(s(x), s(s(x))), s(s(0)))
+1(s(x), y) → +1(x, y)
F(s(x)) → *1(s(x), s(s(x)))
F(s(x)) → -1(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0))))
F(s(x)) → F(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))
TWICE(s(x)) → TWICE(x)
F(s(x)) → *1(s(s(x)), s(s(x)))

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
f(s(x)) → f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
f(s(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
QDP
              ↳ DependencyGraphProof

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

*1(x, s(y)) → +1(x, *(x, y))
*1(x, s(y)) → *1(x, y)
-1(s(x), s(y)) → -1(x, y)
F(s(x)) → +1(*(s(x), s(s(x))), s(s(0)))
+1(s(x), y) → +1(x, y)
F(s(x)) → -1(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0))))
F(s(x)) → *1(s(x), s(s(x)))
F(s(x)) → F(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))
TWICE(s(x)) → TWICE(x)
F(s(x)) → *1(s(s(x)), s(s(x)))

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
f(s(x)) → f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
f(s(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 5 SCCs with 5 less nodes.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
QDP
                    ↳ QDPOrderProof
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP

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:

+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
f(s(x)) → f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
f(s(x0))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


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.
none
Used ordering: Combined order from the following AFS and order.
-1(x1, x2)  =  -1(x2)
s(x1)  =  s(x1)

Lexicographic path order with status [19].
Quasi-Precedence:
s1 > -^11

Status:
-^11: [1]
s1: [1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ PisEmptyProof
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
f(s(x)) → f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
f(s(x0))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
QDP
                    ↳ QDPOrderProof
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP

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

TWICE(s(x)) → TWICE(x)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
f(s(x)) → f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
f(s(x0))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


TWICE(s(x)) → TWICE(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
TWICE(x1)  =  TWICE(x1)
s(x1)  =  s(x1)

Lexicographic path order with status [19].
Quasi-Precedence:
[TWICE1, s1]

Status:
s1: [1]
TWICE1: [1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ PisEmptyProof
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
f(s(x)) → f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
f(s(x0))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
QDP
                    ↳ QDPOrderProof
                  ↳ QDP
                  ↳ QDP

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

+1(s(x), y) → +1(x, y)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
f(s(x)) → f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
f(s(x0))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


+1(s(x), y) → +1(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
+1(x1, x2)  =  +1(x1)
s(x1)  =  s(x1)

Lexicographic path order with status [19].
Quasi-Precedence:
s1 > +^11

Status:
+^11: [1]
s1: [1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ PisEmptyProof
                  ↳ QDP
                  ↳ QDP

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
f(s(x)) → f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
f(s(x0))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
QDP
                    ↳ QDPOrderProof
                  ↳ QDP

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

*1(x, s(y)) → *1(x, y)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
f(s(x)) → f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
f(s(x0))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


*1(x, s(y)) → *1(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
*1(x1, x2)  =  *1(x2)
s(x1)  =  s(x1)

Lexicographic path order with status [19].
Quasi-Precedence:
s1 > *^11

Status:
*^11: [1]
s1: [1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ PisEmptyProof
                  ↳ QDP

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
f(s(x)) → f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
f(s(x0))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
QDP

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

F(s(x)) → F(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
f(s(x)) → f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
f(s(x0))

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