Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

Q is empty.


QTRS
  ↳ Overlay + Local Confluence

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
QTRS
      ↳ DependencyPairsProof

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)


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(s(x), y) → +1(*(x, y), y)
*1(s(x), y) → *1(x, y)
*1(p(x), y) → *1(x, y)
*1(p(x), y) → +1(*(x, y), minus(y))
MINUS(p(x)) → MINUS(x)
MINUS(s(x)) → MINUS(x)
+1(s(x), y) → +1(x, y)
+1(p(x), y) → +1(x, y)
*1(p(x), y) → MINUS(y)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
QDP
          ↳ EdgeDeletionProof

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

*1(s(x), y) → +1(*(x, y), y)
*1(s(x), y) → *1(x, y)
*1(p(x), y) → *1(x, y)
*1(p(x), y) → +1(*(x, y), minus(y))
MINUS(p(x)) → MINUS(x)
MINUS(s(x)) → MINUS(x)
+1(s(x), y) → +1(x, y)
+1(p(x), y) → +1(x, y)
*1(p(x), y) → MINUS(y)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
QDP
              ↳ DependencyGraphProof

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

*1(s(x), y) → +1(*(x, y), y)
*1(s(x), y) → *1(x, y)
*1(p(x), y) → *1(x, y)
MINUS(s(x)) → MINUS(x)
MINUS(p(x)) → MINUS(x)
*1(p(x), y) → +1(*(x, y), minus(y))
+1(s(x), y) → +1(x, y)
+1(p(x), y) → +1(x, y)
*1(p(x), y) → MINUS(y)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 3 less nodes.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
QDP
                    ↳ QDPOrderProof
                  ↳ QDP
                  ↳ QDP

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

MINUS(s(x)) → MINUS(x)
MINUS(p(x)) → MINUS(x)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)

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.


MINUS(s(x)) → MINUS(x)
The remaining pairs can at least be oriented weakly.

MINUS(p(x)) → MINUS(x)
Used ordering: Combined order from the following AFS and order.
MINUS(x1)  =  MINUS(x1)
s(x1)  =  s(x1)
p(x1)  =  x1

Recursive Path Order [2].
Precedence:
[MINUS1, s1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ QDPOrderProof
                  ↳ QDP
                  ↳ QDP

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

MINUS(p(x)) → MINUS(x)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)

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.


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

Recursive Path Order [2].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ 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))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
QDP
                    ↳ QDPOrderProof
                  ↳ QDP

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

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

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)

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.

+1(p(x), y) → +1(x, y)
Used ordering: Combined order from the following AFS and order.
+1(x1, x2)  =  +1(x1, x2)
s(x1)  =  s(x1)
p(x1)  =  x1

Recursive Path Order [2].
Precedence:
[+^12, s1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ QDPOrderProof
                  ↳ QDP

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

+1(p(x), y) → +1(x, y)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)

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(p(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)
p(x1)  =  p(x1)

Recursive Path Order [2].
Precedence:
p1 > +^11


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ 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))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
QDP
                    ↳ QDPOrderProof

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

*1(s(x), y) → *1(x, y)
*1(p(x), y) → *1(x, y)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)

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.

*1(p(x), y) → *1(x, y)
Used ordering: Combined order from the following AFS and order.
*1(x1, x2)  =  *1(x1, x2)
s(x1)  =  s(x1)
p(x1)  =  x1

Recursive Path Order [2].
Precedence:
[*^12, s1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ QDPOrderProof

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

*1(p(x), y) → *1(x, y)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)

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(p(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)
p(x1)  =  p(x1)

Recursive Path Order [2].
Precedence:
p1 > *^11


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof
QDP
                            ↳ PisEmptyProof

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)

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