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:

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

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

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

p(s(x)) → x
fact(0) → s(0)
fact(s(x)) → *(s(x), fact(p(s(x))))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
+(x, 0) → x
+(x, s(y)) → s(+(x, 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:

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

The set Q consists of the following terms:

p(s(x0))
fact(0)
fact(s(x0))
*(0, x0)
*(s(x0), x1)
+(x0, 0)
+(x0, s(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(x, s(y)) → +1(x, y)
*1(s(x), y) → *1(x, y)
FACT(s(x)) → P(s(x))
FACT(s(x)) → *1(s(x), fact(p(s(x))))
FACT(s(x)) → FACT(p(s(x)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

p(s(x0))
fact(0)
fact(s(x0))
*(0, x0)
*(s(x0), x1)
+(x0, 0)
+(x0, s(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(x, s(y)) → +1(x, y)
*1(s(x), y) → *1(x, y)
FACT(s(x)) → P(s(x))
FACT(s(x)) → *1(s(x), fact(p(s(x))))
FACT(s(x)) → FACT(p(s(x)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

p(s(x0))
fact(0)
fact(s(x0))
*(0, x0)
*(s(x0), x1)
+(x0, 0)
+(x0, s(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(x, s(y)) → +1(x, y)
FACT(s(x)) → P(s(x))
*1(s(x), y) → *1(x, y)
FACT(s(x)) → *1(s(x), fact(p(s(x))))
FACT(s(x)) → FACT(p(s(x)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

p(s(x0))
fact(0)
fact(s(x0))
*(0, x0)
*(s(x0), x1)
+(x0, 0)
+(x0, s(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:

+1(x, s(y)) → +1(x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

p(s(x0))
fact(0)
fact(s(x0))
*(0, x0)
*(s(x0), x1)
+(x0, 0)
+(x0, s(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(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)  =  x2
s(x1)  =  s(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
                        ↳ PisEmptyProof
                  ↳ QDP
                  ↳ QDP

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

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

The set Q consists of the following terms:

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

p(s(x0))
fact(0)
fact(s(x0))
*(0, x0)
*(s(x0), x1)
+(x0, 0)
+(x0, s(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.
none
Used ordering: Combined order from the following AFS and order.
*1(x1, x2)  =  x1
s(x1)  =  s(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
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof
                  ↳ QDP

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

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

The set Q consists of the following terms:

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

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

FACT(s(x)) → FACT(p(s(x)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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