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:

fac(0) → 1
fac(s(x)) → *(s(x), fac(x))
floop(0, y) → y
floop(s(x), y) → floop(x, *(s(x), y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
1s(0)
fac(0) → s(0)

Q is empty.


QTRS
  ↳ Overlay + Local Confluence

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

fac(0) → 1
fac(s(x)) → *(s(x), fac(x))
floop(0, y) → y
floop(s(x), y) → floop(x, *(s(x), y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
1s(0)
fac(0) → s(0)

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:

fac(0) → 1
fac(s(x)) → *(s(x), fac(x))
floop(0, y) → y
floop(s(x), y) → floop(x, *(s(x), y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
1s(0)
fac(0) → s(0)

The set Q consists of the following terms:

fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1


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, y), x)
FLOOP(s(x), y) → *1(s(x), y)
FAC(s(x)) → FAC(x)
+1(x, s(y)) → +1(x, y)
FLOOP(s(x), y) → FLOOP(x, *(s(x), y))
FAC(s(x)) → *1(s(x), fac(x))
*1(x, s(y)) → *1(x, y)
FAC(0) → 11

The TRS R consists of the following rules:

fac(0) → 1
fac(s(x)) → *(s(x), fac(x))
floop(0, y) → y
floop(s(x), y) → floop(x, *(s(x), y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
1s(0)
fac(0) → s(0)

The set Q consists of the following terms:

fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1

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(x, s(y)) → +1(*(x, y), x)
FLOOP(s(x), y) → *1(s(x), y)
FAC(s(x)) → FAC(x)
+1(x, s(y)) → +1(x, y)
FLOOP(s(x), y) → FLOOP(x, *(s(x), y))
FAC(s(x)) → *1(s(x), fac(x))
*1(x, s(y)) → *1(x, y)
FAC(0) → 11

The TRS R consists of the following rules:

fac(0) → 1
fac(s(x)) → *(s(x), fac(x))
floop(0, y) → y
floop(s(x), y) → floop(x, *(s(x), y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
1s(0)
fac(0) → s(0)

The set Q consists of the following terms:

fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1

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(x, s(y)) → +1(*(x, y), x)
FAC(s(x)) → FAC(x)
FLOOP(s(x), y) → *1(s(x), y)
FAC(s(x)) → *1(s(x), fac(x))
FLOOP(s(x), y) → FLOOP(x, *(s(x), y))
+1(x, s(y)) → +1(x, y)
*1(x, s(y)) → *1(x, y)
FAC(0) → 11

The TRS R consists of the following rules:

fac(0) → 1
fac(s(x)) → *(s(x), fac(x))
floop(0, y) → y
floop(s(x), y) → floop(x, *(s(x), y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
1s(0)
fac(0) → s(0)

The set Q consists of the following terms:

fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 4 less nodes.

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

fac(0) → 1
fac(s(x)) → *(s(x), fac(x))
floop(0, y) → y
floop(s(x), y) → floop(x, *(s(x), y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
1s(0)
fac(0) → s(0)

The set Q consists of the following terms:

fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1

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(x1, x2)
s(x1)  =  s(x1)

Lexicographic path order with status [19].
Precedence:
s1 > +^12

Status:
s1: multiset
+^12: [1,2]

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
                  ↳ QDP

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

fac(0) → 1
fac(s(x)) → *(s(x), fac(x))
floop(0, y) → y
floop(s(x), y) → floop(x, *(s(x), y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
1s(0)
fac(0) → s(0)

The set Q consists of the following terms:

fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1

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
                  ↳ 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:

fac(0) → 1
fac(s(x)) → *(s(x), fac(x))
floop(0, y) → y
floop(s(x), y) → floop(x, *(s(x), y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
1s(0)
fac(0) → s(0)

The set Q consists of the following terms:

fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1

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(x1, x2)
s(x1)  =  s(x1)

Lexicographic path order with status [19].
Precedence:
s1 > *^12

Status:
*^12: [1,2]
s1: multiset

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
                  ↳ QDP

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

fac(0) → 1
fac(s(x)) → *(s(x), fac(x))
floop(0, y) → y
floop(s(x), y) → floop(x, *(s(x), y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
1s(0)
fac(0) → s(0)

The set Q consists of the following terms:

fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1

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
                  ↳ QDP

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

FLOOP(s(x), y) → FLOOP(x, *(s(x), y))

The TRS R consists of the following rules:

fac(0) → 1
fac(s(x)) → *(s(x), fac(x))
floop(0, y) → y
floop(s(x), y) → floop(x, *(s(x), y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
1s(0)
fac(0) → s(0)

The set Q consists of the following terms:

fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1

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.


FLOOP(s(x), y) → FLOOP(x, *(s(x), y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
FLOOP(x1, x2)  =  FLOOP(x1)
s(x1)  =  s(x1)
*(x1, x2)  =  *
0  =  0
+(x1, x2)  =  +

Lexicographic path order with status [19].
Precedence:
FLOOP1 > s1
* > s1
0 > s1
+ > s1

Status:
FLOOP1: [1]
+: []
s1: [1]
0: multiset
*: []

The following usable rules [14] were oriented: none



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

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

fac(0) → 1
fac(s(x)) → *(s(x), fac(x))
floop(0, y) → y
floop(s(x), y) → floop(x, *(s(x), y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
1s(0)
fac(0) → s(0)

The set Q consists of the following terms:

fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1

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
QDP
                    ↳ QDPOrderProof

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

FAC(s(x)) → FAC(x)

The TRS R consists of the following rules:

fac(0) → 1
fac(s(x)) → *(s(x), fac(x))
floop(0, y) → y
floop(s(x), y) → floop(x, *(s(x), y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
1s(0)
fac(0) → s(0)

The set Q consists of the following terms:

fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1

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.


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

Lexicographic path order with status [19].
Precedence:
s1 > FAC1

Status:
FAC1: [1]
s1: multiset

The following usable rules [14] were oriented: none



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

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

fac(0) → 1
fac(s(x)) → *(s(x), fac(x))
floop(0, y) → y
floop(s(x), y) → floop(x, *(s(x), y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
1s(0)
fac(0) → s(0)

The set Q consists of the following terms:

fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1

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