Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 PisEmptyProof (⇔)
    ↳11 TRUE
  ↳12 QDP
    ↳13 QDPOrderProof (⇔)
    ↳14 QDP
    ↳15 PisEmptyProof (⇔)
    ↳16 TRUE
  ↳17 QDP
    ↳18 QDPOrderProof (⇔)
    ↳19 QDP
    ↳20 PisEmptyProof (⇔)
    ↳21 TRUE
  ↳22 QDP
    ↳23 QDPOrderProof (⇔)
    ↳24 QDP
    ↳25 PisEmptyProof (⇔)
    ↳26 TRUE

(0) Obligation:

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))
1 → s(0)
fac(0) → s(0)

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

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

(2) Obligation:

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))
1 → s(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

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 4 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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))
1 → s(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.

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


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.
Used ordering: Combined order from the following AFS and order.
+1(x1, x2)  =  x2
s(x1)  =  s(x1)
fac(x1)  =  fac(x1)
0  =  0
1  =  1
*(x1, x2)  =  *(x1, x2)
floop(x1, x2)  =  floop(x1, x2)
+(x1, x2)  =  +(x1, x2)

Lexicographic path order with status [LPO].
Quasi-Precedence:
fac1 > 1 > s1
fac1 > 1 > 0
fac1 > *2 > +2 > s1
floop2 > *2 > +2 > s1

Status:
floop2: [1,2]


The following usable rules [FROCOS05] were oriented:

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))
1 → s(0)
fac(0) → s(0)

(9) Obligation:

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))
1 → s(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.

(10) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(11) TRUE

(12) Obligation:

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))
1 → s(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.

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


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.
Used ordering: Combined order from the following AFS and order.
*1(x1, x2)  =  x2
s(x1)  =  s(x1)
fac(x1)  =  fac(x1)
0  =  0
1  =  1
*(x1, x2)  =  *(x1, x2)
floop(x1, x2)  =  floop(x1, x2)
+(x1, x2)  =  +(x1, x2)

Lexicographic path order with status [LPO].
Quasi-Precedence:
fac1 > 1 > s1
fac1 > 1 > 0
fac1 > *2 > +2 > s1
floop2 > *2 > +2 > s1

Status:
floop2: [1,2]


The following usable rules [FROCOS05] were oriented:

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))
1 → s(0)
fac(0) → s(0)

(14) Obligation:

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))
1 → s(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.

(15) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(16) TRUE

(17) Obligation:

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))
1 → s(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.

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


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.
Used ordering: Combined order from the following AFS and order.
FLOOP(x1, x2)  =  x1
s(x1)  =  s(x1)
fac(x1)  =  fac(x1)
0  =  0
1  =  1
*(x1, x2)  =  *(x1, x2)
floop(x1, x2)  =  floop(x1, x2)
+(x1, x2)  =  +(x1, x2)

Lexicographic path order with status [LPO].
Quasi-Precedence:
fac1 > 1 > s1
fac1 > 1 > 0
fac1 > *2 > +2 > s1
floop2 > *2 > +2 > s1

Status:
floop2: [1,2]


The following usable rules [FROCOS05] were oriented:

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))
1 → s(0)
fac(0) → s(0)

(19) Obligation:

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))
1 → s(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.

(20) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(21) TRUE

(22) Obligation:

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))
1 → s(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.

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


FAC(s(x)) → FAC(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FAC(x1)  =  x1
s(x1)  =  s(x1)
fac(x1)  =  fac(x1)
0  =  0
1  =  1
*(x1, x2)  =  *(x1, x2)
floop(x1, x2)  =  floop(x1, x2)
+(x1, x2)  =  +(x1, x2)

Lexicographic path order with status [LPO].
Quasi-Precedence:
fac1 > 1 > s1
fac1 > 1 > 0
fac1 > *2 > +2 > s1
floop2 > *2 > +2 > s1

Status:
floop2: [1,2]


The following usable rules [FROCOS05] were oriented:

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))
1 → s(0)
fac(0) → s(0)

(24) Obligation:

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))
1 → s(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.

(25) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(26) TRUE