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

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

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

(2) 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)

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

(3) DependencyGraphProof (EQUIVALENT transformation)

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

(4) Complex Obligation (AND)

(5) 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)

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

(6) 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: Polynomial interpretation [POLO]:

POL(+1(x1, x2)) = x2   
POL(s(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(7) 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)

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) 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)

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

(11) 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: Polynomial interpretation [POLO]:

POL(*1(x1, x2)) = x2   
POL(s(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(12) 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)

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) 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)

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

(16) 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: Polynomial interpretation [POLO]:

POL(*(x1, x2)) = 0   
POL(+(x1, x2)) = 0   
POL(0) = 0   
POL(FLOOP(x1, x2)) = x1   
POL(s(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(17) 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)

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

(18) PisEmptyProof (EQUIVALENT transformation)

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

(19) TRUE

(20) 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)

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

(21) 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: Polynomial interpretation [POLO]:

POL(FAC(x1)) = x1   
POL(s(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(22) 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)

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

(23) PisEmptyProof (EQUIVALENT transformation)

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

(24) TRUE