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 QDPSizeChangeProof (⇔)
        ↳7 TRUE
    ↳8 QDP
        ↳9 QDPSizeChangeProof (⇔)
        ↳10 TRUE
    ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔)
        ↳13 TRUE
    ↳14 QDP
        ↳15 QDPSizeChangeProof (⇔)
        ↳16 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) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
s(x1)  =  s(x1)

From the DPs we obtained the following set of size-change graphs:

  • +1(x, s(y)) → +1(x, y) (allowed arguments on rhs = {1, 2})
    The graph contains the following edges 1 >= 1, 2 > 2

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(7) TRUE

(8) 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.

(9) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
s(x1)  =  s(x1)

From the DPs we obtained the following set of size-change graphs:

  • *1(x, s(y)) → *1(x, y) (allowed arguments on rhs = {1, 2})
    The graph contains the following edges 1 >= 1, 2 > 2

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(10) TRUE

(11) 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.

(12) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
s(x1)  =  s(x1)

From the DPs we obtained the following set of size-change graphs:

  • FLOOP(s(x), y) → FLOOP(x, *(s(x), y)) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 > 1

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(13) TRUE

(14) 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.

(15) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
s(x1)  =  s(x1)

From the DPs we obtained the following set of size-change graphs:

  • FAC(s(x)) → FAC(x) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 > 1

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(16) TRUE