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

(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)

Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(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)

Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

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

Recursive Path Order [RPO].
Precedence:

s₁ > FLOOP₁
s₁ > +₂
*₁ > 0
*₁ > +₂

The following usable rules [FROCOS05] were oriented: none

(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: Recursive Path Order [RPO].
Precedence:

s₁ > FAC₁

The following usable rules [FROCOS05] were oriented: none

(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