Termination w.r.t. Q of the given QTRS could not be shown:

(0) Obligation:

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

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, 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:

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, s(0))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
p(s(0))
p(s(s(x0)))
fac(0, x0)
fac(s(x0), x1)
factorial(x0)

(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:

PLUS(s(x), y) → PLUS(p(s(x)), y)
PLUS(s(x), y) → P(s(x))
TIMES(s(x), y) → PLUS(y, times(p(s(x)), y))
TIMES(s(x), y) → TIMES(p(s(x)), y)
TIMES(s(x), y) → P(s(x))
P(s(s(x))) → P(s(x))
FAC(s(x), y) → FAC(p(s(x)), times(s(x), y))
FAC(s(x), y) → P(s(x))
FAC(s(x), y) → TIMES(s(x), y)
FACTORIAL(x) → FAC(x, s(0))

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, s(0))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
p(s(0))
p(s(s(x0)))
fac(0, x0)
fac(s(x0), x1)
factorial(x0)

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 6 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

P(s(s(x))) → P(s(x))

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, s(0))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
p(s(0))
p(s(s(x0)))
fac(0, x0)
fac(s(x0), x1)
factorial(x0)

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.

P(s(s(x))) → P(s(x))

The remaining pairs can at least be oriented weakly.
Used ordering: Recursive Path Order [RPO].
Precedence:

[P₁, s₁]

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, s(0))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
p(s(0))
p(s(s(x0)))
fac(0, x0)
fac(s(x0), x1)
factorial(x0)

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:

PLUS(s(x), y) → PLUS(p(s(x)), y)

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, s(0))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
p(s(0))
p(s(s(x0)))
fac(0, x0)
fac(s(x0), x1)
factorial(x0)

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

(13) Obligation:

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

TIMES(s(x), y) → TIMES(p(s(x)), y)

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, s(0))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
p(s(0))
p(s(s(x0)))
fac(0, x0)
fac(s(x0), x1)
factorial(x0)

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

(14) Obligation:

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

FAC(s(x), y) → FAC(p(s(x)), times(s(x), y))

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, s(0))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
p(s(0))
p(s(s(x0)))
fac(0, x0)
fac(s(x0), x1)
factorial(x0)

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