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:

times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(x, 0) → x
plus(0, x) → x
plus(x, s(y)) → s(plus(x, y))
plus(s(x), y) → s(plus(x, y))

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:

times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(x, 0) → x
plus(0, x) → x
plus(x, s(y)) → s(plus(x, y))
plus(s(x), y) → s(plus(x, y))

The set Q consists of the following terms:

times(x0, 0)
times(x0, s(x1))
plus(x0, 0)
plus(0, x0)
plus(x0, s(x1))
plus(s(x0), x1)

(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:

TIMES(x, s(y)) → PLUS(times(x, y), x)
TIMES(x, s(y)) → TIMES(x, y)
PLUS(x, s(y)) → PLUS(x, y)
PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(x, 0) → x
plus(0, x) → x
plus(x, s(y)) → s(plus(x, y))
plus(s(x), y) → s(plus(x, y))

The set Q consists of the following terms:

times(x0, 0)
times(x0, s(x1))
plus(x0, 0)
plus(0, x0)
plus(x0, s(x1))
plus(s(x0), x1)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.

(6) Complex Obligation (AND)

(7) Obligation:

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

PLUS(s(x), y) → PLUS(x, y)
PLUS(x, s(y)) → PLUS(x, y)

The TRS R consists of the following rules:

times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(x, 0) → x
plus(0, x) → x
plus(x, s(y)) → s(plus(x, y))
plus(s(x), y) → s(plus(x, y))

The set Q consists of the following terms:

times(x0, 0)
times(x0, s(x1))
plus(x0, 0)
plus(0, x0)
plus(x0, s(x1))
plus(s(x0), x1)

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.

PLUS(s(x), y) → PLUS(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PLUS(x1, x2)  =  PLUS(x1)
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:

[PLUS₁, s₁]

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

PLUS(x, s(y)) → PLUS(x, y)

The TRS R consists of the following rules:

times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(x, 0) → x
plus(0, x) → x
plus(x, s(y)) → s(plus(x, y))
plus(s(x), y) → s(plus(x, y))

The set Q consists of the following terms:

times(x0, 0)
times(x0, s(x1))
plus(x0, 0)
plus(0, x0)
plus(x0, s(x1))
plus(s(x0), x1)

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

(10) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PLUS(x, s(y)) → PLUS(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PLUS(x1, x2)  =  PLUS(x2)
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:

s₁ > PLUS₁

The following usable rules [FROCOS05] were oriented: none

(11) Obligation:

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

times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(x, 0) → x
plus(0, x) → x
plus(x, s(y)) → s(plus(x, y))
plus(s(x), y) → s(plus(x, y))

The set Q consists of the following terms:

times(x0, 0)
times(x0, s(x1))
plus(x0, 0)
plus(0, x0)
plus(x0, s(x1))
plus(s(x0), x1)

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

(12) PisEmptyProof (EQUIVALENT transformation)

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

(13) TRUE

(14) Obligation:

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

TIMES(x, s(y)) → TIMES(x, y)

The TRS R consists of the following rules:

times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(x, 0) → x
plus(0, x) → x
plus(x, s(y)) → s(plus(x, y))
plus(s(x), y) → s(plus(x, y))

The set Q consists of the following terms:

times(x0, 0)
times(x0, s(x1))
plus(x0, 0)
plus(0, x0)
plus(x0, s(x1))
plus(s(x0), x1)

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

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

TIMES(x, s(y)) → TIMES(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TIMES(x1, x2)  =  TIMES(x2)
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:

s₁ > TIMES₁

The following usable rules [FROCOS05] were oriented: none

(16) Obligation:

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

times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(x, 0) → x
plus(0, x) → x
plus(x, s(y)) → s(plus(x, y))
plus(s(x), y) → s(plus(x, y))

The set Q consists of the following terms:

times(x0, 0)
times(x0, s(x1))
plus(x0, 0)
plus(0, x0)
plus(x0, s(x1))
plus(s(x0), x1)

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

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) TRUE