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

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:

TIMES(x, plus(y, s(z))) → PLUS(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
TIMES(x, plus(y, s(z))) → TIMES(x, plus(y, times(s(z), 0)))
TIMES(x, plus(y, s(z))) → PLUS(y, times(s(z), 0))
TIMES(x, plus(y, s(z))) → TIMES(s(z), 0)
TIMES(x, plus(y, s(z))) → TIMES(x, s(z))
TIMES(x, s(y)) → PLUS(times(x, y), x)
TIMES(x, s(y)) → TIMES(x, y)
PLUS(x, s(y)) → PLUS(x, y)

The TRS R consists of the following rules:

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

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 2 SCCs with 4 less nodes.

(4) Complex Obligation (AND)

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

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.

PLUS(x, s(y)) → PLUS(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(PLUS(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:

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

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:

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

The TRS R consists of the following rules:

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

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.

TIMES(x, plus(y, s(z))) → TIMES(x, s(z))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(TIMES(x₁, x₂)) = x₂   
POL(plus(x₁, x₂)) = 1 + x₁ + x₂   
POL(s(x₁)) = x₁   
POL(times(x₁, x₂)) = 0   

The following usable rules [FROCOS05] were oriented:

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

(12) Obligation:

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

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

The TRS R consists of the following rules:

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

Q is empty.
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.

TIMES(x, s(y)) → TIMES(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(TIMES(x₁, x₂)) = x₂   
POL(plus(x₁, x₂)) = x₁ + x₂   
POL(s(x₁)) = 1 + x₁   
POL(times(x₁, x₂)) = x₁   

The following usable rules [FROCOS05] were oriented:

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

(14) Obligation:

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

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

The TRS R consists of the following rules:

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

Q is empty.
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, plus(y, s(z))) → TIMES(x, plus(y, times(s(z), 0)))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(TIMES(x₁, x₂)) = x₂   
POL(plus(x₁, x₂)) = x₁ + x₂   
POL(s(x₁)) = 1 + x₁   
POL(times(x₁, x₂)) = 0   

The following usable rules [FROCOS05] were oriented:

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

(16) Obligation:

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

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

Q is empty.
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