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:

double(0) → 0
double(s(x)) → s(s(double(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))
double(x) → +(x, x)

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:

DOUBLE(s(x)) → DOUBLE(x)
+¹(x, s(y)) → +¹(x, y)
+¹(s(x), y) → +¹(x, y)
DOUBLE(x) → +¹(x, x)

The TRS R consists of the following rules:

double(0) → 0
double(s(x)) → s(s(double(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))
double(x) → +(x, x)

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 1 less node.

(4) Complex Obligation (AND)

(5) Obligation:

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

+¹(s(x), y) → +¹(x, y)
+¹(x, s(y)) → +¹(x, y)

The TRS R consists of the following rules:

double(0) → 0
double(s(x)) → s(s(double(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))
double(x) → +(x, x)

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.

+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)
double(x1)  =  double(x1)
0  =  0
+(x1, x2)  =  +(x1, x2)

Lexicographic path order with status [LPO].
Quasi-Precedence:

double₁ > +₂ > s₁
0 > s₁

Status:

s₁: [1]
double₁: [1]
0: []
+₂: [1,2]

The following usable rules [FROCOS05] were oriented:

double(0) → 0
double(s(x)) → s(s(double(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))
double(x) → +(x, x)

(7) Obligation:

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

+¹(s(x), y) → +¹(x, y)

The TRS R consists of the following rules:

double(0) → 0
double(s(x)) → s(s(double(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))
double(x) → +(x, x)

Q is empty.
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(s(x), 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)  =  x1
s(x1)  =  s(x1)
double(x1)  =  double(x1)
0  =  0
+(x1, x2)  =  +(x1, x2)

Lexicographic path order with status [LPO].
Quasi-Precedence:

double₁ > 0 > s₁
double₁ > +₂ > s₁

Status:

s₁: [1]
double₁: [1]
0: []
+₂: [1,2]

The following usable rules [FROCOS05] were oriented:

double(0) → 0
double(s(x)) → s(s(double(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))
double(x) → +(x, x)

(9) Obligation:

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

double(0) → 0
double(s(x)) → s(s(double(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))
double(x) → +(x, x)

Q is empty.
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:

DOUBLE(s(x)) → DOUBLE(x)

The TRS R consists of the following rules:

double(0) → 0
double(s(x)) → s(s(double(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))
double(x) → +(x, x)

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.

DOUBLE(s(x)) → DOUBLE(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Quasi-Precedence:

[double₁, 0] > +₂ > s₁

Status:

DOUBLE₁: [1]
s₁: [1]
double₁: [1]
0: []
+₂: [1,2]

The following usable rules [FROCOS05] were oriented:

double(0) → 0
double(s(x)) → s(s(double(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))
double(x) → +(x, x)

(14) Obligation:

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

double(0) → 0
double(s(x)) → s(s(double(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))
double(x) → +(x, x)

Q is empty.
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