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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 QDPOrderProof (⇔)
    ↳7 QDP
    ↳8 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 PisEmptyProof (⇔)
    ↳11 TRUE
  ↳12 QDP
    ↳13 QDPOrderProof (⇔)
    ↳14 QDP
    ↳15 PisEmptyProof (⇔)
    ↳16 TRUE

(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)  =  +1(x2)
s(x1)  =  s(x1)
double(x1)  =  double(x1)
0  =  0
+(x1, x2)  =  +(x1, x2)

Recursive Path Order [RPO].
Precedence:
+^11 > s1
double1 > +2 > s1
0 > s1

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)  =  +1(x1)
s(x1)  =  s(x1)
double(x1)  =  double(x1)
0  =  0
+(x1, x2)  =  +(x1, x2)

Recursive Path Order [RPO].
Precedence:
double1 > +2 > s1 > +^11

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: Combined order from the following AFS and order.
DOUBLE(x1)  =  x1
s(x1)  =  s(x1)
double(x1)  =  double(x1)
0  =  0
+(x1, x2)  =  +(x1, x2)

Recursive Path Order [RPO].
Precedence:
double1 > +2 > s1

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