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 PisEmptyProof (⇔)
    ↳9 TRUE
  ↳10 QDP
    ↳11 QDPOrderProof (⇔)
    ↳12 QDP
    ↳13 PisEmptyProof (⇔)
    ↳14 TRUE
  ↳15 QDP
    ↳16 QDPOrderProof (⇔)
    ↳17 QDP
    ↳18 PisEmptyProof (⇔)
    ↳19 TRUE

(0) Obligation:

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(minus(x, y), double(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:

MINUS(s(x), s(y)) → MINUS(x, y)
DOUBLE(s(x)) → DOUBLE(x)
PLUS(s(x), y) → PLUS(x, y)
PLUS(s(x), y) → PLUS(x, s(y))
PLUS(s(x), y) → PLUS(minus(x, y), double(y))
PLUS(s(x), y) → MINUS(x, y)
PLUS(s(x), y) → DOUBLE(y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(minus(x, y), double(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 3 SCCs with 2 less nodes.

(4) Complex Obligation (AND)

(5) 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:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(minus(x, y), double(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.


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)
minus(x1, x2)  =  x1
0  =  0
double(x1)  =  double(x1)
plus(x1, x2)  =  plus(x1, x2)

Recursive path order with status [RPO].
Quasi-Precedence:
plus2 > double1 > s1

Status:
s1: [1]
0: multiset
double1: multiset
plus2: [1,2]


The following usable rules [FROCOS05] were oriented:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(minus(x, y), double(y)))

(7) Obligation:

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(minus(x, y), double(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:

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(minus(x, y), double(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.


MINUS(s(x), s(y)) → MINUS(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MINUS(x1, x2)  =  MINUS(x2)
s(x1)  =  s(x1)
minus(x1, x2)  =  minus(x1)
0  =  0
double(x1)  =  double(x1)
plus(x1, x2)  =  plus(x1, x2)

Recursive path order with status [RPO].
Quasi-Precedence:
MINUS1 > minus1
plus2 > double1 > s1 > minus1
plus2 > double1 > 0 > minus1

Status:
MINUS1: [1]
s1: [1]
minus1: multiset
0: multiset
double1: [1]
plus2: [1,2]


The following usable rules [FROCOS05] were oriented:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(minus(x, y), double(y)))

(12) Obligation:

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(minus(x, y), double(y)))

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) Obligation:

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

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

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(minus(x, y), double(y)))

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

(16) 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, s(y))
PLUS(s(x), y) → PLUS(x, y)
PLUS(s(x), y) → PLUS(minus(x, y), double(y))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PLUS(x1, x2)  =  x1
s(x1)  =  s(x1)
minus(x1, x2)  =  minus(x1)
double(x1)  =  double(x1)
0  =  0
plus(x1, x2)  =  plus(x1, x2)

Recursive path order with status [RPO].
Quasi-Precedence:
plus2 > [double1, 0] > s1 > minus1

Status:
s1: multiset
minus1: multiset
double1: [1]
0: multiset
plus2: [1,2]


The following usable rules [FROCOS05] were oriented:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(minus(x, y), double(y)))

(17) Obligation:

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(minus(x, y), double(y)))

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

(18) PisEmptyProof (EQUIVALENT transformation)

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

(19) TRUE