Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

DOUBLE(s(x)) → DOUBLE(x)
PLUS(s(x), y) → PLUS(x, y)
PLUS(s(x), y) → MINUS(x, y)
PLUS(s(x), y) → DOUBLE(y)
PLUS(s(x), y) → PLUS(x, s(y))
PLUS(s(plus(x, y)), z) → PLUS(plus(x, y), z)
MINUS(s(x), s(y)) → MINUS(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)))
plus(s(plus(x, y)), z) → s(plus(plus(x, y), z))

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ EdgeDeletionProof

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

DOUBLE(s(x)) → DOUBLE(x)
PLUS(s(x), y) → PLUS(x, y)
PLUS(s(x), y) → MINUS(x, y)
PLUS(s(x), y) → DOUBLE(y)
PLUS(s(x), y) → PLUS(x, s(y))
PLUS(s(plus(x, y)), z) → PLUS(plus(x, y), z)
MINUS(s(x), s(y)) → MINUS(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)))
plus(s(plus(x, y)), z) → s(plus(plus(x, y), z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
QDP
          ↳ DependencyGraphProof

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

DOUBLE(s(x)) → DOUBLE(x)
PLUS(s(x), y) → PLUS(x, y)
PLUS(s(x), y) → MINUS(x, y)
PLUS(s(x), y) → PLUS(x, s(y))
PLUS(s(x), y) → DOUBLE(y)
PLUS(s(plus(x, y)), z) → PLUS(plus(x, y), z)
MINUS(s(x), s(y)) → MINUS(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)))
plus(s(plus(x, y)), z) → s(plus(plus(x, y), z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
QDP
                ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


DOUBLE(s(x)) → DOUBLE(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
DOUBLE(x1)  =  DOUBLE(x1)
s(x1)  =  s(x1)

Lexicographic Path Order [19].
Precedence:
[DOUBLE1, s1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ PisEmptyProof
              ↳ QDP
              ↳ QDP

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
QDP
                ↳ QDPOrderProof
              ↳ QDP

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


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.
none
Used ordering: Combined order from the following AFS and order.
MINUS(x1, x2)  =  MINUS(x1)
s(x1)  =  s(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ PisEmptyProof
              ↳ QDP

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
QDP
                ↳ QDPOrderProof

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

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


PLUS(s(x), y) → PLUS(x, y)
PLUS(s(x), y) → PLUS(x, s(y))
PLUS(s(plus(x, y)), z) → PLUS(plus(x, y), z)
PLUS(s(x), y) → PLUS(minus(x, y), double(y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
PLUS(x1, x2)  =  PLUS(x1, x2)
s(x1)  =  s(x1)
plus(x1, x2)  =  plus(x1, x2)
minus(x1, x2)  =  minus(x1)
double(x1)  =  double(x1)
0  =  0

Lexicographic Path Order [19].
Precedence:
plus2 > PLUS2 > double1 > s1 > minus1
0 > minus1


The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ PisEmptyProof

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.