Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

f(true, x, y) → f(and(gt(x, y), gt(y, s(s(0)))), plus(s(0), x), double(y))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
and(x, true) → x
and(x, false) → false
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
double(0) → 0
double(s(x)) → s(s(double(x)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(true, x, y) → f(and(gt(x, y), gt(y, s(s(0)))), plus(s(0), x), double(y))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
and(x, true) → x
and(x, false) → false
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
double(0) → 0
double(s(x)) → s(s(double(x)))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

F(true, x, y) → PLUS(s(0), x)
DOUBLE(s(x)) → DOUBLE(x)
GT(s(u), s(v)) → GT(u, v)
F(true, x, y) → GT(y, s(s(0)))
F(true, x, y) → GT(x, y)
PLUS(n, s(m)) → PLUS(n, m)
F(true, x, y) → DOUBLE(y)
F(true, x, y) → AND(gt(x, y), gt(y, s(s(0))))
F(true, x, y) → F(and(gt(x, y), gt(y, s(s(0)))), plus(s(0), x), double(y))

The TRS R consists of the following rules:

f(true, x, y) → f(and(gt(x, y), gt(y, s(s(0)))), plus(s(0), x), double(y))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
and(x, true) → x
and(x, false) → false
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
double(0) → 0
double(s(x)) → s(s(double(x)))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

F(true, x, y) → PLUS(s(0), x)
DOUBLE(s(x)) → DOUBLE(x)
GT(s(u), s(v)) → GT(u, v)
F(true, x, y) → GT(y, s(s(0)))
F(true, x, y) → GT(x, y)
PLUS(n, s(m)) → PLUS(n, m)
F(true, x, y) → DOUBLE(y)
F(true, x, y) → AND(gt(x, y), gt(y, s(s(0))))
F(true, x, y) → F(and(gt(x, y), gt(y, s(s(0)))), plus(s(0), x), double(y))

The TRS R consists of the following rules:

f(true, x, y) → f(and(gt(x, y), gt(y, s(s(0)))), plus(s(0), x), double(y))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
and(x, true) → x
and(x, false) → false
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
double(0) → 0
double(s(x)) → s(s(double(x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 4 SCCs with 5 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ 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:

f(true, x, y) → f(and(gt(x, y), gt(y, s(s(0)))), plus(s(0), x), double(y))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
and(x, true) → x
and(x, false) → false
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
double(0) → 0
double(s(x)) → s(s(double(x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


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: Polynomial interpretation [25,35]:

POL(DOUBLE(x1)) = (2)x_1   
POL(s(x1)) = 1/4 + (7/2)x_1   
The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented: none



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

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

f(true, x, y) → f(and(gt(x, y), gt(y, s(s(0)))), plus(s(0), x), double(y))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
and(x, true) → x
and(x, false) → false
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
double(0) → 0
double(s(x)) → s(s(double(x)))

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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

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

PLUS(n, s(m)) → PLUS(n, m)

The TRS R consists of the following rules:

f(true, x, y) → f(and(gt(x, y), gt(y, s(s(0)))), plus(s(0), x), double(y))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
and(x, true) → x
and(x, false) → false
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
double(0) → 0
double(s(x)) → s(s(double(x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


PLUS(n, s(m)) → PLUS(n, m)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(PLUS(x1, x2)) = (2)x_2   
POL(s(x1)) = 1/4 + (7/2)x_1   
The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented: none



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

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

f(true, x, y) → f(and(gt(x, y), gt(y, s(s(0)))), plus(s(0), x), double(y))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
and(x, true) → x
and(x, false) → false
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
double(0) → 0
double(s(x)) → s(s(double(x)))

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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

GT(s(u), s(v)) → GT(u, v)

The TRS R consists of the following rules:

f(true, x, y) → f(and(gt(x, y), gt(y, s(s(0)))), plus(s(0), x), double(y))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
and(x, true) → x
and(x, false) → false
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
double(0) → 0
double(s(x)) → s(s(double(x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


GT(s(u), s(v)) → GT(u, v)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(s(x1)) = 1/2 + (13/4)x_1   
POL(GT(x1, x2)) = (15/4)x_2   
The value of delta used in the strict ordering is 15/8.
The following usable rules [17] were oriented: none



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

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

f(true, x, y) → f(and(gt(x, y), gt(y, s(s(0)))), plus(s(0), x), double(y))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
and(x, true) → x
and(x, false) → false
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
double(0) → 0
double(s(x)) → s(s(double(x)))

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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

F(true, x, y) → F(and(gt(x, y), gt(y, s(s(0)))), plus(s(0), x), double(y))

The TRS R consists of the following rules:

f(true, x, y) → f(and(gt(x, y), gt(y, s(s(0)))), plus(s(0), x), double(y))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
and(x, true) → x
and(x, false) → false
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
double(0) → 0
double(s(x)) → s(s(double(x)))

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