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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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, z) → GT(x, plus(y, z))
F(true, x, y, z) → F(gt(x, plus(y, z)), x, s(y), z)
F(true, x, y, z) → F(gt(x, plus(y, z)), x, y, s(z))
GT(s(u), s(v)) → GT(u, v)
PLUS(n, s(m)) → PLUS(n, m)
F(true, x, y, z) → PLUS(y, z)

The TRS R consists of the following rules:

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

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, z) → GT(x, plus(y, z))
F(true, x, y, z) → F(gt(x, plus(y, z)), x, s(y), z)
F(true, x, y, z) → F(gt(x, plus(y, z)), x, y, s(z))
GT(s(u), s(v)) → GT(u, v)
PLUS(n, s(m)) → PLUS(n, m)
F(true, x, y, z) → PLUS(y, z)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ 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, z) → f(gt(x, plus(y, z)), x, s(y), z)
f(true, x, y, z) → f(gt(x, plus(y, z)), x, y, s(z))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

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)) = 5/4 + (15/4)x_1   
POL(GT(x1, x2)) = x_1 + (13/4)x_2   
The value of delta used in the strict ordering is 85/16.
The following usable rules [17] were oriented: none



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

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

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

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

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

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

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

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

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

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

F(true, x, y, z) → F(gt(x, plus(y, z)), x, s(y), z)
F(true, x, y, z) → F(gt(x, plus(y, z)), x, y, s(z))

The TRS R consists of the following rules:

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

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