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:

minus(x, y) → cond(gt(x, y), x, y)
cond(false, x, y) → 0
cond(true, x, y) → s(minus(x, s(y)))
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:

minus(x, y) → cond(gt(x, y), x, y)
cond(false, x, y) → 0
cond(true, x, y) → s(minus(x, s(y)))
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:

MINUS(x, y) → COND(gt(x, y), x, y)
MINUS(x, y) → GT(x, y)
GT(s(u), s(v)) → GT(u, v)
COND(true, x, y) → MINUS(x, s(y))

The TRS R consists of the following rules:

minus(x, y) → cond(gt(x, y), x, y)
cond(false, x, y) → 0
cond(true, x, y) → s(minus(x, s(y)))
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:

MINUS(x, y) → COND(gt(x, y), x, y)
MINUS(x, y) → GT(x, y)
GT(s(u), s(v)) → GT(u, v)
COND(true, x, y) → MINUS(x, s(y))

The TRS R consists of the following rules:

minus(x, y) → cond(gt(x, y), x, y)
cond(false, x, y) → 0
cond(true, x, y) → s(minus(x, s(y)))
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 2 SCCs with 1 less node.

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

minus(x, y) → cond(gt(x, y), x, y)
cond(false, x, y) → 0
cond(true, x, y) → s(minus(x, s(y)))
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)) = 1 + (4)x_1   
POL(GT(x1, x2)) = (3)x_2   
The value of delta used in the strict ordering is 3.
The following usable rules [17] were oriented: none



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

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

minus(x, y) → cond(gt(x, y), x, y)
cond(false, x, y) → 0
cond(true, x, y) → s(minus(x, s(y)))
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

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

MINUS(x, y) → COND(gt(x, y), x, y)
COND(true, x, y) → MINUS(x, s(y))

The TRS R consists of the following rules:

minus(x, y) → cond(gt(x, y), x, y)
cond(false, x, y) → 0
cond(true, x, y) → s(minus(x, s(y)))
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.