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(equal(min(x, y), y), x, y)
cond(true, x, y) → s(minus(x, s(y)))
min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

minus(x, y) → cond(equal(min(x, y), y), x, y)
cond(true, x, y) → s(minus(x, s(y)))
min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)

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) → MIN(x, y)
EQUAL(s(x), s(y)) → EQUAL(x, y)
MIN(s(u), s(v)) → MIN(u, v)
COND(true, x, y) → MINUS(x, s(y))
MINUS(x, y) → EQUAL(min(x, y), y)
MINUS(x, y) → COND(equal(min(x, y), y), x, y)

The TRS R consists of the following rules:

minus(x, y) → cond(equal(min(x, y), y), x, y)
cond(true, x, y) → s(minus(x, s(y)))
min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)

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) → MIN(x, y)
EQUAL(s(x), s(y)) → EQUAL(x, y)
MIN(s(u), s(v)) → MIN(u, v)
COND(true, x, y) → MINUS(x, s(y))
MINUS(x, y) → EQUAL(min(x, y), y)
MINUS(x, y) → COND(equal(min(x, y), y), x, y)

The TRS R consists of the following rules:

minus(x, y) → cond(equal(min(x, y), y), x, y)
cond(true, x, y) → s(minus(x, s(y)))
min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)

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:

EQUAL(s(x), s(y)) → EQUAL(x, y)

The TRS R consists of the following rules:

minus(x, y) → cond(equal(min(x, y), y), x, y)
cond(true, x, y) → s(minus(x, s(y)))
min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)

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.


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

POL(EQUAL(x1, x2)) = (1/2)x_1   
POL(s(x1)) = 1/2 + (4)x_1   
The value of delta used in the strict ordering is 1/4.
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:

minus(x, y) → cond(equal(min(x, y), y), x, y)
cond(true, x, y) → s(minus(x, s(y)))
min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)

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:

MIN(s(u), s(v)) → MIN(u, v)

The TRS R consists of the following rules:

minus(x, y) → cond(equal(min(x, y), y), x, y)
cond(true, x, y) → s(minus(x, s(y)))
min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)

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.


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

POL(MIN(x1, x2)) = (1/2)x_1   
POL(s(x1)) = 1/2 + (4)x_1   
The value of delta used in the strict ordering is 1/4.
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:

minus(x, y) → cond(equal(min(x, y), y), x, y)
cond(true, x, y) → s(minus(x, s(y)))
min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)

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:

COND(true, x, y) → MINUS(x, s(y))
MINUS(x, y) → COND(equal(min(x, y), y), x, y)

The TRS R consists of the following rules:

minus(x, y) → cond(equal(min(x, y), y), x, y)
cond(true, x, y) → s(minus(x, s(y)))
min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)

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