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

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

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

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

Q is empty.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof

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

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

The set Q consists of the following terms:

minus(x0, x1)
cond(x0, x1, x0)
min(0, x0)
min(x0, 0)
min(s(x0), s(x1))


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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(x0, x1)
cond(x0, x1, x0)
min(0, x0)
min(x0, 0)
min(s(x0), s(x1))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(x0, x1)
cond(x0, x1, x0)
min(0, x0)
min(x0, 0)
min(s(x0), s(x1))

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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
              ↳ 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(min(x, y), x, y)
cond(y, x, y) → s(minus(x, s(y)))
min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))

The set Q consists of the following terms:

minus(x0, x1)
cond(x0, x1, x0)
min(0, x0)
min(x0, 0)
min(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
              ↳ QDP

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

MIN(s(u), s(v)) → MIN(u, v)

R is empty.
The set Q consists of the following terms:

minus(x0, x1)
cond(x0, x1, x0)
min(0, x0)
min(x0, 0)
min(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

minus(x0, x1)
cond(x0, x1, x0)
min(0, x0)
min(x0, 0)
min(s(x0), s(x1))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ QDPSizeChangeProof
              ↳ QDP

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

MIN(s(u), s(v)) → MIN(u, v)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: 



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                ↳ UsableRulesProof

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(x0, x1)
cond(x0, x1, x0)
min(0, x0)
min(x0, 0)
min(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                ↳ UsableRulesProof

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

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

The TRS R consists of the following rules:

min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))

The set Q consists of the following terms:

minus(x0, x1)
cond(x0, x1, x0)
min(0, x0)
min(x0, 0)
min(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

minus(x0, x1)
cond(x0, x1, x0)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                ↳ UsableRulesProof

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

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

The TRS R consists of the following rules:

min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule MINUS(x, y) → COND(min(x, y), x, y) at position [] we obtained the following new rules:

MINUS(s(x0), s(x1)) → COND(s(min(x0, x1)), s(x0), s(x1))
MINUS(x0, 0) → COND(0, x0, 0)
MINUS(0, x0) → COND(0, 0, x0)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                ↳ UsableRulesProof

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

MINUS(0, x0) → COND(0, 0, x0)
MINUS(x0, 0) → COND(0, x0, 0)
COND(y, x, y) → MINUS(x, s(y))
MINUS(s(x0), s(x1)) → COND(s(min(x0, x1)), s(x0), s(x1))

The TRS R consists of the following rules:

min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Instantiation
                ↳ UsableRulesProof

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

MINUS(0, x0) → COND(0, 0, x0)
COND(y, x, y) → MINUS(x, s(y))
MINUS(s(x0), s(x1)) → COND(s(min(x0, x1)), s(x0), s(x1))

The TRS R consists of the following rules:

min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule COND(y, x, y) → MINUS(x, s(y)) we obtained the following new rules:

COND(0, 0, 0) → MINUS(0, s(0))
COND(s(y_0), s(z0), s(y_0)) → MINUS(s(z0), s(s(y_0)))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Instantiation
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                ↳ UsableRulesProof

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

COND(0, 0, 0) → MINUS(0, s(0))
MINUS(0, x0) → COND(0, 0, x0)
COND(s(y_0), s(z0), s(y_0)) → MINUS(s(z0), s(s(y_0)))
MINUS(s(x0), s(x1)) → COND(s(min(x0, x1)), s(x0), s(x1))

The TRS R consists of the following rules:

min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Instantiation
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ AND
                                        ↳ QDP
                                          ↳ UsableRulesProof
                                        ↳ QDP
                ↳ UsableRulesProof

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

COND(0, 0, 0) → MINUS(0, s(0))
MINUS(0, x0) → COND(0, 0, x0)

The TRS R consists of the following rules:

min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Instantiation
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ AND
                                        ↳ QDP
                                          ↳ UsableRulesProof
                                            ↳ QDP
                                              ↳ QReductionProof
                                        ↳ QDP
                ↳ UsableRulesProof

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

COND(0, 0, 0) → MINUS(0, s(0))
MINUS(0, x0) → COND(0, 0, x0)

R is empty.
The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Instantiation
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ AND
                                        ↳ QDP
                                          ↳ UsableRulesProof
                                            ↳ QDP
                                              ↳ QReductionProof
                                                ↳ QDP
                                                  ↳ Instantiation
                                        ↳ QDP
                ↳ UsableRulesProof

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

COND(0, 0, 0) → MINUS(0, s(0))
MINUS(0, x0) → COND(0, 0, x0)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule MINUS(0, x0) → COND(0, 0, x0) we obtained the following new rules:

MINUS(0, s(0)) → COND(0, 0, s(0))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Instantiation
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ AND
                                        ↳ QDP
                                          ↳ UsableRulesProof
                                            ↳ QDP
                                              ↳ QReductionProof
                                                ↳ QDP
                                                  ↳ Instantiation
                                                    ↳ QDP
                                                      ↳ DependencyGraphProof
                                        ↳ QDP
                ↳ UsableRulesProof

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

COND(0, 0, 0) → MINUS(0, s(0))
MINUS(0, s(0)) → COND(0, 0, s(0))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 2 less nodes.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Instantiation
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ AND
                                        ↳ QDP
                                        ↳ QDP
                                          ↳ Instantiation
                ↳ UsableRulesProof

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

COND(s(y_0), s(z0), s(y_0)) → MINUS(s(z0), s(s(y_0)))
MINUS(s(x0), s(x1)) → COND(s(min(x0, x1)), s(x0), s(x1))

The TRS R consists of the following rules:

min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule MINUS(s(x0), s(x1)) → COND(s(min(x0, x1)), s(x0), s(x1)) we obtained the following new rules:

MINUS(s(z1), s(s(z0))) → COND(s(min(z1, s(z0))), s(z1), s(s(z0)))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Instantiation
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ AND
                                        ↳ QDP
                                        ↳ QDP
                                          ↳ Instantiation
                                            ↳ QDP
                                              ↳ Instantiation
                ↳ UsableRulesProof

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

MINUS(s(z1), s(s(z0))) → COND(s(min(z1, s(z0))), s(z1), s(s(z0)))
COND(s(y_0), s(z0), s(y_0)) → MINUS(s(z0), s(s(y_0)))

The TRS R consists of the following rules:

min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule COND(s(y_0), s(z0), s(y_0)) → MINUS(s(z0), s(s(y_0))) we obtained the following new rules:

COND(s(s(z1)), s(z0), s(s(z1))) → MINUS(s(z0), s(s(s(z1))))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Instantiation
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ AND
                                        ↳ QDP
                                        ↳ QDP
                                          ↳ Instantiation
                                            ↳ QDP
                                              ↳ Instantiation
                                                ↳ QDP
                                                  ↳ Instantiation
                ↳ UsableRulesProof

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

COND(s(s(z1)), s(z0), s(s(z1))) → MINUS(s(z0), s(s(s(z1))))
MINUS(s(z1), s(s(z0))) → COND(s(min(z1, s(z0))), s(z1), s(s(z0)))

The TRS R consists of the following rules:

min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule MINUS(s(z1), s(s(z0))) → COND(s(min(z1, s(z0))), s(z1), s(s(z0))) we obtained the following new rules:

MINUS(s(z1), s(s(s(z0)))) → COND(s(min(z1, s(s(z0)))), s(z1), s(s(s(z0))))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Instantiation
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ AND
                                        ↳ QDP
                                        ↳ QDP
                                          ↳ Instantiation
                                            ↳ QDP
                                              ↳ Instantiation
                                                ↳ QDP
                                                  ↳ Instantiation
                                                    ↳ QDP
                ↳ UsableRulesProof

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

MINUS(s(z1), s(s(s(z0)))) → COND(s(min(z1, s(s(z0)))), s(z1), s(s(s(z0))))
COND(s(s(z1)), s(z0), s(s(z1))) → MINUS(s(z0), s(s(s(z1))))

The TRS R consists of the following rules:

min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof

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

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

The TRS R consists of the following rules:

min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))

The set Q consists of the following terms:

minus(x0, x1)
cond(x0, x1, x0)
min(0, x0)
min(x0, 0)
min(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

minus(x0, x1)
cond(x0, x1, x0)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ MNOCProof

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

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

The TRS R consists of the following rules:

min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [17] to decrease Q to the empty set.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ MNOCProof
                          ↳ QDP

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

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

The TRS R consists of the following rules:

min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))

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