Termination w.r.t. Q proof of /tmp/tmpNHT6pO/ex05.trs

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:

minus₂(x, x) -> 0
minus₂(x, y) -> cond₃(equal₂(min₂(x, y), y), x, y)
cond₃(true, x, y) -> s₁(minus₂(x, s₁(y)))
min₂(0, v) -> 0
min₂(u, 0) -> 0
min₂(s₁(u), s₁(v)) -> s₁(min₂(u, v))
equal₂(0, 0) -> true
equal₂(s₁(x), 0) -> false
equal₂(0, s₁(y)) -> false
equal₂(s₁(x), s₁(y)) -> equal₂(x, y)

The set Q consists of the following terms:

min₂(s₁(x0), s₁(x1))
equal₂(s₁(x0), 0)
equal₂(0, 0)
minus₂(x0, x1)
min₂(x0, 0)
equal₂(s₁(x0), s₁(x1))
min₂(0, x0)
cond₃(true, x0, x1)
equal₂(0, s₁(x0))

↳ QTRS

  ↳ DependencyPairsProof

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

minus₂(x, x) -> 0
minus₂(x, y) -> cond₃(equal₂(min₂(x, y), y), x, y)
cond₃(true, x, y) -> s₁(minus₂(x, s₁(y)))
min₂(0, v) -> 0
min₂(u, 0) -> 0
min₂(s₁(u), s₁(v)) -> s₁(min₂(u, v))
equal₂(0, 0) -> true
equal₂(s₁(x), 0) -> false
equal₂(0, s₁(y)) -> false
equal₂(s₁(x), s₁(y)) -> equal₂(x, y)

The set Q consists of the following terms:

min₂(s₁(x0), s₁(x1))
equal₂(s₁(x0), 0)
equal₂(0, 0)
minus₂(x0, x1)
min₂(x0, 0)
equal₂(s₁(x0), s₁(x1))
min₂(0, x0)
cond₃(true, x0, x1)
equal₂(0, s₁(x0))

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

EQUAL₂(s₁(x), s₁(y)) -> EQUAL₂(x, y)
MIN₂(s₁(u), s₁(v)) -> MIN₂(u, v)
COND₃(true, x, y) -> MINUS₂(x, s₁(y))
MINUS₂(x, y) -> COND₃(equal₂(min₂(x, y), y), x, y)
MINUS₂(x, y) -> EQUAL₂(min₂(x, y), y)
MINUS₂(x, y) -> MIN₂(x, y)

The TRS R consists of the following rules:

minus₂(x, x) -> 0
minus₂(x, y) -> cond₃(equal₂(min₂(x, y), y), x, y)
cond₃(true, x, y) -> s₁(minus₂(x, s₁(y)))
min₂(0, v) -> 0
min₂(u, 0) -> 0
min₂(s₁(u), s₁(v)) -> s₁(min₂(u, v))
equal₂(0, 0) -> true
equal₂(s₁(x), 0) -> false
equal₂(0, s₁(y)) -> false
equal₂(s₁(x), s₁(y)) -> equal₂(x, y)

The set Q consists of the following terms:

min₂(s₁(x0), s₁(x1))
equal₂(s₁(x0), 0)
equal₂(0, 0)
minus₂(x0, x1)
min₂(x0, 0)
equal₂(s₁(x0), s₁(x1))
min₂(0, x0)
cond₃(true, x0, x1)
equal₂(0, s₁(x0))

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:

EQUAL₂(s₁(x), s₁(y)) -> EQUAL₂(x, y)
MIN₂(s₁(u), s₁(v)) -> MIN₂(u, v)
COND₃(true, x, y) -> MINUS₂(x, s₁(y))
MINUS₂(x, y) -> COND₃(equal₂(min₂(x, y), y), x, y)
MINUS₂(x, y) -> EQUAL₂(min₂(x, y), y)
MINUS₂(x, y) -> MIN₂(x, y)

The TRS R consists of the following rules:

minus₂(x, x) -> 0
minus₂(x, y) -> cond₃(equal₂(min₂(x, y), y), x, y)
cond₃(true, x, y) -> s₁(minus₂(x, s₁(y)))
min₂(0, v) -> 0
min₂(u, 0) -> 0
min₂(s₁(u), s₁(v)) -> s₁(min₂(u, v))
equal₂(0, 0) -> true
equal₂(s₁(x), 0) -> false
equal₂(0, s₁(y)) -> false
equal₂(s₁(x), s₁(y)) -> equal₂(x, y)

The set Q consists of the following terms:

min₂(s₁(x0), s₁(x1))
equal₂(s₁(x0), 0)
equal₂(0, 0)
minus₂(x0, x1)
min₂(x0, 0)
equal₂(s₁(x0), s₁(x1))
min₂(0, x0)
cond₃(true, x0, x1)
equal₂(0, s₁(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

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

EQUAL₂(s₁(x), s₁(y)) -> EQUAL₂(x, y)

The TRS R consists of the following rules:

minus₂(x, x) -> 0
minus₂(x, y) -> cond₃(equal₂(min₂(x, y), y), x, y)
cond₃(true, x, y) -> s₁(minus₂(x, s₁(y)))
min₂(0, v) -> 0
min₂(u, 0) -> 0
min₂(s₁(u), s₁(v)) -> s₁(min₂(u, v))
equal₂(0, 0) -> true
equal₂(s₁(x), 0) -> false
equal₂(0, s₁(y)) -> false
equal₂(s₁(x), s₁(y)) -> equal₂(x, y)

The set Q consists of the following terms:

min₂(s₁(x0), s₁(x1))
equal₂(s₁(x0), 0)
equal₂(0, 0)
minus₂(x0, x1)
min₂(x0, 0)
equal₂(s₁(x0), s₁(x1))
min₂(0, x0)
cond₃(true, x0, x1)
equal₂(0, s₁(x0))

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 [13] we can delete all non-usable rules [14] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

          ↳ QDP

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

EQUAL₂(s₁(x), s₁(y)) -> EQUAL₂(x, y)

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

min₂(s₁(x0), s₁(x1))
equal₂(s₁(x0), 0)
equal₂(0, 0)
minus₂(x0, x1)
min₂(x0, 0)
equal₂(s₁(x0), s₁(x1))
min₂(0, x0)
cond₃(true, x0, x1)
equal₂(0, s₁(x0))

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.

min₂(s₁(x0), s₁(x1))
equal₂(s₁(x0), 0)
equal₂(0, 0)
minus₂(x0, x1)
min₂(x0, 0)
equal₂(s₁(x0), s₁(x1))
min₂(0, x0)
cond₃(true, x0, x1)
equal₂(0, s₁(x0))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

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

EQUAL₂(s₁(x), s₁(y)) -> EQUAL₂(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [16] together with the size-change analysis [27] 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:

EQUAL₂(s₁(x), s₁(y)) -> EQUAL₂(x, y)
The graph contains the following edges 1 > 1, 2 > 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

MIN₂(s₁(u), s₁(v)) -> MIN₂(u, v)

The TRS R consists of the following rules:

minus₂(x, x) -> 0
minus₂(x, y) -> cond₃(equal₂(min₂(x, y), y), x, y)
cond₃(true, x, y) -> s₁(minus₂(x, s₁(y)))
min₂(0, v) -> 0
min₂(u, 0) -> 0
min₂(s₁(u), s₁(v)) -> s₁(min₂(u, v))
equal₂(0, 0) -> true
equal₂(s₁(x), 0) -> false
equal₂(0, s₁(y)) -> false
equal₂(s₁(x), s₁(y)) -> equal₂(x, y)

The set Q consists of the following terms:

min₂(s₁(x0), s₁(x1))
equal₂(s₁(x0), 0)
equal₂(0, 0)
minus₂(x0, x1)
min₂(x0, 0)
equal₂(s₁(x0), s₁(x1))
min₂(0, x0)
cond₃(true, x0, x1)
equal₂(0, s₁(x0))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

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

MIN₂(s₁(u), s₁(v)) -> MIN₂(u, v)

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

min₂(s₁(x0), s₁(x1))
equal₂(s₁(x0), 0)
equal₂(0, 0)
minus₂(x0, x1)
min₂(x0, 0)
equal₂(s₁(x0), s₁(x1))
min₂(0, x0)
cond₃(true, x0, x1)
equal₂(0, s₁(x0))

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.

min₂(s₁(x0), s₁(x1))
equal₂(s₁(x0), 0)
equal₂(0, 0)
minus₂(x0, x1)
min₂(x0, 0)
equal₂(s₁(x0), s₁(x1))
min₂(0, x0)
cond₃(true, x0, x1)
equal₂(0, s₁(x0))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

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

MIN₂(s₁(u), s₁(v)) -> MIN₂(u, v)

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

MIN₂(s₁(u), s₁(v)) -> MIN₂(u, v)
The graph contains the following edges 1 > 1, 2 > 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

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

COND₃(true, x, y) -> MINUS₂(x, s₁(y))
MINUS₂(x, y) -> COND₃(equal₂(min₂(x, y), y), x, y)

The TRS R consists of the following rules:

minus₂(x, x) -> 0
minus₂(x, y) -> cond₃(equal₂(min₂(x, y), y), x, y)
cond₃(true, x, y) -> s₁(minus₂(x, s₁(y)))
min₂(0, v) -> 0
min₂(u, 0) -> 0
min₂(s₁(u), s₁(v)) -> s₁(min₂(u, v))
equal₂(0, 0) -> true
equal₂(s₁(x), 0) -> false
equal₂(0, s₁(y)) -> false
equal₂(s₁(x), s₁(y)) -> equal₂(x, y)

The set Q consists of the following terms:

min₂(s₁(x0), s₁(x1))
equal₂(s₁(x0), 0)
equal₂(0, 0)
minus₂(x0, x1)
min₂(x0, 0)
equal₂(s₁(x0), s₁(x1))
min₂(0, x0)
cond₃(true, x0, x1)
equal₂(0, s₁(x0))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

            ↳ UsableRulesProof

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

COND₃(true, x, y) -> MINUS₂(x, s₁(y))
MINUS₂(x, y) -> COND₃(equal₂(min₂(x, y), y), x, y)

The TRS R consists of the following rules:

min₂(0, v) -> 0
min₂(u, 0) -> 0
min₂(s₁(u), s₁(v)) -> s₁(min₂(u, v))
equal₂(0, 0) -> true
equal₂(s₁(x), 0) -> false
equal₂(0, s₁(y)) -> false
equal₂(s₁(x), s₁(y)) -> equal₂(x, y)

The set Q consists of the following terms:

min₂(s₁(x0), s₁(x1))
equal₂(s₁(x0), 0)
equal₂(0, 0)
minus₂(x0, x1)
min₂(x0, 0)
equal₂(s₁(x0), s₁(x1))
min₂(0, x0)
cond₃(true, x0, x1)
equal₂(0, s₁(x0))

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.

minus₂(x0, x1)
cond₃(true, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

            ↳ UsableRulesProof

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

COND₃(true, x, y) -> MINUS₂(x, s₁(y))
MINUS₂(x, y) -> COND₃(equal₂(min₂(x, y), y), x, y)

The TRS R consists of the following rules:

min₂(0, v) -> 0
min₂(u, 0) -> 0
min₂(s₁(u), s₁(v)) -> s₁(min₂(u, v))
equal₂(0, 0) -> true
equal₂(s₁(x), 0) -> false
equal₂(0, s₁(y)) -> false
equal₂(s₁(x), s₁(y)) -> equal₂(x, y)

The set Q consists of the following terms:

min₂(s₁(x0), s₁(x1))
equal₂(s₁(x0), 0)
equal₂(0, 0)
min₂(x0, 0)
equal₂(s₁(x0), s₁(x1))
min₂(0, x0)
equal₂(0, s₁(x0))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [13] the rule MINUS₂(x, y) -> COND₃(equal₂(min₂(x, y), y), x, y) at position [0] we obtained the following new rules:

MINUS₂(0, x0) -> COND₃(equal₂(0, x0), 0, x0)
MINUS₂(x0, 0) -> COND₃(equal₂(0, 0), x0, 0)
MINUS₂(s₁(x0), s₁(x1)) -> COND₃(equal₂(s₁(min₂(x0, x1)), s₁(x1)), s₁(x0), s₁(x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

            ↳ UsableRulesProof

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

MINUS₂(x0, 0) -> COND₃(equal₂(0, 0), x0, 0)
COND₃(true, x, y) -> MINUS₂(x, s₁(y))
MINUS₂(0, x0) -> COND₃(equal₂(0, x0), 0, x0)
MINUS₂(s₁(x0), s₁(x1)) -> COND₃(equal₂(s₁(min₂(x0, x1)), s₁(x1)), s₁(x0), s₁(x1))

The TRS R consists of the following rules:

min₂(0, v) -> 0
min₂(u, 0) -> 0
min₂(s₁(u), s₁(v)) -> s₁(min₂(u, v))
equal₂(0, 0) -> true
equal₂(s₁(x), 0) -> false
equal₂(0, s₁(y)) -> false
equal₂(s₁(x), s₁(y)) -> equal₂(x, y)

The set Q consists of the following terms:

min₂(s₁(x0), s₁(x1))
equal₂(s₁(x0), 0)
equal₂(0, 0)
min₂(x0, 0)
equal₂(s₁(x0), s₁(x1))
min₂(0, x0)
equal₂(0, s₁(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

            ↳ UsableRulesProof

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

COND₃(true, x, y) -> MINUS₂(x, s₁(y))
MINUS₂(0, x0) -> COND₃(equal₂(0, x0), 0, x0)
MINUS₂(s₁(x0), s₁(x1)) -> COND₃(equal₂(s₁(min₂(x0, x1)), s₁(x1)), s₁(x0), s₁(x1))

The TRS R consists of the following rules:

min₂(0, v) -> 0
min₂(u, 0) -> 0
min₂(s₁(u), s₁(v)) -> s₁(min₂(u, v))
equal₂(0, 0) -> true
equal₂(s₁(x), 0) -> false
equal₂(0, s₁(y)) -> false
equal₂(s₁(x), s₁(y)) -> equal₂(x, y)

The set Q consists of the following terms:

min₂(s₁(x0), s₁(x1))
equal₂(s₁(x0), 0)
equal₂(0, 0)
min₂(x0, 0)
equal₂(s₁(x0), s₁(x1))
min₂(0, x0)
equal₂(0, s₁(x0))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [13] the rule MINUS₂(0, x0) -> COND₃(equal₂(0, x0), 0, x0) at position [0] we obtained the following new rules:

MINUS₂(0, s₁(x0)) -> COND₃(false, 0, s₁(x0))
MINUS₂(0, 0) -> COND₃(true, 0, 0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

            ↳ UsableRulesProof

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

COND₃(true, x, y) -> MINUS₂(x, s₁(y))
MINUS₂(0, s₁(x0)) -> COND₃(false, 0, s₁(x0))
MINUS₂(0, 0) -> COND₃(true, 0, 0)
MINUS₂(s₁(x0), s₁(x1)) -> COND₃(equal₂(s₁(min₂(x0, x1)), s₁(x1)), s₁(x0), s₁(x1))

The TRS R consists of the following rules:

min₂(0, v) -> 0
min₂(u, 0) -> 0
min₂(s₁(u), s₁(v)) -> s₁(min₂(u, v))
equal₂(0, 0) -> true
equal₂(s₁(x), 0) -> false
equal₂(0, s₁(y)) -> false
equal₂(s₁(x), s₁(y)) -> equal₂(x, y)

The set Q consists of the following terms:

min₂(s₁(x0), s₁(x1))
equal₂(s₁(x0), 0)
equal₂(0, 0)
min₂(x0, 0)
equal₂(s₁(x0), s₁(x1))
min₂(0, x0)
equal₂(0, s₁(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Instantiation

            ↳ UsableRulesProof

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

COND₃(true, x, y) -> MINUS₂(x, s₁(y))
MINUS₂(s₁(x0), s₁(x1)) -> COND₃(equal₂(s₁(min₂(x0, x1)), s₁(x1)), s₁(x0), s₁(x1))

The TRS R consists of the following rules:

min₂(0, v) -> 0
min₂(u, 0) -> 0
min₂(s₁(u), s₁(v)) -> s₁(min₂(u, v))
equal₂(0, 0) -> true
equal₂(s₁(x), 0) -> false
equal₂(0, s₁(y)) -> false
equal₂(s₁(x), s₁(y)) -> equal₂(x, y)

The set Q consists of the following terms:

min₂(s₁(x0), s₁(x1))
equal₂(s₁(x0), 0)
equal₂(0, 0)
min₂(x0, 0)
equal₂(s₁(x0), s₁(x1))
min₂(0, x0)
equal₂(0, s₁(x0))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [13] the rule COND₃(true, x, y) -> MINUS₂(x, s₁(y)) we obtained the following new rules:

COND₃(true, s₁(z0), s₁(z1)) -> MINUS₂(s₁(z0), s₁(s₁(z1)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Instantiation

                                      ↳ QDP

                                        ↳ Instantiation

            ↳ UsableRulesProof

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

COND₃(true, s₁(z0), s₁(z1)) -> MINUS₂(s₁(z0), s₁(s₁(z1)))
MINUS₂(s₁(x0), s₁(x1)) -> COND₃(equal₂(s₁(min₂(x0, x1)), s₁(x1)), s₁(x0), s₁(x1))

The TRS R consists of the following rules:

min₂(0, v) -> 0
min₂(u, 0) -> 0
min₂(s₁(u), s₁(v)) -> s₁(min₂(u, v))
equal₂(0, 0) -> true
equal₂(s₁(x), 0) -> false
equal₂(0, s₁(y)) -> false
equal₂(s₁(x), s₁(y)) -> equal₂(x, y)

The set Q consists of the following terms:

min₂(s₁(x0), s₁(x1))
equal₂(s₁(x0), 0)
equal₂(0, 0)
min₂(x0, 0)
equal₂(s₁(x0), s₁(x1))
min₂(0, x0)
equal₂(0, s₁(x0))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [13] the rule MINUS₂(s₁(x0), s₁(x1)) -> COND₃(equal₂(s₁(min₂(x0, x1)), s₁(x1)), s₁(x0), s₁(x1)) we obtained the following new rules:

MINUS₂(s₁(z0), s₁(s₁(z1))) -> COND₃(equal₂(s₁(min₂(z0, s₁(z1))), s₁(s₁(z1))), s₁(z0), s₁(s₁(z1)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Instantiation

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

            ↳ UsableRulesProof

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

COND₃(true, s₁(z0), s₁(z1)) -> MINUS₂(s₁(z0), s₁(s₁(z1)))
MINUS₂(s₁(z0), s₁(s₁(z1))) -> COND₃(equal₂(s₁(min₂(z0, s₁(z1))), s₁(s₁(z1))), s₁(z0), s₁(s₁(z1)))

The TRS R consists of the following rules:

min₂(0, v) -> 0
min₂(u, 0) -> 0
min₂(s₁(u), s₁(v)) -> s₁(min₂(u, v))
equal₂(0, 0) -> true
equal₂(s₁(x), 0) -> false
equal₂(0, s₁(y)) -> false
equal₂(s₁(x), s₁(y)) -> equal₂(x, y)

The set Q consists of the following terms:

min₂(s₁(x0), s₁(x1))
equal₂(s₁(x0), 0)
equal₂(0, 0)
min₂(x0, 0)
equal₂(s₁(x0), s₁(x1))
min₂(0, x0)
equal₂(0, s₁(x0))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

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

COND₃(true, x, y) -> MINUS₂(x, s₁(y))
MINUS₂(x, y) -> COND₃(equal₂(min₂(x, y), y), x, y)

The TRS R consists of the following rules:

min₂(0, v) -> 0
min₂(u, 0) -> 0
min₂(s₁(u), s₁(v)) -> s₁(min₂(u, v))
equal₂(0, 0) -> true
equal₂(s₁(x), 0) -> false
equal₂(0, s₁(y)) -> false
equal₂(s₁(x), s₁(y)) -> equal₂(x, y)

The set Q consists of the following terms:

min₂(s₁(x0), s₁(x1))
equal₂(s₁(x0), 0)
equal₂(0, 0)
minus₂(x0, x1)
min₂(x0, 0)
equal₂(s₁(x0), s₁(x1))
min₂(0, x0)
cond₃(true, x0, x1)
equal₂(0, s₁(x0))

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.

minus₂(x0, x1)
cond₃(true, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

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

COND₃(true, x, y) -> MINUS₂(x, s₁(y))
MINUS₂(x, y) -> COND₃(equal₂(min₂(x, y), y), x, y)

The TRS R consists of the following rules:

min₂(0, v) -> 0
min₂(u, 0) -> 0
min₂(s₁(u), s₁(v)) -> s₁(min₂(u, v))
equal₂(0, 0) -> true
equal₂(s₁(x), 0) -> false
equal₂(0, s₁(y)) -> false
equal₂(s₁(x), s₁(y)) -> equal₂(x, y)

The set Q consists of the following terms:

min₂(s₁(x0), s₁(x1))
equal₂(s₁(x0), 0)
equal₂(0, 0)
min₂(x0, 0)
equal₂(s₁(x0), s₁(x1))
min₂(0, x0)
equal₂(0, s₁(x0))

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