Termination w.r.t. Q proof of /tmp/tmpIH7549/ex14.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:

diff₂(x, y) -> cond1₃(equal₂(x, y), x, y)
cond1₃(true, x, y) -> 0
cond1₃(false, x, y) -> cond2₃(gt₂(x, y), x, y)
cond2₃(true, x, y) -> s₁(diff₂(x, s₁(y)))
cond2₃(false, x, y) -> s₁(diff₂(s₁(x), y))
gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(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:

equal₂(s₁(x0), 0)
equal₂(0, 0)
gt₂(0, x0)
gt₂(s₁(x0), s₁(x1))
cond2₃(false, x0, x1)
equal₂(s₁(x0), s₁(x1))
cond1₃(false, x0, x1)
diff₂(x0, x1)
equal₂(0, s₁(x0))
cond1₃(true, x0, x1)
gt₂(s₁(x0), 0)
cond2₃(true, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

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

diff₂(x, y) -> cond1₃(equal₂(x, y), x, y)
cond1₃(true, x, y) -> 0
cond1₃(false, x, y) -> cond2₃(gt₂(x, y), x, y)
cond2₃(true, x, y) -> s₁(diff₂(x, s₁(y)))
cond2₃(false, x, y) -> s₁(diff₂(s₁(x), y))
gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(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:

equal₂(s₁(x0), 0)
equal₂(0, 0)
gt₂(0, x0)
gt₂(s₁(x0), s₁(x1))
cond2₃(false, x0, x1)
equal₂(s₁(x0), s₁(x1))
cond1₃(false, x0, x1)
diff₂(x0, x1)
equal₂(0, s₁(x0))
cond1₃(true, x0, x1)
gt₂(s₁(x0), 0)
cond2₃(true, x0, x1)

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)
COND2₃(false, x, y) -> DIFF₂(s₁(x), y)
GT₂(s₁(u), s₁(v)) -> GT₂(u, v)
COND1₃(false, x, y) -> COND2₃(gt₂(x, y), x, y)
DIFF₂(x, y) -> EQUAL₂(x, y)
COND1₃(false, x, y) -> GT₂(x, y)
COND2₃(true, x, y) -> DIFF₂(x, s₁(y))
DIFF₂(x, y) -> COND1₃(equal₂(x, y), x, y)

The TRS R consists of the following rules:

diff₂(x, y) -> cond1₃(equal₂(x, y), x, y)
cond1₃(true, x, y) -> 0
cond1₃(false, x, y) -> cond2₃(gt₂(x, y), x, y)
cond2₃(true, x, y) -> s₁(diff₂(x, s₁(y)))
cond2₃(false, x, y) -> s₁(diff₂(s₁(x), y))
gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(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:

equal₂(s₁(x0), 0)
equal₂(0, 0)
gt₂(0, x0)
gt₂(s₁(x0), s₁(x1))
cond2₃(false, x0, x1)
equal₂(s₁(x0), s₁(x1))
cond1₃(false, x0, x1)
diff₂(x0, x1)
equal₂(0, s₁(x0))
cond1₃(true, x0, x1)
gt₂(s₁(x0), 0)
cond2₃(true, x0, x1)

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)
COND2₃(false, x, y) -> DIFF₂(s₁(x), y)
GT₂(s₁(u), s₁(v)) -> GT₂(u, v)
COND1₃(false, x, y) -> COND2₃(gt₂(x, y), x, y)
DIFF₂(x, y) -> EQUAL₂(x, y)
COND1₃(false, x, y) -> GT₂(x, y)
COND2₃(true, x, y) -> DIFF₂(x, s₁(y))
DIFF₂(x, y) -> COND1₃(equal₂(x, y), x, y)

The TRS R consists of the following rules:

diff₂(x, y) -> cond1₃(equal₂(x, y), x, y)
cond1₃(true, x, y) -> 0
cond1₃(false, x, y) -> cond2₃(gt₂(x, y), x, y)
cond2₃(true, x, y) -> s₁(diff₂(x, s₁(y)))
cond2₃(false, x, y) -> s₁(diff₂(s₁(x), y))
gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(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:

equal₂(s₁(x0), 0)
equal₂(0, 0)
gt₂(0, x0)
gt₂(s₁(x0), s₁(x1))
cond2₃(false, x0, x1)
equal₂(s₁(x0), s₁(x1))
cond1₃(false, x0, x1)
diff₂(x0, x1)
equal₂(0, s₁(x0))
cond1₃(true, x0, x1)
gt₂(s₁(x0), 0)
cond2₃(true, x0, x1)

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:

diff₂(x, y) -> cond1₃(equal₂(x, y), x, y)
cond1₃(true, x, y) -> 0
cond1₃(false, x, y) -> cond2₃(gt₂(x, y), x, y)
cond2₃(true, x, y) -> s₁(diff₂(x, s₁(y)))
cond2₃(false, x, y) -> s₁(diff₂(s₁(x), y))
gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(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:

equal₂(s₁(x0), 0)
equal₂(0, 0)
gt₂(0, x0)
gt₂(s₁(x0), s₁(x1))
cond2₃(false, x0, x1)
equal₂(s₁(x0), s₁(x1))
cond1₃(false, x0, x1)
diff₂(x0, x1)
equal₂(0, s₁(x0))
cond1₃(true, x0, x1)
gt₂(s₁(x0), 0)
cond2₃(true, x0, 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 [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:

equal₂(s₁(x0), 0)
equal₂(0, 0)
gt₂(0, x0)
gt₂(s₁(x0), s₁(x1))
cond2₃(false, x0, x1)
equal₂(s₁(x0), s₁(x1))
cond1₃(false, x0, x1)
diff₂(x0, x1)
equal₂(0, s₁(x0))
cond1₃(true, x0, x1)
gt₂(s₁(x0), 0)
cond2₃(true, x0, 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.

equal₂(s₁(x0), 0)
equal₂(0, 0)
gt₂(0, x0)
gt₂(s₁(x0), s₁(x1))
cond2₃(false, x0, x1)
equal₂(s₁(x0), s₁(x1))
cond1₃(false, x0, x1)
diff₂(x0, x1)
equal₂(0, s₁(x0))
cond1₃(true, x0, x1)
gt₂(s₁(x0), 0)
cond2₃(true, x0, x1)

↳ 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:

GT₂(s₁(u), s₁(v)) -> GT₂(u, v)

The TRS R consists of the following rules:

diff₂(x, y) -> cond1₃(equal₂(x, y), x, y)
cond1₃(true, x, y) -> 0
cond1₃(false, x, y) -> cond2₃(gt₂(x, y), x, y)
cond2₃(true, x, y) -> s₁(diff₂(x, s₁(y)))
cond2₃(false, x, y) -> s₁(diff₂(s₁(x), y))
gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(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:

equal₂(s₁(x0), 0)
equal₂(0, 0)
gt₂(0, x0)
gt₂(s₁(x0), s₁(x1))
cond2₃(false, x0, x1)
equal₂(s₁(x0), s₁(x1))
cond1₃(false, x0, x1)
diff₂(x0, x1)
equal₂(0, s₁(x0))
cond1₃(true, x0, x1)
gt₂(s₁(x0), 0)
cond2₃(true, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

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

GT₂(s₁(u), s₁(v)) -> GT₂(u, v)

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

equal₂(s₁(x0), 0)
equal₂(0, 0)
gt₂(0, x0)
gt₂(s₁(x0), s₁(x1))
cond2₃(false, x0, x1)
equal₂(s₁(x0), s₁(x1))
cond1₃(false, x0, x1)
diff₂(x0, x1)
equal₂(0, s₁(x0))
cond1₃(true, x0, x1)
gt₂(s₁(x0), 0)
cond2₃(true, x0, 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.

equal₂(s₁(x0), 0)
equal₂(0, 0)
gt₂(0, x0)
gt₂(s₁(x0), s₁(x1))
cond2₃(false, x0, x1)
equal₂(s₁(x0), s₁(x1))
cond1₃(false, x0, x1)
diff₂(x0, x1)
equal₂(0, s₁(x0))
cond1₃(true, x0, x1)
gt₂(s₁(x0), 0)
cond2₃(true, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

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

GT₂(s₁(u), s₁(v)) -> GT₂(u, v)

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

GT₂(s₁(u), s₁(v)) -> GT₂(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:

COND2₃(false, x, y) -> DIFF₂(s₁(x), y)
COND1₃(false, x, y) -> COND2₃(gt₂(x, y), x, y)
COND2₃(true, x, y) -> DIFF₂(x, s₁(y))
DIFF₂(x, y) -> COND1₃(equal₂(x, y), x, y)

The TRS R consists of the following rules:

diff₂(x, y) -> cond1₃(equal₂(x, y), x, y)
cond1₃(true, x, y) -> 0
cond1₃(false, x, y) -> cond2₃(gt₂(x, y), x, y)
cond2₃(true, x, y) -> s₁(diff₂(x, s₁(y)))
cond2₃(false, x, y) -> s₁(diff₂(s₁(x), y))
gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(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:

equal₂(s₁(x0), 0)
equal₂(0, 0)
gt₂(0, x0)
gt₂(s₁(x0), s₁(x1))
cond2₃(false, x0, x1)
equal₂(s₁(x0), s₁(x1))
cond1₃(false, x0, x1)
diff₂(x0, x1)
equal₂(0, s₁(x0))
cond1₃(true, x0, x1)
gt₂(s₁(x0), 0)
cond2₃(true, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

            ↳ UsableRulesProof

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

COND2₃(false, x, y) -> DIFF₂(s₁(x), y)
COND1₃(false, x, y) -> COND2₃(gt₂(x, y), x, y)
COND2₃(true, x, y) -> DIFF₂(x, s₁(y))
DIFF₂(x, y) -> COND1₃(equal₂(x, y), x, y)

The TRS R consists of the following rules:

gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(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:

equal₂(s₁(x0), 0)
equal₂(0, 0)
gt₂(0, x0)
gt₂(s₁(x0), s₁(x1))
cond2₃(false, x0, x1)
equal₂(s₁(x0), s₁(x1))
cond1₃(false, x0, x1)
diff₂(x0, x1)
equal₂(0, s₁(x0))
cond1₃(true, x0, x1)
gt₂(s₁(x0), 0)
cond2₃(true, x0, 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.

cond2₃(false, x0, x1)
cond1₃(false, x0, x1)
diff₂(x0, x1)
cond1₃(true, x0, x1)
cond2₃(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:

COND2₃(false, x, y) -> DIFF₂(s₁(x), y)
COND1₃(false, x, y) -> COND2₃(gt₂(x, y), x, y)
COND2₃(true, x, y) -> DIFF₂(x, s₁(y))
DIFF₂(x, y) -> COND1₃(equal₂(x, y), x, y)

The TRS R consists of the following rules:

gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(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:

equal₂(s₁(x0), 0)
equal₂(0, 0)
gt₂(0, x0)
gt₂(s₁(x0), s₁(x1))
equal₂(s₁(x0), s₁(x1))
equal₂(0, s₁(x0))
gt₂(s₁(x0), 0)

We have to consider all minimal (P,Q,R)-chains.
By narrowing [13] the rule COND1₃(false, x, y) -> COND2₃(gt₂(x, y), x, y) at position [0] we obtained the following new rules:

COND1₃(false, s₁(x0), 0) -> COND2₃(true, s₁(x0), 0)
COND1₃(false, 0, x0) -> COND2₃(false, 0, x0)
COND1₃(false, s₁(x0), s₁(x1)) -> COND2₃(gt₂(x0, x1), s₁(x0), s₁(x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ Narrowing

            ↳ UsableRulesProof

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

COND1₃(false, 0, x0) -> COND2₃(false, 0, x0)
COND1₃(false, s₁(x0), 0) -> COND2₃(true, s₁(x0), 0)
COND2₃(false, x, y) -> DIFF₂(s₁(x), y)
COND1₃(false, s₁(x0), s₁(x1)) -> COND2₃(gt₂(x0, x1), s₁(x0), s₁(x1))
COND2₃(true, x, y) -> DIFF₂(x, s₁(y))
DIFF₂(x, y) -> COND1₃(equal₂(x, y), x, y)

The TRS R consists of the following rules:

gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(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:

equal₂(s₁(x0), 0)
equal₂(0, 0)
gt₂(0, x0)
gt₂(s₁(x0), s₁(x1))
equal₂(s₁(x0), s₁(x1))
equal₂(0, s₁(x0))
gt₂(s₁(x0), 0)

We have to consider all minimal (P,Q,R)-chains.
By narrowing [13] the rule DIFF₂(x, y) -> COND1₃(equal₂(x, y), x, y) at position [0] we obtained the following new rules:

DIFF₂(s₁(x0), s₁(x1)) -> COND1₃(equal₂(x0, x1), s₁(x0), s₁(x1))
DIFF₂(0, s₁(x0)) -> COND1₃(false, 0, s₁(x0))
DIFF₂(s₁(x0), 0) -> COND1₃(false, s₁(x0), 0)
DIFF₂(0, 0) -> COND1₃(true, 0, 0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

            ↳ UsableRulesProof

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

DIFF₂(s₁(x0), 0) -> COND1₃(false, s₁(x0), 0)
DIFF₂(0, s₁(x0)) -> COND1₃(false, 0, s₁(x0))
COND2₃(false, x, y) -> DIFF₂(s₁(x), y)
COND1₃(false, s₁(x0), 0) -> COND2₃(true, s₁(x0), 0)
COND1₃(false, 0, x0) -> COND2₃(false, 0, x0)
DIFF₂(0, 0) -> COND1₃(true, 0, 0)
COND1₃(false, s₁(x0), s₁(x1)) -> COND2₃(gt₂(x0, x1), s₁(x0), s₁(x1))
DIFF₂(s₁(x0), s₁(x1)) -> COND1₃(equal₂(x0, x1), s₁(x0), s₁(x1))
COND2₃(true, x, y) -> DIFF₂(x, s₁(y))

The TRS R consists of the following rules:

gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(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:

equal₂(s₁(x0), 0)
equal₂(0, 0)
gt₂(0, x0)
gt₂(s₁(x0), s₁(x1))
equal₂(s₁(x0), s₁(x1))
equal₂(0, s₁(x0))
gt₂(s₁(x0), 0)

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

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Instantiation

            ↳ UsableRulesProof

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

COND1₃(false, s₁(x0), 0) -> COND2₃(true, s₁(x0), 0)
DIFF₂(s₁(x0), 0) -> COND1₃(false, s₁(x0), 0)
COND2₃(false, x, y) -> DIFF₂(s₁(x), y)
COND1₃(false, 0, x0) -> COND2₃(false, 0, x0)
DIFF₂(0, s₁(x0)) -> COND1₃(false, 0, s₁(x0))
COND1₃(false, s₁(x0), s₁(x1)) -> COND2₃(gt₂(x0, x1), s₁(x0), s₁(x1))
DIFF₂(s₁(x0), s₁(x1)) -> COND1₃(equal₂(x0, x1), s₁(x0), s₁(x1))
COND2₃(true, x, y) -> DIFF₂(x, s₁(y))

The TRS R consists of the following rules:

gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(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:

equal₂(s₁(x0), 0)
equal₂(0, 0)
gt₂(0, x0)
gt₂(s₁(x0), s₁(x1))
equal₂(s₁(x0), s₁(x1))
equal₂(0, s₁(x0))
gt₂(s₁(x0), 0)

We have to consider all minimal (P,Q,R)-chains.
By instantiating [13] the rule COND2₃(true, x, y) -> DIFF₂(x, s₁(y)) we obtained the following new rules:

COND2₃(true, s₁(z0), s₁(z1)) -> DIFF₂(s₁(z0), s₁(s₁(z1)))
COND2₃(true, s₁(z0), 0) -> DIFF₂(s₁(z0), s₁(0))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Instantiation

                                  ↳ QDP

                                    ↳ DependencyGraphProof

            ↳ UsableRulesProof

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

DIFF₂(0, s₁(x0)) -> COND1₃(false, 0, s₁(x0))
COND1₃(false, 0, x0) -> COND2₃(false, 0, x0)
COND2₃(false, x, y) -> DIFF₂(s₁(x), y)
DIFF₂(s₁(x0), 0) -> COND1₃(false, s₁(x0), 0)
COND1₃(false, s₁(x0), 0) -> COND2₃(true, s₁(x0), 0)
COND1₃(false, s₁(x0), s₁(x1)) -> COND2₃(gt₂(x0, x1), s₁(x0), s₁(x1))
COND2₃(true, s₁(z0), s₁(z1)) -> DIFF₂(s₁(z0), s₁(s₁(z1)))
DIFF₂(s₁(x0), s₁(x1)) -> COND1₃(equal₂(x0, x1), s₁(x0), s₁(x1))
COND2₃(true, s₁(z0), 0) -> DIFF₂(s₁(z0), s₁(0))

The TRS R consists of the following rules:

gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(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:

equal₂(s₁(x0), 0)
equal₂(0, 0)
gt₂(0, x0)
gt₂(s₁(x0), s₁(x1))
equal₂(s₁(x0), s₁(x1))
equal₂(0, s₁(x0))
gt₂(s₁(x0), 0)

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

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Instantiation

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Instantiation

            ↳ UsableRulesProof

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

COND1₃(false, s₁(x0), 0) -> COND2₃(true, s₁(x0), 0)
DIFF₂(s₁(x0), 0) -> COND1₃(false, s₁(x0), 0)
COND2₃(false, x, y) -> DIFF₂(s₁(x), y)
COND1₃(false, s₁(x0), s₁(x1)) -> COND2₃(gt₂(x0, x1), s₁(x0), s₁(x1))
COND2₃(true, s₁(z0), s₁(z1)) -> DIFF₂(s₁(z0), s₁(s₁(z1)))
DIFF₂(s₁(x0), s₁(x1)) -> COND1₃(equal₂(x0, x1), s₁(x0), s₁(x1))
COND2₃(true, s₁(z0), 0) -> DIFF₂(s₁(z0), s₁(0))

The TRS R consists of the following rules:

gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(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:

equal₂(s₁(x0), 0)
equal₂(0, 0)
gt₂(0, x0)
gt₂(s₁(x0), s₁(x1))
equal₂(s₁(x0), s₁(x1))
equal₂(0, s₁(x0))
gt₂(s₁(x0), 0)

We have to consider all minimal (P,Q,R)-chains.
By instantiating [13] the rule COND2₃(false, x, y) -> DIFF₂(s₁(x), y) we obtained the following new rules:

COND2₃(false, s₁(z0), s₁(z1)) -> DIFF₂(s₁(s₁(z0)), s₁(z1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Instantiation

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ DependencyGraphProof

            ↳ UsableRulesProof

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

COND2₃(false, s₁(z0), s₁(z1)) -> DIFF₂(s₁(s₁(z0)), s₁(z1))
DIFF₂(s₁(x0), 0) -> COND1₃(false, s₁(x0), 0)
COND1₃(false, s₁(x0), 0) -> COND2₃(true, s₁(x0), 0)
COND1₃(false, s₁(x0), s₁(x1)) -> COND2₃(gt₂(x0, x1), s₁(x0), s₁(x1))
COND2₃(true, s₁(z0), s₁(z1)) -> DIFF₂(s₁(z0), s₁(s₁(z1)))
DIFF₂(s₁(x0), s₁(x1)) -> COND1₃(equal₂(x0, x1), s₁(x0), s₁(x1))
COND2₃(true, s₁(z0), 0) -> DIFF₂(s₁(z0), s₁(0))

The TRS R consists of the following rules:

gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(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:

equal₂(s₁(x0), 0)
equal₂(0, 0)
gt₂(0, x0)
gt₂(s₁(x0), s₁(x1))
equal₂(s₁(x0), s₁(x1))
equal₂(0, s₁(x0))
gt₂(s₁(x0), 0)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Instantiation

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Instantiation

            ↳ UsableRulesProof

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

COND2₃(false, s₁(z0), s₁(z1)) -> DIFF₂(s₁(s₁(z0)), s₁(z1))
COND1₃(false, s₁(x0), s₁(x1)) -> COND2₃(gt₂(x0, x1), s₁(x0), s₁(x1))
COND2₃(true, s₁(z0), s₁(z1)) -> DIFF₂(s₁(z0), s₁(s₁(z1)))
DIFF₂(s₁(x0), s₁(x1)) -> COND1₃(equal₂(x0, x1), s₁(x0), s₁(x1))

The TRS R consists of the following rules:

gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(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:

equal₂(s₁(x0), 0)
equal₂(0, 0)
gt₂(0, x0)
gt₂(s₁(x0), s₁(x1))
equal₂(s₁(x0), s₁(x1))
equal₂(0, s₁(x0))
gt₂(s₁(x0), 0)

We have to consider all minimal (P,Q,R)-chains.
By instantiating [13] the rule DIFF₂(s₁(x0), s₁(x1)) -> COND1₃(equal₂(x0, x1), s₁(x0), s₁(x1)) we obtained the following new rules:

DIFF₂(s₁(s₁(z0)), s₁(z1)) -> COND1₃(equal₂(s₁(z0), z1), s₁(s₁(z0)), s₁(z1))
DIFF₂(s₁(z0), s₁(s₁(z1))) -> COND1₃(equal₂(z0, s₁(z1)), s₁(z0), s₁(s₁(z1)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Instantiation

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ Instantiation

            ↳ UsableRulesProof

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

COND2₃(false, s₁(z0), s₁(z1)) -> DIFF₂(s₁(s₁(z0)), s₁(z1))
DIFF₂(s₁(s₁(z0)), s₁(z1)) -> COND1₃(equal₂(s₁(z0), z1), s₁(s₁(z0)), s₁(z1))
COND1₃(false, s₁(x0), s₁(x1)) -> COND2₃(gt₂(x0, x1), s₁(x0), s₁(x1))
DIFF₂(s₁(z0), s₁(s₁(z1))) -> COND1₃(equal₂(z0, s₁(z1)), s₁(z0), s₁(s₁(z1)))
COND2₃(true, s₁(z0), s₁(z1)) -> DIFF₂(s₁(z0), s₁(s₁(z1)))

The TRS R consists of the following rules:

gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(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:

equal₂(s₁(x0), 0)
equal₂(0, 0)
gt₂(0, x0)
gt₂(s₁(x0), s₁(x1))
equal₂(s₁(x0), s₁(x1))
equal₂(0, s₁(x0))
gt₂(s₁(x0), 0)

We have to consider all minimal (P,Q,R)-chains.
By instantiating [13] the rule COND1₃(false, s₁(x0), s₁(x1)) -> COND2₃(gt₂(x0, x1), s₁(x0), s₁(x1)) we obtained the following new rules:

COND1₃(false, s₁(s₁(z0)), s₁(z1)) -> COND2₃(gt₂(s₁(z0), z1), s₁(s₁(z0)), s₁(z1))
COND1₃(false, s₁(z0), s₁(s₁(z1))) -> COND2₃(gt₂(z0, s₁(z1)), s₁(z0), s₁(s₁(z1)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Instantiation

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ Instantiation

                                                      ↳ QDP

            ↳ UsableRulesProof

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

COND2₃(false, s₁(z0), s₁(z1)) -> DIFF₂(s₁(s₁(z0)), s₁(z1))
COND1₃(false, s₁(s₁(z0)), s₁(z1)) -> COND2₃(gt₂(s₁(z0), z1), s₁(s₁(z0)), s₁(z1))
DIFF₂(s₁(s₁(z0)), s₁(z1)) -> COND1₃(equal₂(s₁(z0), z1), s₁(s₁(z0)), s₁(z1))
COND1₃(false, s₁(z0), s₁(s₁(z1))) -> COND2₃(gt₂(z0, s₁(z1)), s₁(z0), s₁(s₁(z1)))
DIFF₂(s₁(z0), s₁(s₁(z1))) -> COND1₃(equal₂(z0, s₁(z1)), s₁(z0), s₁(s₁(z1)))
COND2₃(true, s₁(z0), s₁(z1)) -> DIFF₂(s₁(z0), s₁(s₁(z1)))

The TRS R consists of the following rules:

gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(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:

equal₂(s₁(x0), 0)
equal₂(0, 0)
gt₂(0, x0)
gt₂(s₁(x0), s₁(x1))
equal₂(s₁(x0), s₁(x1))
equal₂(0, s₁(x0))
gt₂(s₁(x0), 0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

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

COND2₃(false, x, y) -> DIFF₂(s₁(x), y)
COND1₃(false, x, y) -> COND2₃(gt₂(x, y), x, y)
COND2₃(true, x, y) -> DIFF₂(x, s₁(y))
DIFF₂(x, y) -> COND1₃(equal₂(x, y), x, y)

The TRS R consists of the following rules:

gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(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:

equal₂(s₁(x0), 0)
equal₂(0, 0)
gt₂(0, x0)
gt₂(s₁(x0), s₁(x1))
cond2₃(false, x0, x1)
equal₂(s₁(x0), s₁(x1))
cond1₃(false, x0, x1)
diff₂(x0, x1)
equal₂(0, s₁(x0))
cond1₃(true, x0, x1)
gt₂(s₁(x0), 0)
cond2₃(true, x0, 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.

cond2₃(false, x0, x1)
cond1₃(false, x0, x1)
diff₂(x0, x1)
cond1₃(true, x0, x1)
cond2₃(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:

COND2₃(false, x, y) -> DIFF₂(s₁(x), y)
COND1₃(false, x, y) -> COND2₃(gt₂(x, y), x, y)
COND2₃(true, x, y) -> DIFF₂(x, s₁(y))
DIFF₂(x, y) -> COND1₃(equal₂(x, y), x, y)

The TRS R consists of the following rules:

gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(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:

equal₂(s₁(x0), 0)
equal₂(0, 0)
gt₂(0, x0)
gt₂(s₁(x0), s₁(x1))
equal₂(s₁(x0), s₁(x1))
equal₂(0, s₁(x0))
gt₂(s₁(x0), 0)

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