Termination w.r.t. Q proof of TRS/Rubioenno.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

lt₂(0, s₁(X)) -> true
lt₂(s₁(X), 0) -> false
lt₂(s₁(X), s₁(Y)) -> lt₂(X, Y)
append₂(nil, Y) -> Y
append₂(add₂(N, X), Y) -> add₂(N, append₂(X, Y))
split₂(N, nil) -> pair₂(nil, nil)
split₂(N, add₂(M, Y)) -> f_1₄(split₂(N, Y), N, M, Y)
f_1₄(pair₂(X, Z), N, M, Y) -> f_2₆(lt₂(N, M), N, M, Y, X, Z)
f_2₆(true, N, M, Y, X, Z) -> pair₂(X, add₂(M, Z))
f_2₆(false, N, M, Y, X, Z) -> pair₂(add₂(M, X), Z)
qsort₁(nil) -> nil
qsort₁(add₂(N, X)) -> f_3₃(split₂(N, X), N, X)
f_3₃(pair₂(Y, Z), N, X) -> append₂(qsort₁(Y), add₂(X, qsort₁(Z)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

lt₂(0, s₁(X)) -> true
lt₂(s₁(X), 0) -> false
lt₂(s₁(X), s₁(Y)) -> lt₂(X, Y)
append₂(nil, Y) -> Y
append₂(add₂(N, X), Y) -> add₂(N, append₂(X, Y))
split₂(N, nil) -> pair₂(nil, nil)
split₂(N, add₂(M, Y)) -> f_1₄(split₂(N, Y), N, M, Y)
f_1₄(pair₂(X, Z), N, M, Y) -> f_2₆(lt₂(N, M), N, M, Y, X, Z)
f_2₆(true, N, M, Y, X, Z) -> pair₂(X, add₂(M, Z))
f_2₆(false, N, M, Y, X, Z) -> pair₂(add₂(M, X), Z)
qsort₁(nil) -> nil
qsort₁(add₂(N, X)) -> f_3₃(split₂(N, X), N, X)
f_3₃(pair₂(Y, Z), N, X) -> append₂(qsort₁(Y), add₂(X, qsort₁(Z)))

Q is empty.

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

F_3₃(pair₂(Y, Z), N, X) -> QSORT₁(Y)
F_3₃(pair₂(Y, Z), N, X) -> APPEND₂(qsort₁(Y), add₂(X, qsort₁(Z)))
F_1₄(pair₂(X, Z), N, M, Y) -> F_2₆(lt₂(N, M), N, M, Y, X, Z)
F_1₄(pair₂(X, Z), N, M, Y) -> LT₂(N, M)
QSORT₁(add₂(N, X)) -> F_3₃(split₂(N, X), N, X)
LT₂(s₁(X), s₁(Y)) -> LT₂(X, Y)
SPLIT₂(N, add₂(M, Y)) -> F_1₄(split₂(N, Y), N, M, Y)
APPEND₂(add₂(N, X), Y) -> APPEND₂(X, Y)
F_3₃(pair₂(Y, Z), N, X) -> QSORT₁(Z)
QSORT₁(add₂(N, X)) -> SPLIT₂(N, X)
SPLIT₂(N, add₂(M, Y)) -> SPLIT₂(N, Y)

The TRS R consists of the following rules:

lt₂(0, s₁(X)) -> true
lt₂(s₁(X), 0) -> false
lt₂(s₁(X), s₁(Y)) -> lt₂(X, Y)
append₂(nil, Y) -> Y
append₂(add₂(N, X), Y) -> add₂(N, append₂(X, Y))
split₂(N, nil) -> pair₂(nil, nil)
split₂(N, add₂(M, Y)) -> f_1₄(split₂(N, Y), N, M, Y)
f_1₄(pair₂(X, Z), N, M, Y) -> f_2₆(lt₂(N, M), N, M, Y, X, Z)
f_2₆(true, N, M, Y, X, Z) -> pair₂(X, add₂(M, Z))
f_2₆(false, N, M, Y, X, Z) -> pair₂(add₂(M, X), Z)
qsort₁(nil) -> nil
qsort₁(add₂(N, X)) -> f_3₃(split₂(N, X), N, X)
f_3₃(pair₂(Y, Z), N, X) -> append₂(qsort₁(Y), add₂(X, qsort₁(Z)))

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:

F_3₃(pair₂(Y, Z), N, X) -> QSORT₁(Y)
F_3₃(pair₂(Y, Z), N, X) -> APPEND₂(qsort₁(Y), add₂(X, qsort₁(Z)))
F_1₄(pair₂(X, Z), N, M, Y) -> F_2₆(lt₂(N, M), N, M, Y, X, Z)
F_1₄(pair₂(X, Z), N, M, Y) -> LT₂(N, M)
QSORT₁(add₂(N, X)) -> F_3₃(split₂(N, X), N, X)
LT₂(s₁(X), s₁(Y)) -> LT₂(X, Y)
SPLIT₂(N, add₂(M, Y)) -> F_1₄(split₂(N, Y), N, M, Y)
APPEND₂(add₂(N, X), Y) -> APPEND₂(X, Y)
F_3₃(pair₂(Y, Z), N, X) -> QSORT₁(Z)
QSORT₁(add₂(N, X)) -> SPLIT₂(N, X)
SPLIT₂(N, add₂(M, Y)) -> SPLIT₂(N, Y)

The TRS R consists of the following rules:

lt₂(0, s₁(X)) -> true
lt₂(s₁(X), 0) -> false
lt₂(s₁(X), s₁(Y)) -> lt₂(X, Y)
append₂(nil, Y) -> Y
append₂(add₂(N, X), Y) -> add₂(N, append₂(X, Y))
split₂(N, nil) -> pair₂(nil, nil)
split₂(N, add₂(M, Y)) -> f_1₄(split₂(N, Y), N, M, Y)
f_1₄(pair₂(X, Z), N, M, Y) -> f_2₆(lt₂(N, M), N, M, Y, X, Z)
f_2₆(true, N, M, Y, X, Z) -> pair₂(X, add₂(M, Z))
f_2₆(false, N, M, Y, X, Z) -> pair₂(add₂(M, X), Z)
qsort₁(nil) -> nil
qsort₁(add₂(N, X)) -> f_3₃(split₂(N, X), N, X)
f_3₃(pair₂(Y, Z), N, X) -> append₂(qsort₁(Y), add₂(X, qsort₁(Z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

APPEND₂(add₂(N, X), Y) -> APPEND₂(X, Y)

The TRS R consists of the following rules:

lt₂(0, s₁(X)) -> true
lt₂(s₁(X), 0) -> false
lt₂(s₁(X), s₁(Y)) -> lt₂(X, Y)
append₂(nil, Y) -> Y
append₂(add₂(N, X), Y) -> add₂(N, append₂(X, Y))
split₂(N, nil) -> pair₂(nil, nil)
split₂(N, add₂(M, Y)) -> f_1₄(split₂(N, Y), N, M, Y)
f_1₄(pair₂(X, Z), N, M, Y) -> f_2₆(lt₂(N, M), N, M, Y, X, Z)
f_2₆(true, N, M, Y, X, Z) -> pair₂(X, add₂(M, Z))
f_2₆(false, N, M, Y, X, Z) -> pair₂(add₂(M, X), Z)
qsort₁(nil) -> nil
qsort₁(add₂(N, X)) -> f_3₃(split₂(N, X), N, X)
f_3₃(pair₂(Y, Z), N, X) -> append₂(qsort₁(Y), add₂(X, qsort₁(Z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

APPEND₂(add₂(N, X), Y) -> APPEND₂(X, Y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( APPEND₂(x₁, x₂) ) = x₁ + 2x₂ + 1

POL( add₂(x₁, x₂) ) = 2x₁ + x₂ + 2

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

lt₂(0, s₁(X)) -> true
lt₂(s₁(X), 0) -> false
lt₂(s₁(X), s₁(Y)) -> lt₂(X, Y)
append₂(nil, Y) -> Y
append₂(add₂(N, X), Y) -> add₂(N, append₂(X, Y))
split₂(N, nil) -> pair₂(nil, nil)
split₂(N, add₂(M, Y)) -> f_1₄(split₂(N, Y), N, M, Y)
f_1₄(pair₂(X, Z), N, M, Y) -> f_2₆(lt₂(N, M), N, M, Y, X, Z)
f_2₆(true, N, M, Y, X, Z) -> pair₂(X, add₂(M, Z))
f_2₆(false, N, M, Y, X, Z) -> pair₂(add₂(M, X), Z)
qsort₁(nil) -> nil
qsort₁(add₂(N, X)) -> f_3₃(split₂(N, X), N, X)
f_3₃(pair₂(Y, Z), N, X) -> append₂(qsort₁(Y), add₂(X, qsort₁(Z)))

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

          ↳ QDP

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

LT₂(s₁(X), s₁(Y)) -> LT₂(X, Y)

The TRS R consists of the following rules:

lt₂(0, s₁(X)) -> true
lt₂(s₁(X), 0) -> false
lt₂(s₁(X), s₁(Y)) -> lt₂(X, Y)
append₂(nil, Y) -> Y
append₂(add₂(N, X), Y) -> add₂(N, append₂(X, Y))
split₂(N, nil) -> pair₂(nil, nil)
split₂(N, add₂(M, Y)) -> f_1₄(split₂(N, Y), N, M, Y)
f_1₄(pair₂(X, Z), N, M, Y) -> f_2₆(lt₂(N, M), N, M, Y, X, Z)
f_2₆(true, N, M, Y, X, Z) -> pair₂(X, add₂(M, Z))
f_2₆(false, N, M, Y, X, Z) -> pair₂(add₂(M, X), Z)
qsort₁(nil) -> nil
qsort₁(add₂(N, X)) -> f_3₃(split₂(N, X), N, X)
f_3₃(pair₂(Y, Z), N, X) -> append₂(qsort₁(Y), add₂(X, qsort₁(Z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

LT₂(s₁(X), s₁(Y)) -> LT₂(X, Y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s₁(x₁) ) = 2x₁ + 2

POL( LT₂(x₁, x₂) ) = max{0, 2x₁ + 2x₂ - 1}

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

lt₂(0, s₁(X)) -> true
lt₂(s₁(X), 0) -> false
lt₂(s₁(X), s₁(Y)) -> lt₂(X, Y)
append₂(nil, Y) -> Y
append₂(add₂(N, X), Y) -> add₂(N, append₂(X, Y))
split₂(N, nil) -> pair₂(nil, nil)
split₂(N, add₂(M, Y)) -> f_1₄(split₂(N, Y), N, M, Y)
f_1₄(pair₂(X, Z), N, M, Y) -> f_2₆(lt₂(N, M), N, M, Y, X, Z)
f_2₆(true, N, M, Y, X, Z) -> pair₂(X, add₂(M, Z))
f_2₆(false, N, M, Y, X, Z) -> pair₂(add₂(M, X), Z)
qsort₁(nil) -> nil
qsort₁(add₂(N, X)) -> f_3₃(split₂(N, X), N, X)
f_3₃(pair₂(Y, Z), N, X) -> append₂(qsort₁(Y), add₂(X, qsort₁(Z)))

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

            ↳ QDPOrderProof

          ↳ QDP

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

SPLIT₂(N, add₂(M, Y)) -> SPLIT₂(N, Y)

The TRS R consists of the following rules:

lt₂(0, s₁(X)) -> true
lt₂(s₁(X), 0) -> false
lt₂(s₁(X), s₁(Y)) -> lt₂(X, Y)
append₂(nil, Y) -> Y
append₂(add₂(N, X), Y) -> add₂(N, append₂(X, Y))
split₂(N, nil) -> pair₂(nil, nil)
split₂(N, add₂(M, Y)) -> f_1₄(split₂(N, Y), N, M, Y)
f_1₄(pair₂(X, Z), N, M, Y) -> f_2₆(lt₂(N, M), N, M, Y, X, Z)
f_2₆(true, N, M, Y, X, Z) -> pair₂(X, add₂(M, Z))
f_2₆(false, N, M, Y, X, Z) -> pair₂(add₂(M, X), Z)
qsort₁(nil) -> nil
qsort₁(add₂(N, X)) -> f_3₃(split₂(N, X), N, X)
f_3₃(pair₂(Y, Z), N, X) -> append₂(qsort₁(Y), add₂(X, qsort₁(Z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

SPLIT₂(N, add₂(M, Y)) -> SPLIT₂(N, Y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( add₂(x₁, x₂) ) = 2x₁ + 2x₂ + 2

POL( SPLIT₂(x₁, x₂) ) = x₁ + x₂ + 1

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

lt₂(0, s₁(X)) -> true
lt₂(s₁(X), 0) -> false
lt₂(s₁(X), s₁(Y)) -> lt₂(X, Y)
append₂(nil, Y) -> Y
append₂(add₂(N, X), Y) -> add₂(N, append₂(X, Y))
split₂(N, nil) -> pair₂(nil, nil)
split₂(N, add₂(M, Y)) -> f_1₄(split₂(N, Y), N, M, Y)
f_1₄(pair₂(X, Z), N, M, Y) -> f_2₆(lt₂(N, M), N, M, Y, X, Z)
f_2₆(true, N, M, Y, X, Z) -> pair₂(X, add₂(M, Z))
f_2₆(false, N, M, Y, X, Z) -> pair₂(add₂(M, X), Z)
qsort₁(nil) -> nil
qsort₁(add₂(N, X)) -> f_3₃(split₂(N, X), N, X)
f_3₃(pair₂(Y, Z), N, X) -> append₂(qsort₁(Y), add₂(X, qsort₁(Z)))

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

          ↳ QDP

            ↳ QDPOrderProof

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

F_3₃(pair₂(Y, Z), N, X) -> QSORT₁(Y)
QSORT₁(add₂(N, X)) -> F_3₃(split₂(N, X), N, X)
F_3₃(pair₂(Y, Z), N, X) -> QSORT₁(Z)

The TRS R consists of the following rules:

lt₂(0, s₁(X)) -> true
lt₂(s₁(X), 0) -> false
lt₂(s₁(X), s₁(Y)) -> lt₂(X, Y)
append₂(nil, Y) -> Y
append₂(add₂(N, X), Y) -> add₂(N, append₂(X, Y))
split₂(N, nil) -> pair₂(nil, nil)
split₂(N, add₂(M, Y)) -> f_1₄(split₂(N, Y), N, M, Y)
f_1₄(pair₂(X, Z), N, M, Y) -> f_2₆(lt₂(N, M), N, M, Y, X, Z)
f_2₆(true, N, M, Y, X, Z) -> pair₂(X, add₂(M, Z))
f_2₆(false, N, M, Y, X, Z) -> pair₂(add₂(M, X), Z)
qsort₁(nil) -> nil
qsort₁(add₂(N, X)) -> f_3₃(split₂(N, X), N, X)
f_3₃(pair₂(Y, Z), N, X) -> append₂(qsort₁(Y), add₂(X, qsort₁(Z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

QSORT₁(add₂(N, X)) -> F_3₃(split₂(N, X), N, X)

The remaining pairs can at least be oriented weakly.

F_3₃(pair₂(Y, Z), N, X) -> QSORT₁(Y)
F_3₃(pair₂(Y, Z), N, X) -> QSORT₁(Z)

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( QSORT₁(x₁) ) = max{0, 2x₁ - 1}

POL( lt₂(x₁, x₂) ) = 2x₁ + 2x₂ + 2

POL( f_1₄(x₁, ..., x₄) ) = x₁ + 2x₃ + 2

POL( 0 ) = 2

POL( nil ) = max{0, -2}

POL( pair₂(x₁, x₂) ) = 2x₁ + 2x₂

POL( true ) = 2

POL( F_3₃(x₁, ..., x₃) ) = x₁ + x₂

POL( false ) = 1

POL( add₂(x₁, x₂) ) = x₁ + x₂ + 1

POL( split₂(x₁, x₂) ) = x₁ + 2x₂

POL( s₁(x₁) ) = x₁ + 2

POL( f_2₆(x₁, ..., x₆) ) = 2x₃ + 2x₅ + 2x₆ + 2

The following usable rules [14] were oriented:

f_2₆(false, N, M, Y, X, Z) -> pair₂(add₂(M, X), Z)
split₂(N, add₂(M, Y)) -> f_1₄(split₂(N, Y), N, M, Y)
f_2₆(true, N, M, Y, X, Z) -> pair₂(X, add₂(M, Z))
f_1₄(pair₂(X, Z), N, M, Y) -> f_2₆(lt₂(N, M), N, M, Y, X, Z)
split₂(N, nil) -> pair₂(nil, nil)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

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

F_3₃(pair₂(Y, Z), N, X) -> QSORT₁(Y)
F_3₃(pair₂(Y, Z), N, X) -> QSORT₁(Z)

The TRS R consists of the following rules:

lt₂(0, s₁(X)) -> true
lt₂(s₁(X), 0) -> false
lt₂(s₁(X), s₁(Y)) -> lt₂(X, Y)
append₂(nil, Y) -> Y
append₂(add₂(N, X), Y) -> add₂(N, append₂(X, Y))
split₂(N, nil) -> pair₂(nil, nil)
split₂(N, add₂(M, Y)) -> f_1₄(split₂(N, Y), N, M, Y)
f_1₄(pair₂(X, Z), N, M, Y) -> f_2₆(lt₂(N, M), N, M, Y, X, Z)
f_2₆(true, N, M, Y, X, Z) -> pair₂(X, add₂(M, Z))
f_2₆(false, N, M, Y, X, Z) -> pair₂(add₂(M, X), Z)
qsort₁(nil) -> nil
qsort₁(add₂(N, X)) -> f_3₃(split₂(N, X), N, X)
f_3₃(pair₂(Y, Z), N, X) -> append₂(qsort₁(Y), add₂(X, qsort₁(Z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 2 less nodes.