Termination w.r.t. Q proof of TRS/Rubioelimdupl.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:

eq₂(0, 0) -> true
eq₂(0, s₁(X)) -> false
eq₂(s₁(X), 0) -> false
eq₂(s₁(X), s₁(Y)) -> eq₂(X, Y)
rm₂(N, nil) -> nil
rm₂(N, add₂(M, X)) -> ifrm₃(eq₂(N, M), N, add₂(M, X))
ifrm₃(true, N, add₂(M, X)) -> rm₂(N, X)
ifrm₃(false, N, add₂(M, X)) -> add₂(M, rm₂(N, X))
purge₁(nil) -> nil
purge₁(add₂(N, X)) -> add₂(N, purge₁(rm₂(N, X)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

eq₂(0, 0) -> true
eq₂(0, s₁(X)) -> false
eq₂(s₁(X), 0) -> false
eq₂(s₁(X), s₁(Y)) -> eq₂(X, Y)
rm₂(N, nil) -> nil
rm₂(N, add₂(M, X)) -> ifrm₃(eq₂(N, M), N, add₂(M, X))
ifrm₃(true, N, add₂(M, X)) -> rm₂(N, X)
ifrm₃(false, N, add₂(M, X)) -> add₂(M, rm₂(N, X))
purge₁(nil) -> nil
purge₁(add₂(N, X)) -> add₂(N, purge₁(rm₂(N, X)))

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:

PURGE₁(add₂(N, X)) -> RM₂(N, X)
PURGE₁(add₂(N, X)) -> PURGE₁(rm₂(N, X))
EQ₂(s₁(X), s₁(Y)) -> EQ₂(X, Y)
IFRM₃(true, N, add₂(M, X)) -> RM₂(N, X)
IFRM₃(false, N, add₂(M, X)) -> RM₂(N, X)
RM₂(N, add₂(M, X)) -> EQ₂(N, M)
RM₂(N, add₂(M, X)) -> IFRM₃(eq₂(N, M), N, add₂(M, X))

The TRS R consists of the following rules:

eq₂(0, 0) -> true
eq₂(0, s₁(X)) -> false
eq₂(s₁(X), 0) -> false
eq₂(s₁(X), s₁(Y)) -> eq₂(X, Y)
rm₂(N, nil) -> nil
rm₂(N, add₂(M, X)) -> ifrm₃(eq₂(N, M), N, add₂(M, X))
ifrm₃(true, N, add₂(M, X)) -> rm₂(N, X)
ifrm₃(false, N, add₂(M, X)) -> add₂(M, rm₂(N, X))
purge₁(nil) -> nil
purge₁(add₂(N, X)) -> add₂(N, purge₁(rm₂(N, X)))

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:

PURGE₁(add₂(N, X)) -> RM₂(N, X)
PURGE₁(add₂(N, X)) -> PURGE₁(rm₂(N, X))
EQ₂(s₁(X), s₁(Y)) -> EQ₂(X, Y)
IFRM₃(true, N, add₂(M, X)) -> RM₂(N, X)
IFRM₃(false, N, add₂(M, X)) -> RM₂(N, X)
RM₂(N, add₂(M, X)) -> EQ₂(N, M)
RM₂(N, add₂(M, X)) -> IFRM₃(eq₂(N, M), N, add₂(M, X))

The TRS R consists of the following rules:

eq₂(0, 0) -> true
eq₂(0, s₁(X)) -> false
eq₂(s₁(X), 0) -> false
eq₂(s₁(X), s₁(Y)) -> eq₂(X, Y)
rm₂(N, nil) -> nil
rm₂(N, add₂(M, X)) -> ifrm₃(eq₂(N, M), N, add₂(M, X))
ifrm₃(true, N, add₂(M, X)) -> rm₂(N, X)
ifrm₃(false, N, add₂(M, X)) -> add₂(M, rm₂(N, X))
purge₁(nil) -> nil
purge₁(add₂(N, X)) -> add₂(N, purge₁(rm₂(N, X)))

Q is empty.
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

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

EQ₂(s₁(X), s₁(Y)) -> EQ₂(X, Y)

The TRS R consists of the following rules:

eq₂(0, 0) -> true
eq₂(0, s₁(X)) -> false
eq₂(s₁(X), 0) -> false
eq₂(s₁(X), s₁(Y)) -> eq₂(X, Y)
rm₂(N, nil) -> nil
rm₂(N, add₂(M, X)) -> ifrm₃(eq₂(N, M), N, add₂(M, X))
ifrm₃(true, N, add₂(M, X)) -> rm₂(N, X)
ifrm₃(false, N, add₂(M, X)) -> add₂(M, rm₂(N, X))
purge₁(nil) -> nil
purge₁(add₂(N, X)) -> add₂(N, purge₁(rm₂(N, X)))

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.

EQ₂(s₁(X), s₁(Y)) -> EQ₂(X, Y)

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

POL( EQ₂(x₁, x₂) ) = x₂

POL( s₁(x₁) ) = x₁ + 1

The following usable rules [14] 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:

eq₂(0, 0) -> true
eq₂(0, s₁(X)) -> false
eq₂(s₁(X), 0) -> false
eq₂(s₁(X), s₁(Y)) -> eq₂(X, Y)
rm₂(N, nil) -> nil
rm₂(N, add₂(M, X)) -> ifrm₃(eq₂(N, M), N, add₂(M, X))
ifrm₃(true, N, add₂(M, X)) -> rm₂(N, X)
ifrm₃(false, N, add₂(M, X)) -> add₂(M, rm₂(N, X))
purge₁(nil) -> nil
purge₁(add₂(N, X)) -> add₂(N, purge₁(rm₂(N, X)))

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:

IFRM₃(true, N, add₂(M, X)) -> RM₂(N, X)
IFRM₃(false, N, add₂(M, X)) -> RM₂(N, X)
RM₂(N, add₂(M, X)) -> IFRM₃(eq₂(N, M), N, add₂(M, X))

The TRS R consists of the following rules:

eq₂(0, 0) -> true
eq₂(0, s₁(X)) -> false
eq₂(s₁(X), 0) -> false
eq₂(s₁(X), s₁(Y)) -> eq₂(X, Y)
rm₂(N, nil) -> nil
rm₂(N, add₂(M, X)) -> ifrm₃(eq₂(N, M), N, add₂(M, X))
ifrm₃(true, N, add₂(M, X)) -> rm₂(N, X)
ifrm₃(false, N, add₂(M, X)) -> add₂(M, rm₂(N, X))
purge₁(nil) -> nil
purge₁(add₂(N, X)) -> add₂(N, purge₁(rm₂(N, X)))

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.

IFRM₃(true, N, add₂(M, X)) -> RM₂(N, X)
IFRM₃(false, N, add₂(M, X)) -> RM₂(N, X)

The remaining pairs can at least be oriented weakly.

RM₂(N, add₂(M, X)) -> IFRM₃(eq₂(N, M), N, add₂(M, X))

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

POL( IFRM₃(x₁, ..., x₃) ) = x₃

POL( add₂(x₁, x₂) ) = x₂ + 1

POL( RM₂(x₁, x₂) ) = x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

          ↳ QDP

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

RM₂(N, add₂(M, X)) -> IFRM₃(eq₂(N, M), N, add₂(M, X))

The TRS R consists of the following rules:

eq₂(0, 0) -> true
eq₂(0, s₁(X)) -> false
eq₂(s₁(X), 0) -> false
eq₂(s₁(X), s₁(Y)) -> eq₂(X, Y)
rm₂(N, nil) -> nil
rm₂(N, add₂(M, X)) -> ifrm₃(eq₂(N, M), N, add₂(M, X))
ifrm₃(true, N, add₂(M, X)) -> rm₂(N, X)
ifrm₃(false, N, add₂(M, X)) -> add₂(M, rm₂(N, X))
purge₁(nil) -> nil
purge₁(add₂(N, X)) -> add₂(N, purge₁(rm₂(N, X)))

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 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

PURGE₁(add₂(N, X)) -> PURGE₁(rm₂(N, X))

The TRS R consists of the following rules:

eq₂(0, 0) -> true
eq₂(0, s₁(X)) -> false
eq₂(s₁(X), 0) -> false
eq₂(s₁(X), s₁(Y)) -> eq₂(X, Y)
rm₂(N, nil) -> nil
rm₂(N, add₂(M, X)) -> ifrm₃(eq₂(N, M), N, add₂(M, X))
ifrm₃(true, N, add₂(M, X)) -> rm₂(N, X)
ifrm₃(false, N, add₂(M, X)) -> add₂(M, rm₂(N, X))
purge₁(nil) -> nil
purge₁(add₂(N, X)) -> add₂(N, purge₁(rm₂(N, X)))

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.

PURGE₁(add₂(N, X)) -> PURGE₁(rm₂(N, X))

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

POL( PURGE₁(x₁) ) = x₁

POL( add₂(x₁, x₂) ) = x₂ + 1

POL( rm₂(x₁, x₂) ) = x₂

POL( nil ) = 0

POL( ifrm₃(x₁, ..., x₃) ) = x₃

The following usable rules [14] were oriented:

rm₂(N, nil) -> nil
ifrm₃(true, N, add₂(M, X)) -> rm₂(N, X)
rm₂(N, add₂(M, X)) -> ifrm₃(eq₂(N, M), N, add₂(M, X))
ifrm₃(false, N, add₂(M, X)) -> add₂(M, rm₂(N, X))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

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

eq₂(0, 0) -> true
eq₂(0, s₁(X)) -> false
eq₂(s₁(X), 0) -> false
eq₂(s₁(X), s₁(Y)) -> eq₂(X, Y)
rm₂(N, nil) -> nil
rm₂(N, add₂(M, X)) -> ifrm₃(eq₂(N, M), N, add₂(M, X))
ifrm₃(true, N, add₂(M, X)) -> rm₂(N, X)
ifrm₃(false, N, add₂(M, X)) -> add₂(M, rm₂(N, X))
purge₁(nil) -> nil
purge₁(add₂(N, X)) -> add₂(N, purge₁(rm₂(N, X)))

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.