Termination w.r.t. Q proof of TRS/Rubioma96.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:

and₂(false, false) -> false
and₂(true, false) -> false
and₂(false, true) -> false
and₂(true, true) -> true
eq₂(nil, nil) -> true
eq₂(cons₂(T, L), nil) -> false
eq₂(nil, cons₂(T, L)) -> false
eq₂(cons₂(T, L), cons₂(Tp, Lp)) -> and₂(eq₂(T, Tp), eq₂(L, Lp))
eq₂(var₁(L), var₁(Lp)) -> eq₂(L, Lp)
eq₂(var₁(L), apply₂(T, S)) -> false
eq₂(var₁(L), lambda₂(X, T)) -> false
eq₂(apply₂(T, S), var₁(L)) -> false
eq₂(apply₂(T, S), apply₂(Tp, Sp)) -> and₂(eq₂(T, Tp), eq₂(S, Sp))
eq₂(apply₂(T, S), lambda₂(X, Tp)) -> false
eq₂(lambda₂(X, T), var₁(L)) -> false
eq₂(lambda₂(X, T), apply₂(Tp, Sp)) -> false
eq₂(lambda₂(X, T), lambda₂(Xp, Tp)) -> and₂(eq₂(T, Tp), eq₂(X, Xp))
if₃(true, var₁(K), var₁(L)) -> var₁(K)
if₃(false, var₁(K), var₁(L)) -> var₁(L)
ren₃(var₁(L), var₁(K), var₁(Lp)) -> if₃(eq₂(L, Lp), var₁(K), var₁(Lp))
ren₃(X, Y, apply₂(T, S)) -> apply₂(ren₃(X, Y, T), ren₃(X, Y, S))
ren₃(X, Y, lambda₂(Z, T)) -> lambda₂(var₁(cons₂(X, cons₂(Y, cons₂(lambda₂(Z, T), nil)))), ren₃(X, Y, ren₃(Z, var₁(cons₂(X, cons₂(Y, cons₂(lambda₂(Z, T), nil)))), T)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

and₂(false, false) -> false
and₂(true, false) -> false
and₂(false, true) -> false
and₂(true, true) -> true
eq₂(nil, nil) -> true
eq₂(cons₂(T, L), nil) -> false
eq₂(nil, cons₂(T, L)) -> false
eq₂(cons₂(T, L), cons₂(Tp, Lp)) -> and₂(eq₂(T, Tp), eq₂(L, Lp))
eq₂(var₁(L), var₁(Lp)) -> eq₂(L, Lp)
eq₂(var₁(L), apply₂(T, S)) -> false
eq₂(var₁(L), lambda₂(X, T)) -> false
eq₂(apply₂(T, S), var₁(L)) -> false
eq₂(apply₂(T, S), apply₂(Tp, Sp)) -> and₂(eq₂(T, Tp), eq₂(S, Sp))
eq₂(apply₂(T, S), lambda₂(X, Tp)) -> false
eq₂(lambda₂(X, T), var₁(L)) -> false
eq₂(lambda₂(X, T), apply₂(Tp, Sp)) -> false
eq₂(lambda₂(X, T), lambda₂(Xp, Tp)) -> and₂(eq₂(T, Tp), eq₂(X, Xp))
if₃(true, var₁(K), var₁(L)) -> var₁(K)
if₃(false, var₁(K), var₁(L)) -> var₁(L)
ren₃(var₁(L), var₁(K), var₁(Lp)) -> if₃(eq₂(L, Lp), var₁(K), var₁(Lp))
ren₃(X, Y, apply₂(T, S)) -> apply₂(ren₃(X, Y, T), ren₃(X, Y, S))
ren₃(X, Y, lambda₂(Z, T)) -> lambda₂(var₁(cons₂(X, cons₂(Y, cons₂(lambda₂(Z, T), nil)))), ren₃(X, Y, ren₃(Z, var₁(cons₂(X, cons₂(Y, cons₂(lambda₂(Z, T), nil)))), T)))

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:

EQ₂(var₁(L), var₁(Lp)) -> EQ₂(L, Lp)
REN₃(X, Y, apply₂(T, S)) -> REN₃(X, Y, T)
EQ₂(apply₂(T, S), apply₂(Tp, Sp)) -> EQ₂(T, Tp)
EQ₂(lambda₂(X, T), lambda₂(Xp, Tp)) -> EQ₂(X, Xp)
EQ₂(cons₂(T, L), cons₂(Tp, Lp)) -> EQ₂(L, Lp)
EQ₂(apply₂(T, S), apply₂(Tp, Sp)) -> AND₂(eq₂(T, Tp), eq₂(S, Sp))
EQ₂(cons₂(T, L), cons₂(Tp, Lp)) -> EQ₂(T, Tp)
REN₃(var₁(L), var₁(K), var₁(Lp)) -> EQ₂(L, Lp)
EQ₂(lambda₂(X, T), lambda₂(Xp, Tp)) -> AND₂(eq₂(T, Tp), eq₂(X, Xp))
EQ₂(cons₂(T, L), cons₂(Tp, Lp)) -> AND₂(eq₂(T, Tp), eq₂(L, Lp))
REN₃(X, Y, lambda₂(Z, T)) -> REN₃(X, Y, ren₃(Z, var₁(cons₂(X, cons₂(Y, cons₂(lambda₂(Z, T), nil)))), T))
EQ₂(lambda₂(X, T), lambda₂(Xp, Tp)) -> EQ₂(T, Tp)
EQ₂(apply₂(T, S), apply₂(Tp, Sp)) -> EQ₂(S, Sp)
REN₃(var₁(L), var₁(K), var₁(Lp)) -> IF₃(eq₂(L, Lp), var₁(K), var₁(Lp))
REN₃(X, Y, lambda₂(Z, T)) -> REN₃(Z, var₁(cons₂(X, cons₂(Y, cons₂(lambda₂(Z, T), nil)))), T)
REN₃(X, Y, apply₂(T, S)) -> REN₃(X, Y, S)

The TRS R consists of the following rules:

and₂(false, false) -> false
and₂(true, false) -> false
and₂(false, true) -> false
and₂(true, true) -> true
eq₂(nil, nil) -> true
eq₂(cons₂(T, L), nil) -> false
eq₂(nil, cons₂(T, L)) -> false
eq₂(cons₂(T, L), cons₂(Tp, Lp)) -> and₂(eq₂(T, Tp), eq₂(L, Lp))
eq₂(var₁(L), var₁(Lp)) -> eq₂(L, Lp)
eq₂(var₁(L), apply₂(T, S)) -> false
eq₂(var₁(L), lambda₂(X, T)) -> false
eq₂(apply₂(T, S), var₁(L)) -> false
eq₂(apply₂(T, S), apply₂(Tp, Sp)) -> and₂(eq₂(T, Tp), eq₂(S, Sp))
eq₂(apply₂(T, S), lambda₂(X, Tp)) -> false
eq₂(lambda₂(X, T), var₁(L)) -> false
eq₂(lambda₂(X, T), apply₂(Tp, Sp)) -> false
eq₂(lambda₂(X, T), lambda₂(Xp, Tp)) -> and₂(eq₂(T, Tp), eq₂(X, Xp))
if₃(true, var₁(K), var₁(L)) -> var₁(K)
if₃(false, var₁(K), var₁(L)) -> var₁(L)
ren₃(var₁(L), var₁(K), var₁(Lp)) -> if₃(eq₂(L, Lp), var₁(K), var₁(Lp))
ren₃(X, Y, apply₂(T, S)) -> apply₂(ren₃(X, Y, T), ren₃(X, Y, S))
ren₃(X, Y, lambda₂(Z, T)) -> lambda₂(var₁(cons₂(X, cons₂(Y, cons₂(lambda₂(Z, T), nil)))), ren₃(X, Y, ren₃(Z, var₁(cons₂(X, cons₂(Y, cons₂(lambda₂(Z, T), nil)))), T)))

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:

EQ₂(var₁(L), var₁(Lp)) -> EQ₂(L, Lp)
REN₃(X, Y, apply₂(T, S)) -> REN₃(X, Y, T)
EQ₂(apply₂(T, S), apply₂(Tp, Sp)) -> EQ₂(T, Tp)
EQ₂(lambda₂(X, T), lambda₂(Xp, Tp)) -> EQ₂(X, Xp)
EQ₂(cons₂(T, L), cons₂(Tp, Lp)) -> EQ₂(L, Lp)
EQ₂(apply₂(T, S), apply₂(Tp, Sp)) -> AND₂(eq₂(T, Tp), eq₂(S, Sp))
EQ₂(cons₂(T, L), cons₂(Tp, Lp)) -> EQ₂(T, Tp)
REN₃(var₁(L), var₁(K), var₁(Lp)) -> EQ₂(L, Lp)
EQ₂(lambda₂(X, T), lambda₂(Xp, Tp)) -> AND₂(eq₂(T, Tp), eq₂(X, Xp))
EQ₂(cons₂(T, L), cons₂(Tp, Lp)) -> AND₂(eq₂(T, Tp), eq₂(L, Lp))
REN₃(X, Y, lambda₂(Z, T)) -> REN₃(X, Y, ren₃(Z, var₁(cons₂(X, cons₂(Y, cons₂(lambda₂(Z, T), nil)))), T))
EQ₂(lambda₂(X, T), lambda₂(Xp, Tp)) -> EQ₂(T, Tp)
EQ₂(apply₂(T, S), apply₂(Tp, Sp)) -> EQ₂(S, Sp)
REN₃(var₁(L), var₁(K), var₁(Lp)) -> IF₃(eq₂(L, Lp), var₁(K), var₁(Lp))
REN₃(X, Y, lambda₂(Z, T)) -> REN₃(Z, var₁(cons₂(X, cons₂(Y, cons₂(lambda₂(Z, T), nil)))), T)
REN₃(X, Y, apply₂(T, S)) -> REN₃(X, Y, S)

The TRS R consists of the following rules:

and₂(false, false) -> false
and₂(true, false) -> false
and₂(false, true) -> false
and₂(true, true) -> true
eq₂(nil, nil) -> true
eq₂(cons₂(T, L), nil) -> false
eq₂(nil, cons₂(T, L)) -> false
eq₂(cons₂(T, L), cons₂(Tp, Lp)) -> and₂(eq₂(T, Tp), eq₂(L, Lp))
eq₂(var₁(L), var₁(Lp)) -> eq₂(L, Lp)
eq₂(var₁(L), apply₂(T, S)) -> false
eq₂(var₁(L), lambda₂(X, T)) -> false
eq₂(apply₂(T, S), var₁(L)) -> false
eq₂(apply₂(T, S), apply₂(Tp, Sp)) -> and₂(eq₂(T, Tp), eq₂(S, Sp))
eq₂(apply₂(T, S), lambda₂(X, Tp)) -> false
eq₂(lambda₂(X, T), var₁(L)) -> false
eq₂(lambda₂(X, T), apply₂(Tp, Sp)) -> false
eq₂(lambda₂(X, T), lambda₂(Xp, Tp)) -> and₂(eq₂(T, Tp), eq₂(X, Xp))
if₃(true, var₁(K), var₁(L)) -> var₁(K)
if₃(false, var₁(K), var₁(L)) -> var₁(L)
ren₃(var₁(L), var₁(K), var₁(Lp)) -> if₃(eq₂(L, Lp), var₁(K), var₁(Lp))
ren₃(X, Y, apply₂(T, S)) -> apply₂(ren₃(X, Y, T), ren₃(X, Y, S))
ren₃(X, Y, lambda₂(Z, T)) -> lambda₂(var₁(cons₂(X, cons₂(Y, cons₂(lambda₂(Z, T), nil)))), ren₃(X, Y, ren₃(Z, var₁(cons₂(X, cons₂(Y, cons₂(lambda₂(Z, T), nil)))), T)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

EQ₂(var₁(L), var₁(Lp)) -> EQ₂(L, Lp)
EQ₂(apply₂(T, S), apply₂(Tp, Sp)) -> EQ₂(T, Tp)
EQ₂(lambda₂(X, T), lambda₂(Xp, Tp)) -> EQ₂(X, Xp)
EQ₂(cons₂(T, L), cons₂(Tp, Lp)) -> EQ₂(L, Lp)
EQ₂(lambda₂(X, T), lambda₂(Xp, Tp)) -> EQ₂(T, Tp)
EQ₂(apply₂(T, S), apply₂(Tp, Sp)) -> EQ₂(S, Sp)
EQ₂(cons₂(T, L), cons₂(Tp, Lp)) -> EQ₂(T, Tp)

The TRS R consists of the following rules:

and₂(false, false) -> false
and₂(true, false) -> false
and₂(false, true) -> false
and₂(true, true) -> true
eq₂(nil, nil) -> true
eq₂(cons₂(T, L), nil) -> false
eq₂(nil, cons₂(T, L)) -> false
eq₂(cons₂(T, L), cons₂(Tp, Lp)) -> and₂(eq₂(T, Tp), eq₂(L, Lp))
eq₂(var₁(L), var₁(Lp)) -> eq₂(L, Lp)
eq₂(var₁(L), apply₂(T, S)) -> false
eq₂(var₁(L), lambda₂(X, T)) -> false
eq₂(apply₂(T, S), var₁(L)) -> false
eq₂(apply₂(T, S), apply₂(Tp, Sp)) -> and₂(eq₂(T, Tp), eq₂(S, Sp))
eq₂(apply₂(T, S), lambda₂(X, Tp)) -> false
eq₂(lambda₂(X, T), var₁(L)) -> false
eq₂(lambda₂(X, T), apply₂(Tp, Sp)) -> false
eq₂(lambda₂(X, T), lambda₂(Xp, Tp)) -> and₂(eq₂(T, Tp), eq₂(X, Xp))
if₃(true, var₁(K), var₁(L)) -> var₁(K)
if₃(false, var₁(K), var₁(L)) -> var₁(L)
ren₃(var₁(L), var₁(K), var₁(Lp)) -> if₃(eq₂(L, Lp), var₁(K), var₁(Lp))
ren₃(X, Y, apply₂(T, S)) -> apply₂(ren₃(X, Y, T), ren₃(X, Y, S))
ren₃(X, Y, lambda₂(Z, T)) -> lambda₂(var₁(cons₂(X, cons₂(Y, cons₂(lambda₂(Z, T), nil)))), ren₃(X, Y, ren₃(Z, var₁(cons₂(X, cons₂(Y, cons₂(lambda₂(Z, T), nil)))), T)))

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(var₁(L), var₁(Lp)) -> EQ₂(L, Lp)
EQ₂(apply₂(T, S), apply₂(Tp, Sp)) -> EQ₂(T, Tp)
EQ₂(lambda₂(X, T), lambda₂(Xp, Tp)) -> EQ₂(X, Xp)
EQ₂(cons₂(T, L), cons₂(Tp, Lp)) -> EQ₂(L, Lp)
EQ₂(lambda₂(X, T), lambda₂(Xp, Tp)) -> EQ₂(T, Tp)
EQ₂(apply₂(T, S), apply₂(Tp, Sp)) -> EQ₂(S, Sp)
EQ₂(cons₂(T, L), cons₂(Tp, Lp)) -> EQ₂(T, Tp)

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

POL( var₁(x₁) ) = 3x₁ + 3

POL( EQ₂(x₁, x₂) ) = max{0, 2x₁ - 3}

POL( cons₂(x₁, x₂) ) = 3x₁ + 3x₂ + 3

POL( lambda₂(x₁, x₂) ) = 3x₁ + 3x₂ + 2

POL( apply₂(x₁, x₂) ) = 3x₁ + 3x₂ + 3

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

and₂(false, false) -> false
and₂(true, false) -> false
and₂(false, true) -> false
and₂(true, true) -> true
eq₂(nil, nil) -> true
eq₂(cons₂(T, L), nil) -> false
eq₂(nil, cons₂(T, L)) -> false
eq₂(cons₂(T, L), cons₂(Tp, Lp)) -> and₂(eq₂(T, Tp), eq₂(L, Lp))
eq₂(var₁(L), var₁(Lp)) -> eq₂(L, Lp)
eq₂(var₁(L), apply₂(T, S)) -> false
eq₂(var₁(L), lambda₂(X, T)) -> false
eq₂(apply₂(T, S), var₁(L)) -> false
eq₂(apply₂(T, S), apply₂(Tp, Sp)) -> and₂(eq₂(T, Tp), eq₂(S, Sp))
eq₂(apply₂(T, S), lambda₂(X, Tp)) -> false
eq₂(lambda₂(X, T), var₁(L)) -> false
eq₂(lambda₂(X, T), apply₂(Tp, Sp)) -> false
eq₂(lambda₂(X, T), lambda₂(Xp, Tp)) -> and₂(eq₂(T, Tp), eq₂(X, Xp))
if₃(true, var₁(K), var₁(L)) -> var₁(K)
if₃(false, var₁(K), var₁(L)) -> var₁(L)
ren₃(var₁(L), var₁(K), var₁(Lp)) -> if₃(eq₂(L, Lp), var₁(K), var₁(Lp))
ren₃(X, Y, apply₂(T, S)) -> apply₂(ren₃(X, Y, T), ren₃(X, Y, S))
ren₃(X, Y, lambda₂(Z, T)) -> lambda₂(var₁(cons₂(X, cons₂(Y, cons₂(lambda₂(Z, T), nil)))), ren₃(X, Y, ren₃(Z, var₁(cons₂(X, cons₂(Y, cons₂(lambda₂(Z, T), nil)))), T)))

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

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

REN₃(X, Y, apply₂(T, S)) -> REN₃(X, Y, T)
REN₃(X, Y, lambda₂(Z, T)) -> REN₃(X, Y, ren₃(Z, var₁(cons₂(X, cons₂(Y, cons₂(lambda₂(Z, T), nil)))), T))
REN₃(X, Y, lambda₂(Z, T)) -> REN₃(Z, var₁(cons₂(X, cons₂(Y, cons₂(lambda₂(Z, T), nil)))), T)
REN₃(X, Y, apply₂(T, S)) -> REN₃(X, Y, S)

The TRS R consists of the following rules:

and₂(false, false) -> false
and₂(true, false) -> false
and₂(false, true) -> false
and₂(true, true) -> true
eq₂(nil, nil) -> true
eq₂(cons₂(T, L), nil) -> false
eq₂(nil, cons₂(T, L)) -> false
eq₂(cons₂(T, L), cons₂(Tp, Lp)) -> and₂(eq₂(T, Tp), eq₂(L, Lp))
eq₂(var₁(L), var₁(Lp)) -> eq₂(L, Lp)
eq₂(var₁(L), apply₂(T, S)) -> false
eq₂(var₁(L), lambda₂(X, T)) -> false
eq₂(apply₂(T, S), var₁(L)) -> false
eq₂(apply₂(T, S), apply₂(Tp, Sp)) -> and₂(eq₂(T, Tp), eq₂(S, Sp))
eq₂(apply₂(T, S), lambda₂(X, Tp)) -> false
eq₂(lambda₂(X, T), var₁(L)) -> false
eq₂(lambda₂(X, T), apply₂(Tp, Sp)) -> false
eq₂(lambda₂(X, T), lambda₂(Xp, Tp)) -> and₂(eq₂(T, Tp), eq₂(X, Xp))
if₃(true, var₁(K), var₁(L)) -> var₁(K)
if₃(false, var₁(K), var₁(L)) -> var₁(L)
ren₃(var₁(L), var₁(K), var₁(Lp)) -> if₃(eq₂(L, Lp), var₁(K), var₁(Lp))
ren₃(X, Y, apply₂(T, S)) -> apply₂(ren₃(X, Y, T), ren₃(X, Y, S))
ren₃(X, Y, lambda₂(Z, T)) -> lambda₂(var₁(cons₂(X, cons₂(Y, cons₂(lambda₂(Z, T), nil)))), ren₃(X, Y, ren₃(Z, var₁(cons₂(X, cons₂(Y, cons₂(lambda₂(Z, T), nil)))), T)))

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.

REN₃(X, Y, apply₂(T, S)) -> REN₃(X, Y, T)
REN₃(X, Y, apply₂(T, S)) -> REN₃(X, Y, S)

The remaining pairs can at least be oriented weakly.

REN₃(X, Y, lambda₂(Z, T)) -> REN₃(X, Y, ren₃(Z, var₁(cons₂(X, cons₂(Y, cons₂(lambda₂(Z, T), nil)))), T))
REN₃(X, Y, lambda₂(Z, T)) -> REN₃(Z, var₁(cons₂(X, cons₂(Y, cons₂(lambda₂(Z, T), nil)))), T)

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

POL( var₁(x₁) ) = max{0, -1}

POL( true ) = 1

POL( if₃(x₁, ..., x₃) ) = max{0, -3}

POL( and₂(x₁, x₂) ) = max{0, -3}

POL( ren₃(x₁, ..., x₃) ) = x₃

POL( false ) = 0

POL( lambda₂(x₁, x₂) ) = x₂

POL( eq₂(x₁, x₂) ) = max{0, -3}

POL( REN₃(x₁, ..., x₃) ) = x₃ + 2

POL( nil ) = 2

POL( cons₂(x₁, x₂) ) = max{0, -3}

POL( apply₂(x₁, x₂) ) = x₁ + x₂ + 2

The following usable rules [14] were oriented:

if₃(true, var₁(K), var₁(L)) -> var₁(K)
ren₃(X, Y, lambda₂(Z, T)) -> lambda₂(var₁(cons₂(X, cons₂(Y, cons₂(lambda₂(Z, T), nil)))), ren₃(X, Y, ren₃(Z, var₁(cons₂(X, cons₂(Y, cons₂(lambda₂(Z, T), nil)))), T)))
if₃(false, var₁(K), var₁(L)) -> var₁(L)
ren₃(var₁(L), var₁(K), var₁(Lp)) -> if₃(eq₂(L, Lp), var₁(K), var₁(Lp))
ren₃(X, Y, apply₂(T, S)) -> apply₂(ren₃(X, Y, T), ren₃(X, Y, S))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

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

REN₃(X, Y, lambda₂(Z, T)) -> REN₃(X, Y, ren₃(Z, var₁(cons₂(X, cons₂(Y, cons₂(lambda₂(Z, T), nil)))), T))
REN₃(X, Y, lambda₂(Z, T)) -> REN₃(Z, var₁(cons₂(X, cons₂(Y, cons₂(lambda₂(Z, T), nil)))), T)

The TRS R consists of the following rules:

and₂(false, false) -> false
and₂(true, false) -> false
and₂(false, true) -> false
and₂(true, true) -> true
eq₂(nil, nil) -> true
eq₂(cons₂(T, L), nil) -> false
eq₂(nil, cons₂(T, L)) -> false
eq₂(cons₂(T, L), cons₂(Tp, Lp)) -> and₂(eq₂(T, Tp), eq₂(L, Lp))
eq₂(var₁(L), var₁(Lp)) -> eq₂(L, Lp)
eq₂(var₁(L), apply₂(T, S)) -> false
eq₂(var₁(L), lambda₂(X, T)) -> false
eq₂(apply₂(T, S), var₁(L)) -> false
eq₂(apply₂(T, S), apply₂(Tp, Sp)) -> and₂(eq₂(T, Tp), eq₂(S, Sp))
eq₂(apply₂(T, S), lambda₂(X, Tp)) -> false
eq₂(lambda₂(X, T), var₁(L)) -> false
eq₂(lambda₂(X, T), apply₂(Tp, Sp)) -> false
eq₂(lambda₂(X, T), lambda₂(Xp, Tp)) -> and₂(eq₂(T, Tp), eq₂(X, Xp))
if₃(true, var₁(K), var₁(L)) -> var₁(K)
if₃(false, var₁(K), var₁(L)) -> var₁(L)
ren₃(var₁(L), var₁(K), var₁(Lp)) -> if₃(eq₂(L, Lp), var₁(K), var₁(Lp))
ren₃(X, Y, apply₂(T, S)) -> apply₂(ren₃(X, Y, T), ren₃(X, Y, S))
ren₃(X, Y, lambda₂(Z, T)) -> lambda₂(var₁(cons₂(X, cons₂(Y, cons₂(lambda₂(Z, T), nil)))), ren₃(X, Y, ren₃(Z, var₁(cons₂(X, cons₂(Y, cons₂(lambda₂(Z, T), nil)))), T)))

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.

REN₃(X, Y, lambda₂(Z, T)) -> REN₃(X, Y, ren₃(Z, var₁(cons₂(X, cons₂(Y, cons₂(lambda₂(Z, T), nil)))), T))
REN₃(X, Y, lambda₂(Z, T)) -> REN₃(Z, var₁(cons₂(X, cons₂(Y, cons₂(lambda₂(Z, T), nil)))), T)

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

POL( var₁(x₁) ) = max{0, -3}

POL( true ) = 1

POL( if₃(x₁, ..., x₃) ) = max{0, 2x₃ - 3}

POL( and₂(x₁, x₂) ) = max{0, 2x₂ - 2}

POL( ren₃(x₁, ..., x₃) ) = x₃

POL( false ) = 1

POL( lambda₂(x₁, x₂) ) = x₂ + 2

POL( eq₂(x₁, x₂) ) = max{0, 2x₂ - 3}

POL( REN₃(x₁, ..., x₃) ) = 2x₃

POL( nil ) = 0

POL( cons₂(x₁, x₂) ) = max{0, 3x₁ + 2x₂ - 3}

POL( apply₂(x₁, x₂) ) = max{0, -2}

The following usable rules [14] were oriented:

if₃(true, var₁(K), var₁(L)) -> var₁(K)
ren₃(X, Y, lambda₂(Z, T)) -> lambda₂(var₁(cons₂(X, cons₂(Y, cons₂(lambda₂(Z, T), nil)))), ren₃(X, Y, ren₃(Z, var₁(cons₂(X, cons₂(Y, cons₂(lambda₂(Z, T), nil)))), T)))
if₃(false, var₁(K), var₁(L)) -> var₁(L)
ren₃(var₁(L), var₁(K), var₁(Lp)) -> if₃(eq₂(L, Lp), var₁(K), var₁(Lp))
ren₃(X, Y, apply₂(T, S)) -> apply₂(ren₃(X, Y, T), ren₃(X, Y, S))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

and₂(false, false) -> false
and₂(true, false) -> false
and₂(false, true) -> false
and₂(true, true) -> true
eq₂(nil, nil) -> true
eq₂(cons₂(T, L), nil) -> false
eq₂(nil, cons₂(T, L)) -> false
eq₂(cons₂(T, L), cons₂(Tp, Lp)) -> and₂(eq₂(T, Tp), eq₂(L, Lp))
eq₂(var₁(L), var₁(Lp)) -> eq₂(L, Lp)
eq₂(var₁(L), apply₂(T, S)) -> false
eq₂(var₁(L), lambda₂(X, T)) -> false
eq₂(apply₂(T, S), var₁(L)) -> false
eq₂(apply₂(T, S), apply₂(Tp, Sp)) -> and₂(eq₂(T, Tp), eq₂(S, Sp))
eq₂(apply₂(T, S), lambda₂(X, Tp)) -> false
eq₂(lambda₂(X, T), var₁(L)) -> false
eq₂(lambda₂(X, T), apply₂(Tp, Sp)) -> false
eq₂(lambda₂(X, T), lambda₂(Xp, Tp)) -> and₂(eq₂(T, Tp), eq₂(X, Xp))
if₃(true, var₁(K), var₁(L)) -> var₁(K)
if₃(false, var₁(K), var₁(L)) -> var₁(L)
ren₃(var₁(L), var₁(K), var₁(Lp)) -> if₃(eq₂(L, Lp), var₁(K), var₁(Lp))
ren₃(X, Y, apply₂(T, S)) -> apply₂(ren₃(X, Y, T), ren₃(X, Y, S))
ren₃(X, Y, lambda₂(Z, T)) -> lambda₂(var₁(cons₂(X, cons₂(Y, cons₂(lambda₂(Z, T), nil)))), ren₃(X, Y, ren₃(Z, var₁(cons₂(X, cons₂(Y, cons₂(lambda₂(Z, T), nil)))), T)))

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.