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.