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 interpretation [21]:

POL(EQ2(x1, x2)) = 3·x1 + 3·x2   
POL(apply2(x1, x2)) = 3 + 2·x1 + 2·x2   
POL(cons2(x1, x2)) = 3 + 2·x1 + 2·x2   
POL(lambda2(x1, x2)) = 3 + 2·x1 + 2·x2   
POL(var1(x1)) = 3 + x1   

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 interpretation [21]:

POL(REN3(x1, x2, x3)) = 2·x3   
POL(and2(x1, x2)) = 0   
POL(apply2(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(cons2(x1, x2)) = 2   
POL(eq2(x1, x2)) = 3·x2   
POL(false) = 0   
POL(if3(x1, x2, x3)) = x2   
POL(lambda2(x1, x2)) = x2   
POL(nil) = 0   
POL(ren3(x1, x2, x3)) = x3   
POL(true) = 0   
POL(var1(x1)) = 2   

The following usable rules [14] were oriented:

ren3(var1(L), var1(K), var1(Lp)) -> if3(eq2(L, Lp), var1(K), var1(Lp))
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)
if3(true, var1(K), var1(L)) -> var1(K)
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 interpretation [21]:

POL(REN3(x1, x2, x3)) = 2·x3   
POL(and2(x1, x2)) = 0   
POL(apply2(x1, x2)) = 0   
POL(cons2(x1, x2)) = 0   
POL(eq2(x1, x2)) = x2   
POL(false) = 0   
POL(if3(x1, x2, x3)) = 0   
POL(lambda2(x1, x2)) = 3 + x2   
POL(nil) = 0   
POL(ren3(x1, x2, x3)) = x3   
POL(true) = 0   
POL(var1(x1)) = 0   

The following usable rules [14] were oriented:

ren3(var1(L), var1(K), var1(Lp)) -> if3(eq2(L, Lp), var1(K), var1(Lp))
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)
if3(true, var1(K), var1(L)) -> var1(K)
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.