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
  ↳ Non-Overlap Check

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.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
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)))

The set Q consists of the following terms:

and2(false, false)
and2(true, false)
and2(false, true)
and2(true, true)
eq2(nil, nil)
eq2(cons2(x0, x1), nil)
eq2(nil, cons2(x0, x1))
eq2(cons2(x0, x1), cons2(x2, x3))
eq2(var1(x0), var1(x1))
eq2(var1(x0), apply2(x1, x2))
eq2(var1(x0), lambda2(x1, x2))
eq2(apply2(x0, x1), var1(x2))
eq2(apply2(x0, x1), apply2(x2, x3))
eq2(apply2(x0, x1), lambda2(x2, x3))
eq2(lambda2(x0, x1), var1(x2))
eq2(lambda2(x0, x1), apply2(x2, x3))
eq2(lambda2(x0, x1), lambda2(x2, x3))
if3(true, var1(x0), var1(x1))
if3(false, var1(x0), var1(x1))
ren3(var1(x0), var1(x1), var1(x2))
ren3(x0, x1, apply2(x2, x3))
ren3(x0, x1, lambda2(x2, x3))


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)))

The set Q consists of the following terms:

and2(false, false)
and2(true, false)
and2(false, true)
and2(true, true)
eq2(nil, nil)
eq2(cons2(x0, x1), nil)
eq2(nil, cons2(x0, x1))
eq2(cons2(x0, x1), cons2(x2, x3))
eq2(var1(x0), var1(x1))
eq2(var1(x0), apply2(x1, x2))
eq2(var1(x0), lambda2(x1, x2))
eq2(apply2(x0, x1), var1(x2))
eq2(apply2(x0, x1), apply2(x2, x3))
eq2(apply2(x0, x1), lambda2(x2, x3))
eq2(lambda2(x0, x1), var1(x2))
eq2(lambda2(x0, x1), apply2(x2, x3))
eq2(lambda2(x0, x1), lambda2(x2, x3))
if3(true, var1(x0), var1(x1))
if3(false, var1(x0), var1(x1))
ren3(var1(x0), var1(x1), var1(x2))
ren3(x0, x1, apply2(x2, x3))
ren3(x0, x1, lambda2(x2, x3))

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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)))

The set Q consists of the following terms:

and2(false, false)
and2(true, false)
and2(false, true)
and2(true, true)
eq2(nil, nil)
eq2(cons2(x0, x1), nil)
eq2(nil, cons2(x0, x1))
eq2(cons2(x0, x1), cons2(x2, x3))
eq2(var1(x0), var1(x1))
eq2(var1(x0), apply2(x1, x2))
eq2(var1(x0), lambda2(x1, x2))
eq2(apply2(x0, x1), var1(x2))
eq2(apply2(x0, x1), apply2(x2, x3))
eq2(apply2(x0, x1), lambda2(x2, x3))
eq2(lambda2(x0, x1), var1(x2))
eq2(lambda2(x0, x1), apply2(x2, x3))
eq2(lambda2(x0, x1), lambda2(x2, x3))
if3(true, var1(x0), var1(x1))
if3(false, var1(x0), var1(x1))
ren3(var1(x0), var1(x1), var1(x2))
ren3(x0, x1, apply2(x2, x3))
ren3(x0, x1, lambda2(x2, x3))

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
  ↳ Non-Overlap Check
    ↳ 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)))

The set Q consists of the following terms:

and2(false, false)
and2(true, false)
and2(false, true)
and2(true, true)
eq2(nil, nil)
eq2(cons2(x0, x1), nil)
eq2(nil, cons2(x0, x1))
eq2(cons2(x0, x1), cons2(x2, x3))
eq2(var1(x0), var1(x1))
eq2(var1(x0), apply2(x1, x2))
eq2(var1(x0), lambda2(x1, x2))
eq2(apply2(x0, x1), var1(x2))
eq2(apply2(x0, x1), apply2(x2, x3))
eq2(apply2(x0, x1), lambda2(x2, x3))
eq2(lambda2(x0, x1), var1(x2))
eq2(lambda2(x0, x1), apply2(x2, x3))
eq2(lambda2(x0, x1), lambda2(x2, x3))
if3(true, var1(x0), var1(x1))
if3(false, var1(x0), var1(x1))
ren3(var1(x0), var1(x1), var1(x2))
ren3(x0, x1, apply2(x2, x3))
ren3(x0, x1, lambda2(x2, x3))

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


The following pairs can be strictly oriented and are deleted.


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 by weakly be oriented.

EQ2(var1(L), var1(Lp)) -> EQ2(L, Lp)
Used ordering: Combined order from the following AFS and order.
EQ2(x1, x2)  =  x1
var1(x1)  =  x1
apply2(x1, x2)  =  apply2(x1, x2)
lambda2(x1, x2)  =  lambda2(x1, x2)
cons2(x1, x2)  =  cons2(x1, x2)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ QDPOrderProof
              ↳ QDP

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

EQ2(var1(L), var1(Lp)) -> EQ2(L, Lp)

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)))

The set Q consists of the following terms:

and2(false, false)
and2(true, false)
and2(false, true)
and2(true, true)
eq2(nil, nil)
eq2(cons2(x0, x1), nil)
eq2(nil, cons2(x0, x1))
eq2(cons2(x0, x1), cons2(x2, x3))
eq2(var1(x0), var1(x1))
eq2(var1(x0), apply2(x1, x2))
eq2(var1(x0), lambda2(x1, x2))
eq2(apply2(x0, x1), var1(x2))
eq2(apply2(x0, x1), apply2(x2, x3))
eq2(apply2(x0, x1), lambda2(x2, x3))
eq2(lambda2(x0, x1), var1(x2))
eq2(lambda2(x0, x1), apply2(x2, x3))
eq2(lambda2(x0, x1), lambda2(x2, x3))
if3(true, var1(x0), var1(x1))
if3(false, var1(x0), var1(x1))
ren3(var1(x0), var1(x1), var1(x2))
ren3(x0, x1, apply2(x2, x3))
ren3(x0, x1, lambda2(x2, x3))

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


The following pairs can be strictly oriented and are deleted.


EQ2(var1(L), var1(Lp)) -> EQ2(L, Lp)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
EQ2(x1, x2)  =  EQ1(x1)
var1(x1)  =  var1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ 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)))

The set Q consists of the following terms:

and2(false, false)
and2(true, false)
and2(false, true)
and2(true, true)
eq2(nil, nil)
eq2(cons2(x0, x1), nil)
eq2(nil, cons2(x0, x1))
eq2(cons2(x0, x1), cons2(x2, x3))
eq2(var1(x0), var1(x1))
eq2(var1(x0), apply2(x1, x2))
eq2(var1(x0), lambda2(x1, x2))
eq2(apply2(x0, x1), var1(x2))
eq2(apply2(x0, x1), apply2(x2, x3))
eq2(apply2(x0, x1), lambda2(x2, x3))
eq2(lambda2(x0, x1), var1(x2))
eq2(lambda2(x0, x1), apply2(x2, x3))
eq2(lambda2(x0, x1), lambda2(x2, x3))
if3(true, var1(x0), var1(x1))
if3(false, var1(x0), var1(x1))
ren3(var1(x0), var1(x1), var1(x2))
ren3(x0, x1, apply2(x2, x3))
ren3(x0, x1, lambda2(x2, x3))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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)))

The set Q consists of the following terms:

and2(false, false)
and2(true, false)
and2(false, true)
and2(true, true)
eq2(nil, nil)
eq2(cons2(x0, x1), nil)
eq2(nil, cons2(x0, x1))
eq2(cons2(x0, x1), cons2(x2, x3))
eq2(var1(x0), var1(x1))
eq2(var1(x0), apply2(x1, x2))
eq2(var1(x0), lambda2(x1, x2))
eq2(apply2(x0, x1), var1(x2))
eq2(apply2(x0, x1), apply2(x2, x3))
eq2(apply2(x0, x1), lambda2(x2, x3))
eq2(lambda2(x0, x1), var1(x2))
eq2(lambda2(x0, x1), apply2(x2, x3))
eq2(lambda2(x0, x1), lambda2(x2, x3))
if3(true, var1(x0), var1(x1))
if3(false, var1(x0), var1(x1))
ren3(var1(x0), var1(x1), var1(x2))
ren3(x0, x1, apply2(x2, x3))
ren3(x0, x1, lambda2(x2, x3))

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


The following pairs can be strictly oriented 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 by weakly be oriented.

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: Combined order from the following AFS and order.
REN3(x1, x2, x3)  =  REN2(x2, x3)
apply2(x1, x2)  =  apply2(x1, x2)
lambda2(x1, x2)  =  x2
ren3(x1, x2, x3)  =  x3
var1(x1)  =  var
cons2(x1, x2)  =  x2
nil  =  nil
if3(x1, x2, x3)  =  if
eq2(x1, x2)  =  eq
true  =  true
false  =  false
and2(x1, x2)  =  and

Lexicographic Path Order [19].
Precedence:
apply2 > REN2 > [var, if]
nil > [true, false, and] > [var, if]
eq > [true, false, and] > [var, if]


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, 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)))
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)
and2(false, false) -> false
and2(true, false) -> false
and2(false, true) -> false
and2(true, true) -> true



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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)))

The set Q consists of the following terms:

and2(false, false)
and2(true, false)
and2(false, true)
and2(true, true)
eq2(nil, nil)
eq2(cons2(x0, x1), nil)
eq2(nil, cons2(x0, x1))
eq2(cons2(x0, x1), cons2(x2, x3))
eq2(var1(x0), var1(x1))
eq2(var1(x0), apply2(x1, x2))
eq2(var1(x0), lambda2(x1, x2))
eq2(apply2(x0, x1), var1(x2))
eq2(apply2(x0, x1), apply2(x2, x3))
eq2(apply2(x0, x1), lambda2(x2, x3))
eq2(lambda2(x0, x1), var1(x2))
eq2(lambda2(x0, x1), apply2(x2, x3))
eq2(lambda2(x0, x1), lambda2(x2, x3))
if3(true, var1(x0), var1(x1))
if3(false, var1(x0), var1(x1))
ren3(var1(x0), var1(x1), var1(x2))
ren3(x0, x1, apply2(x2, x3))
ren3(x0, x1, lambda2(x2, x3))

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


The following pairs can be strictly oriented 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 by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
REN3(x1, x2, x3)  =  x3
lambda2(x1, x2)  =  lambda2(x1, x2)
ren3(x1, x2, x3)  =  x3
var1(x1)  =  var
cons2(x1, x2)  =  cons
nil  =  nil
if3(x1, x2, x3)  =  x1
eq2(x1, x2)  =  eq
apply2(x1, x2)  =  apply
true  =  true
false  =  false
and2(x1, x2)  =  x2

Lexicographic Path Order [19].
Precedence:
cons > [lambda2, var, eq, apply, true, false]
nil > [lambda2, var, eq, apply, true, false]


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, 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)))
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)
and2(false, false) -> false
and2(true, false) -> false
and2(false, true) -> false
and2(true, true) -> true



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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)))

The set Q consists of the following terms:

and2(false, false)
and2(true, false)
and2(false, true)
and2(true, true)
eq2(nil, nil)
eq2(cons2(x0, x1), nil)
eq2(nil, cons2(x0, x1))
eq2(cons2(x0, x1), cons2(x2, x3))
eq2(var1(x0), var1(x1))
eq2(var1(x0), apply2(x1, x2))
eq2(var1(x0), lambda2(x1, x2))
eq2(apply2(x0, x1), var1(x2))
eq2(apply2(x0, x1), apply2(x2, x3))
eq2(apply2(x0, x1), lambda2(x2, x3))
eq2(lambda2(x0, x1), var1(x2))
eq2(lambda2(x0, x1), apply2(x2, x3))
eq2(lambda2(x0, x1), lambda2(x2, x3))
if3(true, var1(x0), var1(x1))
if3(false, var1(x0), var1(x1))
ren3(var1(x0), var1(x1), var1(x2))
ren3(x0, x1, apply2(x2, x3))
ren3(x0, x1, lambda2(x2, x3))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.