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(true, y) -> y
and2(false, y) -> false
eq2(nil, nil) -> true
eq2(cons2(t, l), nil) -> false
eq2(nil, cons2(t, l)) -> false
eq2(cons2(t, l), cons2(t', l')) -> and2(eq2(t, t'), eq2(l, l'))
eq2(var1(l), var1(l')) -> eq2(l, l')
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(t', s')) -> and2(eq2(t, t'), eq2(s, s'))
eq2(apply2(t, s), lambda2(x, t)) -> false
eq2(lambda2(x, t), var1(l)) -> false
eq2(lambda2(x, t), apply2(t, s)) -> false
eq2(lambda2(x, t), lambda2(x', t')) -> and2(eq2(x, x'), eq2(t, t'))
if3(true, var1(k), var1(l')) -> var1(k)
if3(false, var1(k), var1(l')) -> var1(l')
ren3(var1(l), var1(k), var1(l')) -> if3(eq2(l, l'), var1(k), var1(l'))
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(true, y) -> y
and2(false, y) -> false
eq2(nil, nil) -> true
eq2(cons2(t, l), nil) -> false
eq2(nil, cons2(t, l)) -> false
eq2(cons2(t, l), cons2(t', l')) -> and2(eq2(t, t'), eq2(l, l'))
eq2(var1(l), var1(l')) -> eq2(l, l')
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(t', s')) -> and2(eq2(t, t'), eq2(s, s'))
eq2(apply2(t, s), lambda2(x, t)) -> false
eq2(lambda2(x, t), var1(l)) -> false
eq2(lambda2(x, t), apply2(t, s)) -> false
eq2(lambda2(x, t), lambda2(x', t')) -> and2(eq2(x, x'), eq2(t, t'))
if3(true, var1(k), var1(l')) -> var1(k)
if3(false, var1(k), var1(l')) -> var1(l')
ren3(var1(l), var1(k), var1(l')) -> if3(eq2(l, l'), var1(k), var1(l'))
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(true, y) -> y
and2(false, y) -> false
eq2(nil, nil) -> true
eq2(cons2(t, l), nil) -> false
eq2(nil, cons2(t, l)) -> false
eq2(cons2(t, l), cons2(t', l')) -> and2(eq2(t, t'), eq2(l, l'))
eq2(var1(l), var1(l')) -> eq2(l, l')
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(t', s')) -> and2(eq2(t, t'), eq2(s, s'))
eq2(apply2(t, s), lambda2(x, t)) -> false
eq2(lambda2(x, t), var1(l)) -> false
eq2(lambda2(x, t), apply2(t, s)) -> false
eq2(lambda2(x, t), lambda2(x', t')) -> and2(eq2(x, x'), eq2(t, t'))
if3(true, var1(k), var1(l')) -> var1(k)
if3(false, var1(k), var1(l')) -> var1(l')
ren3(var1(l), var1(k), var1(l')) -> if3(eq2(l, l'), var1(k), var1(l'))
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(true, x0)
and2(false, x0)
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, x0))
eq2(lambda2(x0, x1), var1(x2))
eq2(lambda2(x0, x1), apply2(x1, x2))
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))


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

EQ2(var1(l), var1(l')) -> EQ2(l, l')
REN3(x, y, apply2(t, s)) -> REN3(x, y, t)
EQ2(apply2(t, s), apply2(t', s')) -> EQ2(t, t')
EQ2(lambda2(x, t), lambda2(x', t')) -> EQ2(x, x')
EQ2(cons2(t, l), cons2(t', l')) -> EQ2(l, l')
EQ2(apply2(t, s), apply2(t', s')) -> AND2(eq2(t, t'), eq2(s, s'))
EQ2(cons2(t, l), cons2(t', l')) -> EQ2(t, t')
REN3(var1(l), var1(k), var1(l')) -> EQ2(l, l')
EQ2(lambda2(x, t), lambda2(x', t')) -> AND2(eq2(x, x'), eq2(t, t'))
EQ2(cons2(t, l), cons2(t', l')) -> AND2(eq2(t, t'), eq2(l, l'))
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(x', t')) -> EQ2(t, t')
EQ2(apply2(t, s), apply2(t', s')) -> EQ2(s, s')
REN3(var1(l), var1(k), var1(l')) -> IF3(eq2(l, l'), var1(k), var1(l'))
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(true, y) -> y
and2(false, y) -> false
eq2(nil, nil) -> true
eq2(cons2(t, l), nil) -> false
eq2(nil, cons2(t, l)) -> false
eq2(cons2(t, l), cons2(t', l')) -> and2(eq2(t, t'), eq2(l, l'))
eq2(var1(l), var1(l')) -> eq2(l, l')
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(t', s')) -> and2(eq2(t, t'), eq2(s, s'))
eq2(apply2(t, s), lambda2(x, t)) -> false
eq2(lambda2(x, t), var1(l)) -> false
eq2(lambda2(x, t), apply2(t, s)) -> false
eq2(lambda2(x, t), lambda2(x', t')) -> and2(eq2(x, x'), eq2(t, t'))
if3(true, var1(k), var1(l')) -> var1(k)
if3(false, var1(k), var1(l')) -> var1(l')
ren3(var1(l), var1(k), var1(l')) -> if3(eq2(l, l'), var1(k), var1(l'))
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(true, x0)
and2(false, x0)
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, x0))
eq2(lambda2(x0, x1), var1(x2))
eq2(lambda2(x0, x1), apply2(x1, x2))
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(l')) -> EQ2(l, l')
REN3(x, y, apply2(t, s)) -> REN3(x, y, t)
EQ2(apply2(t, s), apply2(t', s')) -> EQ2(t, t')
EQ2(lambda2(x, t), lambda2(x', t')) -> EQ2(x, x')
EQ2(cons2(t, l), cons2(t', l')) -> EQ2(l, l')
EQ2(apply2(t, s), apply2(t', s')) -> AND2(eq2(t, t'), eq2(s, s'))
EQ2(cons2(t, l), cons2(t', l')) -> EQ2(t, t')
REN3(var1(l), var1(k), var1(l')) -> EQ2(l, l')
EQ2(lambda2(x, t), lambda2(x', t')) -> AND2(eq2(x, x'), eq2(t, t'))
EQ2(cons2(t, l), cons2(t', l')) -> AND2(eq2(t, t'), eq2(l, l'))
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(x', t')) -> EQ2(t, t')
EQ2(apply2(t, s), apply2(t', s')) -> EQ2(s, s')
REN3(var1(l), var1(k), var1(l')) -> IF3(eq2(l, l'), var1(k), var1(l'))
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(true, y) -> y
and2(false, y) -> false
eq2(nil, nil) -> true
eq2(cons2(t, l), nil) -> false
eq2(nil, cons2(t, l)) -> false
eq2(cons2(t, l), cons2(t', l')) -> and2(eq2(t, t'), eq2(l, l'))
eq2(var1(l), var1(l')) -> eq2(l, l')
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(t', s')) -> and2(eq2(t, t'), eq2(s, s'))
eq2(apply2(t, s), lambda2(x, t)) -> false
eq2(lambda2(x, t), var1(l)) -> false
eq2(lambda2(x, t), apply2(t, s)) -> false
eq2(lambda2(x, t), lambda2(x', t')) -> and2(eq2(x, x'), eq2(t, t'))
if3(true, var1(k), var1(l')) -> var1(k)
if3(false, var1(k), var1(l')) -> var1(l')
ren3(var1(l), var1(k), var1(l')) -> if3(eq2(l, l'), var1(k), var1(l'))
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(true, x0)
and2(false, x0)
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, x0))
eq2(lambda2(x0, x1), var1(x2))
eq2(lambda2(x0, x1), apply2(x1, x2))
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 contains 2 SCCs with 5 less nodes.

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

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

EQ2(var1(l), var1(l')) -> EQ2(l, l')
EQ2(apply2(t, s), apply2(t', s')) -> EQ2(t, t')
EQ2(lambda2(x, t), lambda2(x', t')) -> EQ2(x, x')
EQ2(cons2(t, l), cons2(t', l')) -> EQ2(l, l')
EQ2(lambda2(x, t), lambda2(x', t')) -> EQ2(t, t')
EQ2(apply2(t, s), apply2(t', s')) -> EQ2(s, s')
EQ2(cons2(t, l), cons2(t', l')) -> EQ2(t, t')

The TRS R consists of the following rules:

and2(true, y) -> y
and2(false, y) -> false
eq2(nil, nil) -> true
eq2(cons2(t, l), nil) -> false
eq2(nil, cons2(t, l)) -> false
eq2(cons2(t, l), cons2(t', l')) -> and2(eq2(t, t'), eq2(l, l'))
eq2(var1(l), var1(l')) -> eq2(l, l')
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(t', s')) -> and2(eq2(t, t'), eq2(s, s'))
eq2(apply2(t, s), lambda2(x, t)) -> false
eq2(lambda2(x, t), var1(l)) -> false
eq2(lambda2(x, t), apply2(t, s)) -> false
eq2(lambda2(x, t), lambda2(x', t')) -> and2(eq2(x, x'), eq2(t, t'))
if3(true, var1(k), var1(l')) -> var1(k)
if3(false, var1(k), var1(l')) -> var1(l')
ren3(var1(l), var1(k), var1(l')) -> if3(eq2(l, l'), var1(k), var1(l'))
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(true, x0)
and2(false, x0)
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, x0))
eq2(lambda2(x0, x1), var1(x2))
eq2(lambda2(x0, x1), apply2(x1, x2))
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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

EQ2(apply2(t, s), apply2(t', s')) -> EQ2(t, t')
EQ2(lambda2(x, t), lambda2(x', t')) -> EQ2(x, x')
EQ2(cons2(t, l), cons2(t', l')) -> EQ2(l, l')
EQ2(lambda2(x, t), lambda2(x', t')) -> EQ2(t, t')
EQ2(apply2(t, s), apply2(t', s')) -> EQ2(s, s')
EQ2(cons2(t, l), cons2(t', l')) -> EQ2(t, t')
Used argument filtering: EQ2(x1, x2)  =  x2
var1(x1)  =  x1
apply2(x1, x2)  =  apply2(x1, x2)
lambda2(x1, x2)  =  lambda2(x1, x2)
cons2(x1, x2)  =  cons2(x1, x2)
Used ordering: Quasi Precedence: trivial


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

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

EQ2(var1(l), var1(l')) -> EQ2(l, l')

The TRS R consists of the following rules:

and2(true, y) -> y
and2(false, y) -> false
eq2(nil, nil) -> true
eq2(cons2(t, l), nil) -> false
eq2(nil, cons2(t, l)) -> false
eq2(cons2(t, l), cons2(t', l')) -> and2(eq2(t, t'), eq2(l, l'))
eq2(var1(l), var1(l')) -> eq2(l, l')
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(t', s')) -> and2(eq2(t, t'), eq2(s, s'))
eq2(apply2(t, s), lambda2(x, t)) -> false
eq2(lambda2(x, t), var1(l)) -> false
eq2(lambda2(x, t), apply2(t, s)) -> false
eq2(lambda2(x, t), lambda2(x', t')) -> and2(eq2(x, x'), eq2(t, t'))
if3(true, var1(k), var1(l')) -> var1(k)
if3(false, var1(k), var1(l')) -> var1(l')
ren3(var1(l), var1(k), var1(l')) -> if3(eq2(l, l'), var1(k), var1(l'))
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(true, x0)
and2(false, x0)
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, x0))
eq2(lambda2(x0, x1), var1(x2))
eq2(lambda2(x0, x1), apply2(x1, x2))
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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

EQ2(var1(l), var1(l')) -> EQ2(l, l')
Used argument filtering: EQ2(x1, x2)  =  x2
var1(x1)  =  var1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ QDPAfsSolverProof
QDP
                        ↳ PisEmptyProof
              ↳ QDP

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

and2(true, y) -> y
and2(false, y) -> false
eq2(nil, nil) -> true
eq2(cons2(t, l), nil) -> false
eq2(nil, cons2(t, l)) -> false
eq2(cons2(t, l), cons2(t', l')) -> and2(eq2(t, t'), eq2(l, l'))
eq2(var1(l), var1(l')) -> eq2(l, l')
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(t', s')) -> and2(eq2(t, t'), eq2(s, s'))
eq2(apply2(t, s), lambda2(x, t)) -> false
eq2(lambda2(x, t), var1(l)) -> false
eq2(lambda2(x, t), apply2(t, s)) -> false
eq2(lambda2(x, t), lambda2(x', t')) -> and2(eq2(x, x'), eq2(t, t'))
if3(true, var1(k), var1(l')) -> var1(k)
if3(false, var1(k), var1(l')) -> var1(l')
ren3(var1(l), var1(k), var1(l')) -> if3(eq2(l, l'), var1(k), var1(l'))
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(true, x0)
and2(false, x0)
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, x0))
eq2(lambda2(x0, x1), var1(x2))
eq2(lambda2(x0, x1), apply2(x1, x2))
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
                ↳ QDPAfsSolverProof

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(true, y) -> y
and2(false, y) -> false
eq2(nil, nil) -> true
eq2(cons2(t, l), nil) -> false
eq2(nil, cons2(t, l)) -> false
eq2(cons2(t, l), cons2(t', l')) -> and2(eq2(t, t'), eq2(l, l'))
eq2(var1(l), var1(l')) -> eq2(l, l')
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(t', s')) -> and2(eq2(t, t'), eq2(s, s'))
eq2(apply2(t, s), lambda2(x, t)) -> false
eq2(lambda2(x, t), var1(l)) -> false
eq2(lambda2(x, t), apply2(t, s)) -> false
eq2(lambda2(x, t), lambda2(x', t')) -> and2(eq2(x, x'), eq2(t, t'))
if3(true, var1(k), var1(l')) -> var1(k)
if3(false, var1(k), var1(l')) -> var1(l')
ren3(var1(l), var1(k), var1(l')) -> if3(eq2(l, l'), var1(k), var1(l'))
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(true, x0)
and2(false, x0)
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, x0))
eq2(lambda2(x0, x1), var1(x2))
eq2(lambda2(x0, x1), apply2(x1, x2))
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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

REN3(x, y, apply2(t, s)) -> REN3(x, y, t)
REN3(x, y, apply2(t, s)) -> REN3(x, y, s)
Used argument filtering: REN3(x1, x2, x3)  =  x3
apply2(x1, x2)  =  apply2(x1, x2)
lambda2(x1, x2)  =  x2
ren3(x1, x2, x3)  =  x3
var1(x1)  =  var
if3(x1, x2, x3)  =  if
Used ordering: Quasi Precedence: [var, if]


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

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(true, y) -> y
and2(false, y) -> false
eq2(nil, nil) -> true
eq2(cons2(t, l), nil) -> false
eq2(nil, cons2(t, l)) -> false
eq2(cons2(t, l), cons2(t', l')) -> and2(eq2(t, t'), eq2(l, l'))
eq2(var1(l), var1(l')) -> eq2(l, l')
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(t', s')) -> and2(eq2(t, t'), eq2(s, s'))
eq2(apply2(t, s), lambda2(x, t)) -> false
eq2(lambda2(x, t), var1(l)) -> false
eq2(lambda2(x, t), apply2(t, s)) -> false
eq2(lambda2(x, t), lambda2(x', t')) -> and2(eq2(x, x'), eq2(t, t'))
if3(true, var1(k), var1(l')) -> var1(k)
if3(false, var1(k), var1(l')) -> var1(l')
ren3(var1(l), var1(k), var1(l')) -> if3(eq2(l, l'), var1(k), var1(l'))
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(true, x0)
and2(false, x0)
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, x0))
eq2(lambda2(x0, x1), var1(x2))
eq2(lambda2(x0, x1), apply2(x1, x2))
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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly 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 argument filtering: REN3(x1, x2, x3)  =  x3
lambda2(x1, x2)  =  lambda1(x2)
ren3(x1, x2, x3)  =  x3
var1(x1)  =  var
if3(x1, x2, x3)  =  if
apply2(x1, x2)  =  apply
Used ordering: Quasi Precedence: [var, if]


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ QDPAfsSolverProof
QDP
                        ↳ PisEmptyProof

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

and2(true, y) -> y
and2(false, y) -> false
eq2(nil, nil) -> true
eq2(cons2(t, l), nil) -> false
eq2(nil, cons2(t, l)) -> false
eq2(cons2(t, l), cons2(t', l')) -> and2(eq2(t, t'), eq2(l, l'))
eq2(var1(l), var1(l')) -> eq2(l, l')
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(t', s')) -> and2(eq2(t, t'), eq2(s, s'))
eq2(apply2(t, s), lambda2(x, t)) -> false
eq2(lambda2(x, t), var1(l)) -> false
eq2(lambda2(x, t), apply2(t, s)) -> false
eq2(lambda2(x, t), lambda2(x', t')) -> and2(eq2(x, x'), eq2(t, t'))
if3(true, var1(k), var1(l')) -> var1(k)
if3(false, var1(k), var1(l')) -> var1(l')
ren3(var1(l), var1(k), var1(l')) -> if3(eq2(l, l'), var1(k), var1(l'))
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(true, x0)
and2(false, x0)
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, x0))
eq2(lambda2(x0, x1), var1(x2))
eq2(lambda2(x0, x1), apply2(x1, x2))
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.