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:

del1(.2(x, .2(y, z))) -> f4(=2(x, y), x, y, z)
f4(true, x, y, z) -> del1(.2(y, z))
f4(false, x, y, z) -> .2(x, del1(.2(y, z)))
=2(nil, nil) -> true
=2(.2(x, y), nil) -> false
=2(nil, .2(y, z)) -> false
=2(.2(x, y), .2(u, v)) -> and2(=2(x, u), =2(y, v))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

del1(.2(x, .2(y, z))) -> f4(=2(x, y), x, y, z)
f4(true, x, y, z) -> del1(.2(y, z))
f4(false, x, y, z) -> .2(x, del1(.2(y, z)))
=2(nil, nil) -> true
=2(.2(x, y), nil) -> false
=2(nil, .2(y, z)) -> false
=2(.2(x, y), .2(u, v)) -> and2(=2(x, u), =2(y, v))

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:

del1(.2(x, .2(y, z))) -> f4(=2(x, y), x, y, z)
f4(true, x, y, z) -> del1(.2(y, z))
f4(false, x, y, z) -> .2(x, del1(.2(y, z)))
=2(nil, nil) -> true
=2(.2(x, y), nil) -> false
=2(nil, .2(y, z)) -> false
=2(.2(x, y), .2(u, v)) -> and2(=2(x, u), =2(y, v))

The set Q consists of the following terms:

del1(.2(x0, .2(x1, x2)))
f4(true, x0, x1, x2)
f4(false, x0, x1, x2)
=2(nil, nil)
=2(.2(x0, x1), nil)
=2(nil, .2(x0, x1))
=2(.2(x0, x1), .2(u, v))


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

F4(false, x, y, z) -> DEL1(.2(y, z))
=12(.2(x, y), .2(u, v)) -> =12(x, u)
F4(true, x, y, z) -> DEL1(.2(y, z))
DEL1(.2(x, .2(y, z))) -> F4(=2(x, y), x, y, z)
DEL1(.2(x, .2(y, z))) -> =12(x, y)
=12(.2(x, y), .2(u, v)) -> =12(y, v)

The TRS R consists of the following rules:

del1(.2(x, .2(y, z))) -> f4(=2(x, y), x, y, z)
f4(true, x, y, z) -> del1(.2(y, z))
f4(false, x, y, z) -> .2(x, del1(.2(y, z)))
=2(nil, nil) -> true
=2(.2(x, y), nil) -> false
=2(nil, .2(y, z)) -> false
=2(.2(x, y), .2(u, v)) -> and2(=2(x, u), =2(y, v))

The set Q consists of the following terms:

del1(.2(x0, .2(x1, x2)))
f4(true, x0, x1, x2)
f4(false, x0, x1, x2)
=2(nil, nil)
=2(.2(x0, x1), nil)
=2(nil, .2(x0, x1))
=2(.2(x0, x1), .2(u, v))

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:

F4(false, x, y, z) -> DEL1(.2(y, z))
=12(.2(x, y), .2(u, v)) -> =12(x, u)
F4(true, x, y, z) -> DEL1(.2(y, z))
DEL1(.2(x, .2(y, z))) -> F4(=2(x, y), x, y, z)
DEL1(.2(x, .2(y, z))) -> =12(x, y)
=12(.2(x, y), .2(u, v)) -> =12(y, v)

The TRS R consists of the following rules:

del1(.2(x, .2(y, z))) -> f4(=2(x, y), x, y, z)
f4(true, x, y, z) -> del1(.2(y, z))
f4(false, x, y, z) -> .2(x, del1(.2(y, z)))
=2(nil, nil) -> true
=2(.2(x, y), nil) -> false
=2(nil, .2(y, z)) -> false
=2(.2(x, y), .2(u, v)) -> and2(=2(x, u), =2(y, v))

The set Q consists of the following terms:

del1(.2(x0, .2(x1, x2)))
f4(true, x0, x1, x2)
f4(false, x0, x1, x2)
=2(nil, nil)
=2(.2(x0, x1), nil)
=2(nil, .2(x0, x1))
=2(.2(x0, x1), .2(u, v))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 3 less nodes.

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

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

F4(false, x, y, z) -> DEL1(.2(y, z))
F4(true, x, y, z) -> DEL1(.2(y, z))
DEL1(.2(x, .2(y, z))) -> F4(=2(x, y), x, y, z)

The TRS R consists of the following rules:

del1(.2(x, .2(y, z))) -> f4(=2(x, y), x, y, z)
f4(true, x, y, z) -> del1(.2(y, z))
f4(false, x, y, z) -> .2(x, del1(.2(y, z)))
=2(nil, nil) -> true
=2(.2(x, y), nil) -> false
=2(nil, .2(y, z)) -> false
=2(.2(x, y), .2(u, v)) -> and2(=2(x, u), =2(y, v))

The set Q consists of the following terms:

del1(.2(x0, .2(x1, x2)))
f4(true, x0, x1, x2)
f4(false, x0, x1, x2)
=2(nil, nil)
=2(.2(x0, x1), nil)
=2(nil, .2(x0, x1))
=2(.2(x0, x1), .2(u, v))

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.

DEL1(.2(x, .2(y, z))) -> F4(=2(x, y), x, y, z)
Used argument filtering: F4(x1, x2, x3, x4)  =  F1(x4)
DEL1(x1)  =  x1
.2(x1, x2)  =  .1(x2)
=2(x1, x2)  =  =
nil  =  nil
true  =  true
false  =  false
v  =  v
and2(x1, x2)  =  and
u  =  u
Used ordering: Quasi Precedence: [F_1, ._1] [=, true, false, and] 


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

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

F4(false, x, y, z) -> DEL1(.2(y, z))
F4(true, x, y, z) -> DEL1(.2(y, z))

The TRS R consists of the following rules:

del1(.2(x, .2(y, z))) -> f4(=2(x, y), x, y, z)
f4(true, x, y, z) -> del1(.2(y, z))
f4(false, x, y, z) -> .2(x, del1(.2(y, z)))
=2(nil, nil) -> true
=2(.2(x, y), nil) -> false
=2(nil, .2(y, z)) -> false
=2(.2(x, y), .2(u, v)) -> and2(=2(x, u), =2(y, v))

The set Q consists of the following terms:

del1(.2(x0, .2(x1, x2)))
f4(true, x0, x1, x2)
f4(false, x0, x1, x2)
=2(nil, nil)
=2(.2(x0, x1), nil)
=2(nil, .2(x0, x1))
=2(.2(x0, x1), .2(u, v))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 2 less nodes.