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))


Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 1 SCC with 3 less nodes.

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

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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


DEL1(.2(x, .2(y, z))) -> F4(=2(x, y), x, y, z)
The remaining pairs can at least by weakly be oriented.

F4(false, x, y, z) -> DEL1(.2(y, z))
F4(true, x, y, z) -> DEL1(.2(y, z))
Used ordering: Combined order from the following AFS and order.
F4(x1, x2, x3, x4)  =  F1(x4)
false  =  false
DEL1(x1)  =  x1
.2(x1, x2)  =  .1(x2)
true  =  true
=2(x1, x2)  =  =
nil  =  nil
u  =  u
v  =  v
and2(x1, x2)  =  x2

Lexicographic Path Order [19].
Precedence:
[=, u] > v > [F1, false, .1, true]
nil > [F1, false, .1, true]


The following usable rules [14] were oriented:

=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))



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ 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 [13,14,18] contains 0 SCCs with 2 less nodes.