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
  ↳ 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))

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:

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))

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:

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))

Q is empty.
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
  ↳ 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))

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.


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

DEL1(.2(x, .2(y, z))) -> F4(=2(x, y), x, y, z)
Used ordering: Polynomial interpretation [21]:

POL(.2(x1, x2)) = 2·x1 + 2·x2   
POL(=2(x1, x2)) = 1 + x1 + x2   
POL(DEL1(x1)) = 2 + x1   
POL(F4(x1, x2, x3, x4)) = 2·x1 + 2·x3 + 2·x4   
POL(and2(x1, x2)) = x2   
POL(false) = 2   
POL(nil) = 1   
POL(true) = 2   
POL(u) = 0   
POL(v) = 0   

The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof

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

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))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.