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:

del(.(x, .(y, z))) → f(=(x, y), x, y, z)
f(true, x, y, z) → del(.(y, z))
f(false, x, y, z) → .(x, del(.(y, z)))
=(nil, nil) → true
=(.(x, y), nil) → false
=(nil, .(y, z)) → false
=(.(x, y), .(u, v)) → and(=(x, u), =(y, v))

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

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

del(.(x, .(y, z))) → f(=(x, y), x, y, z)
f(true, x, y, z) → del(.(y, z))
f(false, x, y, z) → .(x, del(.(y, z)))
=(nil, nil) → true
=(.(x, y), nil) → false
=(nil, .(y, z)) → false
=(.(x, y), .(u, v)) → and(=(x, u), =(y, v))

Q is empty.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof

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

del(.(x, .(y, z))) → f(=(x, y), x, y, z)
f(true, x, y, z) → del(.(y, z))
f(false, x, y, z) → .(x, del(.(y, z)))
=(nil, nil) → true
=(.(x, y), nil) → false
=(nil, .(y, z)) → false
=(.(x, y), .(u, v)) → and(=(x, u), =(y, v))

The set Q consists of the following terms:

del(.(x0, .(x1, x2)))
f(true, x0, x1, x2)
f(false, x0, x1, x2)
=(nil, nil)
=(.(x0, x1), nil)
=(nil, .(x0, x1))
=(.(x0, x1), .(u, v))


Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

F(false, x, y, z) → DEL(.(y, z))
=1(.(x, y), .(u, v)) → =1(x, u)
DEL(.(x, .(y, z))) → =1(x, y)
F(true, x, y, z) → DEL(.(y, z))
=1(.(x, y), .(u, v)) → =1(y, v)
DEL(.(x, .(y, z))) → F(=(x, y), x, y, z)

The TRS R consists of the following rules:

del(.(x, .(y, z))) → f(=(x, y), x, y, z)
f(true, x, y, z) → del(.(y, z))
f(false, x, y, z) → .(x, del(.(y, z)))
=(nil, nil) → true
=(.(x, y), nil) → false
=(nil, .(y, z)) → false
=(.(x, y), .(u, v)) → and(=(x, u), =(y, v))

The set Q consists of the following terms:

del(.(x0, .(x1, x2)))
f(true, x0, x1, x2)
f(false, x0, x1, x2)
=(nil, nil)
=(.(x0, x1), nil)
=(nil, .(x0, x1))
=(.(x0, x1), .(u, v))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

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

F(false, x, y, z) → DEL(.(y, z))
=1(.(x, y), .(u, v)) → =1(x, u)
DEL(.(x, .(y, z))) → =1(x, y)
F(true, x, y, z) → DEL(.(y, z))
=1(.(x, y), .(u, v)) → =1(y, v)
DEL(.(x, .(y, z))) → F(=(x, y), x, y, z)

The TRS R consists of the following rules:

del(.(x, .(y, z))) → f(=(x, y), x, y, z)
f(true, x, y, z) → del(.(y, z))
f(false, x, y, z) → .(x, del(.(y, z)))
=(nil, nil) → true
=(.(x, y), nil) → false
=(nil, .(y, z)) → false
=(.(x, y), .(u, v)) → and(=(x, u), =(y, v))

The set Q consists of the following terms:

del(.(x0, .(x1, x2)))
f(true, x0, x1, x2)
f(false, x0, x1, x2)
=(nil, nil)
=(.(x0, x1), nil)
=(nil, .(x0, x1))
=(.(x0, x1), .(u, v))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 3 less nodes.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof

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

F(false, x, y, z) → DEL(.(y, z))
F(true, x, y, z) → DEL(.(y, z))
DEL(.(x, .(y, z))) → F(=(x, y), x, y, z)

The TRS R consists of the following rules:

del(.(x, .(y, z))) → f(=(x, y), x, y, z)
f(true, x, y, z) → del(.(y, z))
f(false, x, y, z) → .(x, del(.(y, z)))
=(nil, nil) → true
=(.(x, y), nil) → false
=(nil, .(y, z)) → false
=(.(x, y), .(u, v)) → and(=(x, u), =(y, v))

The set Q consists of the following terms:

del(.(x0, .(x1, x2)))
f(true, x0, x1, x2)
f(false, x0, x1, x2)
=(nil, nil)
=(.(x0, x1), nil)
=(nil, .(x0, x1))
=(.(x0, x1), .(u, v))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof

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

F(false, x, y, z) → DEL(.(y, z))
F(true, x, y, z) → DEL(.(y, z))
DEL(.(x, .(y, z))) → F(=(x, y), x, y, z)

The TRS R consists of the following rules:

=(nil, nil) → true
=(.(x, y), nil) → false
=(nil, .(y, z)) → false
=(.(x, y), .(u, v)) → and(=(x, u), =(y, v))

The set Q consists of the following terms:

del(.(x0, .(x1, x2)))
f(true, x0, x1, x2)
f(false, x0, x1, x2)
=(nil, nil)
=(.(x0, x1), nil)
=(nil, .(x0, x1))
=(.(x0, x1), .(u, v))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

del(.(x0, .(x1, x2)))
f(true, x0, x1, x2)
f(false, x0, x1, x2)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof
                    ↳ QDP
                      ↳ UsableRulesReductionPairsProof

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

F(false, x, y, z) → DEL(.(y, z))
F(true, x, y, z) → DEL(.(y, z))
DEL(.(x, .(y, z))) → F(=(x, y), x, y, z)

The TRS R consists of the following rules:

=(nil, nil) → true
=(.(x, y), nil) → false
=(nil, .(y, z)) → false
=(.(x, y), .(u, v)) → and(=(x, u), =(y, v))

The set Q consists of the following terms:

=(nil, nil)
=(.(x0, x1), nil)
=(nil, .(x0, x1))
=(.(x0, x1), .(u, v))

We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

The following rules are removed from R:

=(nil, nil) → true
=(.(x, y), nil) → false
=(nil, .(y, z)) → false
=(.(x, y), .(u, v)) → and(=(x, u), =(y, v))
Used ordering: POLO with Polynomial interpretation [25]:

POL(.(x1, x2)) = 2·x1 + 2·x2   
POL(=(x1, x2)) = x1 + x2   
POL(DEL(x1)) = x1   
POL(F(x1, x2, x3, x4)) = x1 + x2 + 2·x3 + 2·x4   
POL(and(x1, x2)) = 1 + x1 + x2   
POL(false) = 0   
POL(nil) = 0   
POL(true) = 0   
POL(u) = 2   
POL(v) = 0   



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof
                    ↳ QDP
                      ↳ UsableRulesReductionPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof

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

F(false, x, y, z) → DEL(.(y, z))
F(true, x, y, z) → DEL(.(y, z))
DEL(.(x, .(y, z))) → F(=(x, y), x, y, z)

R is empty.
The set Q consists of the following terms:

=(nil, nil)
=(.(x0, x1), nil)
=(nil, .(x0, x1))
=(.(x0, x1), .(u, v))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 3 less nodes.