Termination w.r.t. Q proof of /tmp/tmpqk7pRL/LengthOfFiniteLists_nosorts-noand

Termination w.r.t. Q of the following Term Rewriting System could be disproven:

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

zeros → cons(0, n__zeros)
U11(tt, L) → U12(tt, activate(L))
U12(tt, L) → s(length(activate(L)))
length(nil) → 0
length(cons(N, L)) → U11(tt, activate(L))
zeros → n__zeros
activate(n__zeros) → zeros
activate(X) → X

Q is empty.

↳ QTRS

  ↳ RRRPoloQTRSProof

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

zeros → cons(0, n__zeros)
U11(tt, L) → U12(tt, activate(L))
U12(tt, L) → s(length(activate(L)))
length(nil) → 0
length(cons(N, L)) → U11(tt, activate(L))
zeros → n__zeros
activate(n__zeros) → zeros
activate(X) → X

Q is empty.

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

zeros → cons(0, n__zeros)
U11(tt, L) → U12(tt, activate(L))
U12(tt, L) → s(length(activate(L)))
length(nil) → 0
length(cons(N, L)) → U11(tt, activate(L))
zeros → n__zeros
activate(n__zeros) → zeros
activate(X) → X

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

length(nil) → 0

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(U11(x₁, x₂)) = 2·x₁ + 2·x₂
POL(U12(x₁, x₂)) = 2·x₁ + 2·x₂
POL(activate(x₁)) = x₁
POL(cons(x₁, x₂)) = 2·x₁ + 2·x₂
POL(length(x₁)) = 2·x₁
POL(n__zeros) = 0
POL(nil) = 1
POL(s(x₁)) = x₁
POL(tt) = 0
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

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

zeros → cons(0, n__zeros)
U11(tt, L) → U12(tt, activate(L))
U12(tt, L) → s(length(activate(L)))
length(cons(N, L)) → U11(tt, activate(L))
zeros → n__zeros
activate(n__zeros) → zeros
activate(X) → X

Q is empty.

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

U12¹(tt, L) → ACTIVATE(L)
LENGTH(cons(N, L)) → ACTIVATE(L)
U12¹(tt, L) → LENGTH(activate(L))
U11¹(tt, L) → U12¹(tt, activate(L))
U11¹(tt, L) → ACTIVATE(L)
LENGTH(cons(N, L)) → U11¹(tt, activate(L))
ACTIVATE(n__zeros) → ZEROS

The TRS R consists of the following rules:

zeros → cons(0, n__zeros)
U11(tt, L) → U12(tt, activate(L))
U12(tt, L) → s(length(activate(L)))
length(cons(N, L)) → U11(tt, activate(L))
zeros → n__zeros
activate(n__zeros) → zeros
activate(X) → X

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

U12¹(tt, L) → ACTIVATE(L)
LENGTH(cons(N, L)) → ACTIVATE(L)
U12¹(tt, L) → LENGTH(activate(L))
U11¹(tt, L) → U12¹(tt, activate(L))
U11¹(tt, L) → ACTIVATE(L)
LENGTH(cons(N, L)) → U11¹(tt, activate(L))
ACTIVATE(n__zeros) → ZEROS

The TRS R consists of the following rules:

zeros → cons(0, n__zeros)
U11(tt, L) → U12(tt, activate(L))
U12(tt, L) → s(length(activate(L)))
length(cons(N, L)) → U11(tt, activate(L))
zeros → n__zeros
activate(n__zeros) → zeros
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 4 less nodes.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ UsableRulesProof

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

U12¹(tt, L) → LENGTH(activate(L))
U11¹(tt, L) → U12¹(tt, activate(L))
LENGTH(cons(N, L)) → U11¹(tt, activate(L))

The TRS R consists of the following rules:

zeros → cons(0, n__zeros)
U11(tt, L) → U12(tt, activate(L))
U12(tt, L) → s(length(activate(L)))
length(cons(N, L)) → U11(tt, activate(L))
zeros → n__zeros
activate(n__zeros) → zeros
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ UsableRulesProof

                ↳ QDP

                  ↳ Narrowing

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

U12¹(tt, L) → LENGTH(activate(L))
U11¹(tt, L) → U12¹(tt, activate(L))
LENGTH(cons(N, L)) → U11¹(tt, activate(L))

The TRS R consists of the following rules:

activate(n__zeros) → zeros
activate(X) → X
zeros → cons(0, n__zeros)
zeros → n__zeros

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule U12¹(tt, L) → LENGTH(activate(L)) at position [0] we obtained the following new rules:

U12¹(tt, n__zeros) → LENGTH(zeros)
U12¹(tt, x0) → LENGTH(x0)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ UsableRulesProof

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

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

U11¹(tt, L) → U12¹(tt, activate(L))
U12¹(tt, n__zeros) → LENGTH(zeros)
LENGTH(cons(N, L)) → U11¹(tt, activate(L))
U12¹(tt, x0) → LENGTH(x0)

The TRS R consists of the following rules:

activate(n__zeros) → zeros
activate(X) → X
zeros → cons(0, n__zeros)
zeros → n__zeros

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule U12¹(tt, n__zeros) → LENGTH(zeros) at position [0] we obtained the following new rules:

U12¹(tt, n__zeros) → LENGTH(cons(0, n__zeros))
U12¹(tt, n__zeros) → LENGTH(n__zeros)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ UsableRulesProof

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ DependencyGraphProof

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

U12¹(tt, n__zeros) → LENGTH(cons(0, n__zeros))
U12¹(tt, n__zeros) → LENGTH(n__zeros)
U11¹(tt, L) → U12¹(tt, activate(L))
LENGTH(cons(N, L)) → U11¹(tt, activate(L))
U12¹(tt, x0) → LENGTH(x0)

The TRS R consists of the following rules:

activate(n__zeros) → zeros
activate(X) → X
zeros → cons(0, n__zeros)
zeros → n__zeros

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ UsableRulesProof

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ NonTerminationProof

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

U12¹(tt, n__zeros) → LENGTH(cons(0, n__zeros))
U11¹(tt, L) → U12¹(tt, activate(L))
LENGTH(cons(N, L)) → U11¹(tt, activate(L))
U12¹(tt, x0) → LENGTH(x0)

The TRS R consists of the following rules:

activate(n__zeros) → zeros
activate(X) → X
zeros → cons(0, n__zeros)
zeros → n__zeros

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

The TRS P consists of the following rules:

U12¹(tt, n__zeros) → LENGTH(cons(0, n__zeros))
U11¹(tt, L) → U12¹(tt, activate(L))
LENGTH(cons(N, L)) → U11¹(tt, activate(L))
U12¹(tt, x0) → LENGTH(x0)

The TRS R consists of the following rules:

activate(n__zeros) → zeros
activate(X) → X
zeros → cons(0, n__zeros)
zeros → n__zeros

s = LENGTH(cons(N, n__zeros)) evaluates to t =LENGTH(cons(0, n__zeros))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

Semiunifier: [ ]
Matcher: [N / 0]

Rewriting sequence

LENGTH(cons(N, n__zeros)) → U11¹(tt, activate(n__zeros))
with rule LENGTH(cons(N', L)) → U11¹(tt, activate(L)) at position [] and matcher [L / n__zeros, N' / N]

U11¹(tt, activate(n__zeros)) → U11¹(tt, n__zeros)
with rule activate(X) → X at position [1] and matcher [X / n__zeros]

U11¹(tt, n__zeros) → U12¹(tt, activate(n__zeros))
with rule U11¹(tt, L) → U12¹(tt, activate(L)) at position [] and matcher [L / n__zeros]

U12¹(tt, activate(n__zeros)) → U12¹(tt, n__zeros)
with rule activate(X) → X at position [1] and matcher [X / n__zeros]

U12¹(tt, n__zeros) → LENGTH(cons(0, n__zeros))
with rule U12¹(tt, n__zeros) → LENGTH(cons(0, n__zeros))

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence

All these steps are and every following step will be a correct step w.r.t to Q.