NO Termination w.r.t. Q proof of ../tpdb/TRS/TRCSR/LengthOfFiniteLists_nosorts-noand_Z.trs
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)) = x1 + 2·x2   
POL(U12(x1, x2)) = 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) = 2   
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:

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

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

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

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

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

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



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ Narrowing
                    ↳ QDP
                      ↳ NonTerminationProof

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

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


s = LENGTH(activate(activate(n__zeros))) evaluates to t =LENGTH(activate(activate(n__zeros)))

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



Rewriting sequence

LENGTH(activate(activate(n__zeros))) → LENGTH(activate(n__zeros))
with rule activate(X) → X at position [0] and matcher [X / activate(n__zeros)]

LENGTH(activate(n__zeros)) → LENGTH(zeros)
with rule activate(n__zeros) → zeros at position [0] and matcher [ ]

LENGTH(zeros) → LENGTH(cons(0, n__zeros))
with rule zeros → cons(0, n__zeros) at position [0] and matcher [ ]

LENGTH(cons(0, 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 / 0]

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

U121(tt, activate(activate(n__zeros))) → LENGTH(activate(activate(n__zeros)))
with rule U121(tt, x0) → LENGTH(x0)

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.