Termination w.r.t. Q of the given QTRS could be disproven:

0 QTRS
↳1 QTRSRRRProof (⇔, 0 ms)
↳2 QTRS
↳3 QTRSRRRProof (⇔, 0 ms)
↳4 QTRS
↳5 QTRSRRRProof (⇔, 0 ms)
↳6 QTRS
↳7 QTRSRRRProof (⇔, 12 ms)
↳8 QTRS
↳9 DependencyPairsProof (⇔, 0 ms)
↳10 QDP
↳11 DependencyGraphProof (⇔, 0 ms)
↳12 QDP
↳13 NonLoopProof (⇔, 28 ms)
↳14 NO

(0) Obligation:

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

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

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(activate(x1)) = x1   
POL(and(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(cons(x1, x2)) = 2·x1 + 2·x2   
POL(length(x1)) = 2·x1   
POL(n__zeros) = 0   
POL(nil) = 0   
POL(s(x1)) = 2·x1   
POL(tt) = 0   
POL(zeros) = 0   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

and(tt, X) → activate(X)


(2) Obligation:

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

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

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(activate(x1)) = x1   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(length(x1)) = 2·x1   
POL(n__zeros) = 0   
POL(nil) = 2   
POL(s(x1)) = x1   
POL(zeros) = 0   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

length(nil) → 0


(4) Obligation:

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

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

Q is empty.

(5) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(activate(x1)) = 2·x1   
POL(cons(x1, x2)) = x1 + 2·x2   
POL(length(x1)) = 2·x1   
POL(n__zeros) = 1   
POL(s(x1)) = x1   
POL(zeros) = 2   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

zeros → n__zeros


(6) Obligation:

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

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

Q is empty.

(7) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(activate(x1)) = 1 + x1   
POL(cons(x1, x2)) = 1 + 2·x1 + x2   
POL(length(x1)) = 2·x1   
POL(n__zeros) = 0   
POL(s(x1)) = x1   
POL(zeros) = 1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

activate(X) → X


(8) Obligation:

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

zeros → cons(0, n__zeros)
length(cons(N, L)) → s(length(activate(L)))
activate(n__zeros) → zeros

Q is empty.

(9) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(10) Obligation:

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

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

The TRS R consists of the following rules:

zeros → cons(0, n__zeros)
length(cons(N, L)) → s(length(activate(L)))
activate(n__zeros) → zeros

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

(11) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(12) Obligation:

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

LENGTH(cons(N, L)) → LENGTH(activate(L))

The TRS R consists of the following rules:

zeros → cons(0, n__zeros)
length(cons(N, L)) → s(length(activate(L)))
activate(n__zeros) → zeros

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

(13) NonLoopProof (EQUIVALENT transformation)

By Theorem 8 [NONLOOP] we deduce infiniteness of the QDP.
We apply the theorem with m = 1, b = 0,
σ' = [ ], and μ' = [x0 / 0] on the rule
LENGTH(cons(0, n__zeros))[ ]n[ ] → LENGTH(cons(0, n__zeros))[ ]n[x0 / 0]
This rule is correct for the QDP as the following derivation shows:

intermediate steps: Equivalent (Simplify mu) - Instantiate mu
LENGTH(cons(x0, n__zeros))[ ]n[ ] → LENGTH(cons(0, n__zeros))[ ]n[ ]
    by Narrowing at position: [0]
        LENGTH(cons(x0, n__zeros))[ ]n[ ] → LENGTH(zeros)[ ]n[ ]
            by Narrowing at position: [0]
                intermediate steps: Instantiation - Instantiation
                LENGTH(cons(N, L))[ ]n[ ] → LENGTH(activate(L))[ ]n[ ]
                    by OriginalRule from TRS P

                activate(n__zeros)[ ]n[ ] → zeros[ ]n[ ]
                    by OriginalRule from TRS R

        zeros[ ]n[ ] → cons(0, n__zeros)[ ]n[ ]
            by OriginalRule from TRS R

(14) NO