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

(0) Obligation:

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

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
nats → adx(zeros)
zeros → cons(0, n__zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → activate(L)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
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(adx(x₁)) = 2·x₁   
POL(cons(x₁, x₂)) = x₁ + 2·x₂   
POL(head(x₁)) = 1 + x₁   
POL(incr(x₁)) = x₁   
POL(n__adx(x₁)) = 2·x₁   
POL(n__incr(x₁)) = x₁   
POL(n__zeros) = 0   
POL(nats) = 1   
POL(nil) = 0   
POL(s(x₁)) = x₁   
POL(tail(x₁)) = 2 + x₁   
POL(zeros) = 0   

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

nats → adx(zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → activate(L)

(2) Obligation:

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

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
zeros → cons(0, n__zeros)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
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(adx(x₁)) = 2·x₁   
POL(cons(x₁, x₂)) = 2·x₁ + 2·x₂   
POL(incr(x₁)) = x₁   
POL(n__adx(x₁)) = 2·x₁   
POL(n__incr(x₁)) = 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:

adx(nil) → nil

(4) Obligation:

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

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
zeros → cons(0, n__zeros)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
activate(n__zeros) → zeros
activate(X) → X

Q is empty.

(5) DependencyPairsProof (EQUIVALENT transformation)

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

(6) Obligation:

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

INCR(cons(X, L)) → ACTIVATE(L)
ADX(cons(X, L)) → INCR(cons(X, n__adx(activate(L))))
ADX(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__incr(X)) → INCR(X)
ACTIVATE(n__adx(X)) → ADX(X)
ACTIVATE(n__zeros) → ZEROS

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
zeros → cons(0, n__zeros)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
activate(n__zeros) → zeros
activate(X) → X

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

(7) DependencyGraphProof (EQUIVALENT transformation)

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

(8) Obligation:

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

ACTIVATE(n__incr(X)) → INCR(X)
INCR(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__adx(X)) → ADX(X)
ADX(cons(X, L)) → INCR(cons(X, n__adx(activate(L))))
ADX(cons(X, L)) → ACTIVATE(L)

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
zeros → cons(0, n__zeros)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
activate(n__zeros) → zeros
activate(X) → X

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

(9) 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
ADX(cons(0, n__zeros))[ ]ⁿ[ ] → ADX(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
ADX(cons(x0, n__zeros))[ ]ⁿ[ ] → ADX(cons(0, n__zeros))[ ]ⁿ[ ]
    by Narrowing at position: [0]
        ADX(cons(x0, n__zeros))[ ]ⁿ[ ] → ADX(zeros)[ ]ⁿ[ ]
            by Narrowing at position: [0]
                intermediate steps: Instantiation
                ADX(cons(x0, x1))[ ]ⁿ[ ] → ADX(activate(x1))[ ]ⁿ[ ]
                    by Narrowing at position: []
                        intermediate steps: Instantiation
                        ADX(cons(X, L))[ ]ⁿ[ ] → INCR(cons(X, n__adx(activate(L))))[ ]ⁿ[ ]
                            by OriginalRule from TRS P

                        intermediate steps: Instantiation - Instantiation - Instantiation
                        INCR(cons(x0, n__adx(y0)))[ ]ⁿ[ ] → ADX(y0)[ ]ⁿ[ ]
                            by Narrowing at position: []
                                intermediate steps: Instantiation - Instantiation
                                INCR(cons(X, L))[ ]ⁿ[ ] → ACTIVATE(L)[ ]ⁿ[ ]
                                    by OriginalRule from TRS P

                                intermediate steps: Instantiation
                                ACTIVATE(n__adx(X))[ ]ⁿ[ ] → ADX(X)[ ]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

(10) NO