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:

nats → adx(zeros)
zeros → cons(n__0, n__zeros)
incr(cons(X, Y)) → cons(n__s(activate(X)), n__incr(activate(Y)))
adx(cons(X, Y)) → incr(cons(activate(X), n__adx(activate(Y))))
hd(cons(X, Y)) → activate(X)
tl(cons(X, Y)) → activate(Y)
0 → n__0
zeros → n__zeros
s(X) → n__s(X)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
activate(n__0) → 0
activate(n__zeros) → zeros
activate(n__s(X)) → s(X)
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
activate(X) → X

Q is empty.


↳ QTRS
  ↳ RRRPoloQTRSProof

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

nats → adx(zeros)
zeros → cons(n__0, n__zeros)
incr(cons(X, Y)) → cons(n__s(activate(X)), n__incr(activate(Y)))
adx(cons(X, Y)) → incr(cons(activate(X), n__adx(activate(Y))))
hd(cons(X, Y)) → activate(X)
tl(cons(X, Y)) → activate(Y)
0 → n__0
zeros → n__zeros
s(X) → n__s(X)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
activate(n__0) → 0
activate(n__zeros) → zeros
activate(n__s(X)) → s(X)
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
activate(X) → X

Q is empty.

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

nats → adx(zeros)
zeros → cons(n__0, n__zeros)
incr(cons(X, Y)) → cons(n__s(activate(X)), n__incr(activate(Y)))
adx(cons(X, Y)) → incr(cons(activate(X), n__adx(activate(Y))))
hd(cons(X, Y)) → activate(X)
tl(cons(X, Y)) → activate(Y)
0 → n__0
zeros → n__zeros
s(X) → n__s(X)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
activate(n__0) → 0
activate(n__zeros) → zeros
activate(n__s(X)) → s(X)
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
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:

tl(cons(X, Y)) → activate(Y)
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(activate(x1)) = x1   
POL(adx(x1)) = 2·x1   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(hd(x1)) = 2·x1   
POL(incr(x1)) = x1   
POL(n__0) = 0   
POL(n__adx(x1)) = 2·x1   
POL(n__incr(x1)) = x1   
POL(n__s(x1)) = x1   
POL(n__zeros) = 0   
POL(nats) = 0   
POL(s(x1)) = x1   
POL(tl(x1)) = 1 + 2·x1   
POL(zeros) = 0   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof

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

nats → adx(zeros)
zeros → cons(n__0, n__zeros)
incr(cons(X, Y)) → cons(n__s(activate(X)), n__incr(activate(Y)))
adx(cons(X, Y)) → incr(cons(activate(X), n__adx(activate(Y))))
hd(cons(X, Y)) → activate(X)
0 → n__0
zeros → n__zeros
s(X) → n__s(X)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
activate(n__0) → 0
activate(n__zeros) → zeros
activate(n__s(X)) → s(X)
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
activate(X) → X

Q is empty.

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

nats → adx(zeros)
zeros → cons(n__0, n__zeros)
incr(cons(X, Y)) → cons(n__s(activate(X)), n__incr(activate(Y)))
adx(cons(X, Y)) → incr(cons(activate(X), n__adx(activate(Y))))
hd(cons(X, Y)) → activate(X)
0 → n__0
zeros → n__zeros
s(X) → n__s(X)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
activate(n__0) → 0
activate(n__zeros) → zeros
activate(n__s(X)) → s(X)
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
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:

hd(cons(X, Y)) → activate(X)
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(activate(x1)) = x1   
POL(adx(x1)) = 2·x1   
POL(cons(x1, x2)) = 2·x1 + 2·x2   
POL(hd(x1)) = 2 + 2·x1   
POL(incr(x1)) = x1   
POL(n__0) = 0   
POL(n__adx(x1)) = 2·x1   
POL(n__incr(x1)) = x1   
POL(n__s(x1)) = x1   
POL(n__zeros) = 0   
POL(nats) = 0   
POL(s(x1)) = x1   
POL(zeros) = 0   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof

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

nats → adx(zeros)
zeros → cons(n__0, n__zeros)
incr(cons(X, Y)) → cons(n__s(activate(X)), n__incr(activate(Y)))
adx(cons(X, Y)) → incr(cons(activate(X), n__adx(activate(Y))))
0 → n__0
zeros → n__zeros
s(X) → n__s(X)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
activate(n__0) → 0
activate(n__zeros) → zeros
activate(n__s(X)) → s(X)
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
activate(X) → X

Q is empty.

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

nats → adx(zeros)
zeros → cons(n__0, n__zeros)
incr(cons(X, Y)) → cons(n__s(activate(X)), n__incr(activate(Y)))
adx(cons(X, Y)) → incr(cons(activate(X), n__adx(activate(Y))))
0 → n__0
zeros → n__zeros
s(X) → n__s(X)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
activate(n__0) → 0
activate(n__zeros) → zeros
activate(n__s(X)) → s(X)
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
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:

nats → adx(zeros)
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(activate(x1)) = x1   
POL(adx(x1)) = x1   
POL(cons(x1, x2)) = x1 + 2·x2   
POL(incr(x1)) = x1   
POL(n__0) = 0   
POL(n__adx(x1)) = x1   
POL(n__incr(x1)) = x1   
POL(n__s(x1)) = x1   
POL(n__zeros) = 0   
POL(nats) = 2   
POL(s(x1)) = x1   
POL(zeros) = 0   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof

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

zeros → cons(n__0, n__zeros)
incr(cons(X, Y)) → cons(n__s(activate(X)), n__incr(activate(Y)))
adx(cons(X, Y)) → incr(cons(activate(X), n__adx(activate(Y))))
0 → n__0
zeros → n__zeros
s(X) → n__s(X)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
activate(n__0) → 0
activate(n__zeros) → zeros
activate(n__s(X)) → s(X)
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
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:

ADX(cons(X, Y)) → ACTIVATE(X)
INCR(cons(X, Y)) → ACTIVATE(Y)
ACTIVATE(n__incr(X)) → INCR(X)
ADX(cons(X, Y)) → ACTIVATE(Y)
ACTIVATE(n__0) → 01
ADX(cons(X, Y)) → INCR(cons(activate(X), n__adx(activate(Y))))
ACTIVATE(n__s(X)) → S(X)
ACTIVATE(n__adx(X)) → ADX(X)
INCR(cons(X, Y)) → ACTIVATE(X)
ACTIVATE(n__zeros) → ZEROS

The TRS R consists of the following rules:

zeros → cons(n__0, n__zeros)
incr(cons(X, Y)) → cons(n__s(activate(X)), n__incr(activate(Y)))
adx(cons(X, Y)) → incr(cons(activate(X), n__adx(activate(Y))))
0 → n__0
zeros → n__zeros
s(X) → n__s(X)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
activate(n__0) → 0
activate(n__zeros) → zeros
activate(n__s(X)) → s(X)
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
activate(X) → X

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

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

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

ADX(cons(X, Y)) → ACTIVATE(X)
INCR(cons(X, Y)) → ACTIVATE(Y)
ACTIVATE(n__incr(X)) → INCR(X)
ADX(cons(X, Y)) → ACTIVATE(Y)
ACTIVATE(n__0) → 01
ADX(cons(X, Y)) → INCR(cons(activate(X), n__adx(activate(Y))))
ACTIVATE(n__s(X)) → S(X)
ACTIVATE(n__adx(X)) → ADX(X)
INCR(cons(X, Y)) → ACTIVATE(X)
ACTIVATE(n__zeros) → ZEROS

The TRS R consists of the following rules:

zeros → cons(n__0, n__zeros)
incr(cons(X, Y)) → cons(n__s(activate(X)), n__incr(activate(Y)))
adx(cons(X, Y)) → incr(cons(activate(X), n__adx(activate(Y))))
0 → n__0
zeros → n__zeros
s(X) → n__s(X)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
activate(n__0) → 0
activate(n__zeros) → zeros
activate(n__s(X)) → s(X)
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
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 3 less nodes.

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

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

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

The TRS R consists of the following rules:

zeros → cons(n__0, n__zeros)
incr(cons(X, Y)) → cons(n__s(activate(X)), n__incr(activate(Y)))
adx(cons(X, Y)) → incr(cons(activate(X), n__adx(activate(Y))))
0 → n__0
zeros → n__zeros
s(X) → n__s(X)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
activate(n__0) → 0
activate(n__zeros) → zeros
activate(n__s(X)) → s(X)
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

ADX(cons(X, Y)) → ACTIVATE(X)
ADX(cons(X, Y)) → ACTIVATE(Y)


Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(ACTIVATE(x1)) = 2·x1   
POL(ADX(x1)) = 2 + 2·x1   
POL(INCR(x1)) = 2·x1   
POL(activate(x1)) = x1   
POL(adx(x1)) = 1 + x1   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(incr(x1)) = x1   
POL(n__0) = 0   
POL(n__adx(x1)) = 1 + x1   
POL(n__incr(x1)) = x1   
POL(n__s(x1)) = x1   
POL(n__zeros) = 0   
POL(s(x1)) = x1   
POL(zeros) = 0   



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ RuleRemovalProof
                        ↳ QDP
                          ↳ NonTerminationProof

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

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

The TRS R consists of the following rules:

zeros → cons(n__0, n__zeros)
incr(cons(X, Y)) → cons(n__s(activate(X)), n__incr(activate(Y)))
adx(cons(X, Y)) → incr(cons(activate(X), n__adx(activate(Y))))
0 → n__0
zeros → n__zeros
s(X) → n__s(X)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
activate(n__0) → 0
activate(n__zeros) → zeros
activate(n__s(X)) → s(X)
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
activate(X) → X

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:

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

The TRS R consists of the following rules:

zeros → cons(n__0, n__zeros)
incr(cons(X, Y)) → cons(n__s(activate(X)), n__incr(activate(Y)))
adx(cons(X, Y)) → incr(cons(activate(X), n__adx(activate(Y))))
0 → n__0
zeros → n__zeros
s(X) → n__s(X)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
activate(n__0) → 0
activate(n__zeros) → zeros
activate(n__s(X)) → s(X)
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
activate(X) → X


s = ACTIVATE(n__adx(activate(n__zeros))) evaluates to t =ACTIVATE(n__adx(activate(n__zeros)))

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



Rewriting sequence

ACTIVATE(n__adx(activate(n__zeros))) → ACTIVATE(n__adx(zeros))
with rule activate(n__zeros) → zeros at position [0,0] and matcher [ ]

ACTIVATE(n__adx(zeros)) → ACTIVATE(n__adx(cons(n__0, n__zeros)))
with rule zeros → cons(n__0, n__zeros) at position [0,0] and matcher [ ]

ACTIVATE(n__adx(cons(n__0, n__zeros))) → ADX(cons(n__0, n__zeros))
with rule ACTIVATE(n__adx(X)) → ADX(X) at position [] and matcher [X / cons(n__0, n__zeros)]

ADX(cons(n__0, n__zeros)) → INCR(cons(activate(n__0), n__adx(activate(n__zeros))))
with rule ADX(cons(X', Y')) → INCR(cons(activate(X'), n__adx(activate(Y')))) at position [] and matcher [Y' / n__zeros, X' / n__0]

INCR(cons(activate(n__0), n__adx(activate(n__zeros)))) → ACTIVATE(n__adx(activate(n__zeros)))
with rule INCR(cons(X, Y)) → ACTIVATE(Y)

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.