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(0, zeros)
incr(cons(X, Y)) → cons(s(X), incr(Y))
adx(cons(X, Y)) → incr(cons(X, adx(Y)))
hd(cons(X, Y)) → X
tl(cons(X, Y)) → Y

Q is empty.


↳ QTRS
  ↳ RRRPoloQTRSProof

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

nats → adx(zeros)
zeros → cons(0, zeros)
incr(cons(X, Y)) → cons(s(X), incr(Y))
adx(cons(X, Y)) → incr(cons(X, adx(Y)))
hd(cons(X, Y)) → X
tl(cons(X, Y)) → Y

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(0, zeros)
incr(cons(X, Y)) → cons(s(X), incr(Y))
adx(cons(X, Y)) → incr(cons(X, adx(Y)))
hd(cons(X, Y)) → X
tl(cons(X, Y)) → Y

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

POL(0) = 0   
POL(adx(x1)) = x1   
POL(cons(x1, x2)) = 2·x1 + 2·x2   
POL(hd(x1)) = 2 + 2·x1   
POL(incr(x1)) = x1   
POL(nats) = 2   
POL(s(x1)) = x1   
POL(tl(x1)) = 1 + x1   
POL(zeros) = 0   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ Overlay + Local Confluence

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

zeros → cons(0, zeros)
incr(cons(X, Y)) → cons(s(X), incr(Y))
adx(cons(X, Y)) → incr(cons(X, adx(Y)))

Q is empty.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ QTRS
          ↳ DependencyPairsProof

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

zeros → cons(0, zeros)
incr(cons(X, Y)) → cons(s(X), incr(Y))
adx(cons(X, Y)) → incr(cons(X, adx(Y)))

The set Q consists of the following terms:

zeros
incr(cons(x0, x1))
adx(cons(x0, x1))


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)) → ADX(Y)
ZEROS → ZEROS
ADX(cons(X, Y)) → INCR(cons(X, adx(Y)))
INCR(cons(X, Y)) → INCR(Y)

The TRS R consists of the following rules:

zeros → cons(0, zeros)
incr(cons(X, Y)) → cons(s(X), incr(Y))
adx(cons(X, Y)) → incr(cons(X, adx(Y)))

The set Q consists of the following terms:

zeros
incr(cons(x0, x1))
adx(cons(x0, x1))

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

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof

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

ADX(cons(X, Y)) → ADX(Y)
ZEROS → ZEROS
ADX(cons(X, Y)) → INCR(cons(X, adx(Y)))
INCR(cons(X, Y)) → INCR(Y)

The TRS R consists of the following rules:

zeros → cons(0, zeros)
incr(cons(X, Y)) → cons(s(X), incr(Y))
adx(cons(X, Y)) → incr(cons(X, adx(Y)))

The set Q consists of the following terms:

zeros
incr(cons(x0, x1))
adx(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 1 less node.

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ UsableRulesProof
                  ↳ QDP
                  ↳ QDP

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

INCR(cons(X, Y)) → INCR(Y)

The TRS R consists of the following rules:

zeros → cons(0, zeros)
incr(cons(X, Y)) → cons(s(X), incr(Y))
adx(cons(X, Y)) → incr(cons(X, adx(Y)))

The set Q consists of the following terms:

zeros
incr(cons(x0, x1))
adx(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ QReductionProof
                  ↳ QDP
                  ↳ QDP

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

INCR(cons(X, Y)) → INCR(Y)

R is empty.
The set Q consists of the following terms:

zeros
incr(cons(x0, x1))
adx(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

zeros
incr(cons(x0, x1))
adx(cons(x0, x1))



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ QReductionProof
                          ↳ QDP
                            ↳ QDPSizeChangeProof
                  ↳ QDP
                  ↳ QDP

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

INCR(cons(X, Y)) → INCR(Y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: 



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesProof
                  ↳ QDP

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

ADX(cons(X, Y)) → ADX(Y)

The TRS R consists of the following rules:

zeros → cons(0, zeros)
incr(cons(X, Y)) → cons(s(X), incr(Y))
adx(cons(X, Y)) → incr(cons(X, adx(Y)))

The set Q consists of the following terms:

zeros
incr(cons(x0, x1))
adx(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ QReductionProof
                  ↳ QDP

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

ADX(cons(X, Y)) → ADX(Y)

R is empty.
The set Q consists of the following terms:

zeros
incr(cons(x0, x1))
adx(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

zeros
incr(cons(x0, x1))
adx(cons(x0, x1))



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ QReductionProof
                          ↳ QDP
                            ↳ QDPSizeChangeProof
                  ↳ QDP

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

ADX(cons(X, Y)) → ADX(Y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: 



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesProof

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

ZEROS → ZEROS

The TRS R consists of the following rules:

zeros → cons(0, zeros)
incr(cons(X, Y)) → cons(s(X), incr(Y))
adx(cons(X, Y)) → incr(cons(X, adx(Y)))

The set Q consists of the following terms:

zeros
incr(cons(x0, x1))
adx(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ QReductionProof

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

ZEROS → ZEROS

R is empty.
The set Q consists of the following terms:

zeros
incr(cons(x0, x1))
adx(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

zeros
incr(cons(x0, x1))
adx(cons(x0, x1))



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ QReductionProof
                          ↳ QDP
                            ↳ NonTerminationProof

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

ZEROS → ZEROS

R is empty.
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 semiunifying a rule from P directly.

The TRS P consists of the following rules:

ZEROS → ZEROS

The TRS R consists of the following rules:none


s = ZEROS evaluates to t =ZEROS

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



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from ZEROS to ZEROS.