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:

and(true, X) → X
and(false, Y) → false
if(true, X, Y) → X
if(false, X, Y) → Y
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))

The set Q consists of the following terms:

from(x0)
first(s(x0), cons(x1, x2))
add(0, x0)
if(true, x0, x1)
and(true, x0)
and(false, x0)
first(0, x0)
add(s(x0), x1)
if(false, x0, x1)



QTRS
  ↳ DependencyPairsProof

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

and(true, X) → X
and(false, Y) → false
if(true, X, Y) → X
if(false, X, Y) → Y
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))

The set Q consists of the following terms:

from(x0)
first(s(x0), cons(x1, x2))
add(0, x0)
if(true, x0, x1)
and(true, x0)
and(false, x0)
first(0, x0)
add(s(x0), x1)
if(false, 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:

FIRST(s(X), cons(Y, Z)) → FIRST(X, Z)
ADD(s(X), Y) → ADD(X, Y)
FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

and(true, X) → X
and(false, Y) → false
if(true, X, Y) → X
if(false, X, Y) → Y
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))

The set Q consists of the following terms:

from(x0)
first(s(x0), cons(x1, x2))
add(0, x0)
if(true, x0, x1)
and(true, x0)
and(false, x0)
first(0, x0)
add(s(x0), x1)
if(false, x0, x1)

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ DependencyGraphProof

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

FIRST(s(X), cons(Y, Z)) → FIRST(X, Z)
ADD(s(X), Y) → ADD(X, Y)
FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

and(true, X) → X
and(false, Y) → false
if(true, X, Y) → X
if(false, X, Y) → Y
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))

The set Q consists of the following terms:

from(x0)
first(s(x0), cons(x1, x2))
add(0, x0)
if(true, x0, x1)
and(true, x0)
and(false, x0)
first(0, x0)
add(s(x0), x1)
if(false, x0, x1)

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ UsableRulesProof
          ↳ QDP
          ↳ QDP

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

FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

and(true, X) → X
and(false, Y) → false
if(true, X, Y) → X
if(false, X, Y) → Y
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))

The set Q consists of the following terms:

from(x0)
first(s(x0), cons(x1, x2))
add(0, x0)
if(true, x0, x1)
and(true, x0)
and(false, x0)
first(0, x0)
add(s(x0), x1)
if(false, 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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ QReductionProof
          ↳ QDP
          ↳ QDP

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

FROM(X) → FROM(s(X))

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

from(x0)
first(s(x0), cons(x1, x2))
add(0, x0)
if(true, x0, x1)
and(true, x0)
and(false, x0)
first(0, x0)
add(s(x0), x1)
if(false, 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.

from(x0)
first(s(x0), cons(x1, x2))
add(0, x0)
if(true, x0, x1)
and(true, x0)
and(false, x0)
first(0, x0)
add(s(x0), x1)
if(false, x0, x1)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QReductionProof
QDP
                    ↳ Instantiation
          ↳ QDP
          ↳ QDP

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

FROM(X) → FROM(s(X))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule FROM(X) → FROM(s(X)) we obtained the following new rules:

FROM(s(z0)) → FROM(s(s(z0)))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QReductionProof
                  ↳ QDP
                    ↳ Instantiation
QDP
                        ↳ Instantiation
          ↳ QDP
          ↳ QDP

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

FROM(s(z0)) → FROM(s(s(z0)))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule FROM(s(z0)) → FROM(s(s(z0))) we obtained the following new rules:

FROM(s(s(z0))) → FROM(s(s(s(z0))))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QReductionProof
                  ↳ QDP
                    ↳ Instantiation
                      ↳ QDP
                        ↳ Instantiation
QDP
                            ↳ NonTerminationProof
          ↳ QDP
          ↳ QDP

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

FROM(s(s(z0))) → FROM(s(s(s(z0))))

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:

FROM(s(s(z0))) → FROM(s(s(s(z0))))

The TRS R consists of the following rules:none


s = FROM(s(s(z0))) evaluates to t =FROM(s(s(s(z0))))

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 FROM(s(s(z0))) to FROM(s(s(s(z0)))).





↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
QDP
            ↳ UsableRulesProof
          ↳ QDP

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

FIRST(s(X), cons(Y, Z)) → FIRST(X, Z)

The TRS R consists of the following rules:

and(true, X) → X
and(false, Y) → false
if(true, X, Y) → X
if(false, X, Y) → Y
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))

The set Q consists of the following terms:

from(x0)
first(s(x0), cons(x1, x2))
add(0, x0)
if(true, x0, x1)
and(true, x0)
and(false, x0)
first(0, x0)
add(s(x0), x1)
if(false, 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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ QReductionProof
          ↳ QDP

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

FIRST(s(X), cons(Y, Z)) → FIRST(X, Z)

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

from(x0)
first(s(x0), cons(x1, x2))
add(0, x0)
if(true, x0, x1)
and(true, x0)
and(false, x0)
first(0, x0)
add(s(x0), x1)
if(false, 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.

from(x0)
first(s(x0), cons(x1, x2))
add(0, x0)
if(true, x0, x1)
and(true, x0)
and(false, x0)
first(0, x0)
add(s(x0), x1)
if(false, x0, x1)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QReductionProof
QDP
                    ↳ QDPSizeChangeProof
          ↳ QDP

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

FIRST(s(X), cons(Y, Z)) → FIRST(X, Z)

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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
QDP
            ↳ UsableRulesProof

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

ADD(s(X), Y) → ADD(X, Y)

The TRS R consists of the following rules:

and(true, X) → X
and(false, Y) → false
if(true, X, Y) → X
if(false, X, Y) → Y
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))

The set Q consists of the following terms:

from(x0)
first(s(x0), cons(x1, x2))
add(0, x0)
if(true, x0, x1)
and(true, x0)
and(false, x0)
first(0, x0)
add(s(x0), x1)
if(false, 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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ QReductionProof

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

ADD(s(X), Y) → ADD(X, Y)

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

from(x0)
first(s(x0), cons(x1, x2))
add(0, x0)
if(true, x0, x1)
and(true, x0)
and(false, x0)
first(0, x0)
add(s(x0), x1)
if(false, 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.

from(x0)
first(s(x0), cons(x1, x2))
add(0, x0)
if(true, x0, x1)
and(true, x0)
and(false, x0)
first(0, x0)
add(s(x0), x1)
if(false, x0, x1)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QReductionProof
QDP
                    ↳ QDPSizeChangeProof

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

ADD(s(X), Y) → ADD(X, 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: