Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

empty(nil) → true
empty(cons(x, l)) → false
head(cons(x, l)) → x
tail(nil) → nil
tail(cons(x, l)) → l
rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
last(x, l) → if(empty(l), x, l)
if(true, x, l) → x
if(false, x, l) → last(head(l), tail(l))
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

Q is empty.


QTRS
  ↳ Overlay + Local Confluence

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

empty(nil) → true
empty(cons(x, l)) → false
head(cons(x, l)) → x
tail(nil) → nil
tail(cons(x, l)) → l
rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
last(x, l) → if(empty(l), x, l)
if(true, x, l) → x
if(false, x, l) → last(head(l), tail(l))
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

Q is empty.

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

↳ QTRS
  ↳ Overlay + Local Confluence
QTRS
      ↳ DependencyPairsProof

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

empty(nil) → true
empty(cons(x, l)) → false
head(cons(x, l)) → x
tail(nil) → nil
tail(cons(x, l)) → l
rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
last(x, l) → if(empty(l), x, l)
if(true, x, l) → x
if(false, x, l) → last(head(l), tail(l))
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

The set Q consists of the following terms:

empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
rev(nil)
rev(cons(x0, x1))
last(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
rev2(x0, nil)
rev2(x0, cons(x1, x2))


Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

IF(false, x, l) → TAIL(l)
IF(false, x, l) → HEAD(l)
REV(cons(x, l)) → REV2(x, l)
IF(false, x, l) → LAST(head(l), tail(l))
REV2(x, cons(y, l)) → REV2(y, l)
REV2(x, cons(y, l)) → REV(cons(x, rev2(y, l)))
LAST(x, l) → EMPTY(l)
LAST(x, l) → IF(empty(l), x, l)

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(x, l)) → false
head(cons(x, l)) → x
tail(nil) → nil
tail(cons(x, l)) → l
rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
last(x, l) → if(empty(l), x, l)
if(true, x, l) → x
if(false, x, l) → last(head(l), tail(l))
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

The set Q consists of the following terms:

empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
rev(nil)
rev(cons(x0, x1))
last(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
rev2(x0, nil)
rev2(x0, cons(x1, x2))

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

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

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

IF(false, x, l) → TAIL(l)
IF(false, x, l) → HEAD(l)
REV(cons(x, l)) → REV2(x, l)
IF(false, x, l) → LAST(head(l), tail(l))
REV2(x, cons(y, l)) → REV2(y, l)
REV2(x, cons(y, l)) → REV(cons(x, rev2(y, l)))
LAST(x, l) → EMPTY(l)
LAST(x, l) → IF(empty(l), x, l)

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(x, l)) → false
head(cons(x, l)) → x
tail(nil) → nil
tail(cons(x, l)) → l
rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
last(x, l) → if(empty(l), x, l)
if(true, x, l) → x
if(false, x, l) → last(head(l), tail(l))
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

The set Q consists of the following terms:

empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
rev(nil)
rev(cons(x0, x1))
last(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
rev2(x0, nil)
rev2(x0, cons(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 3 less nodes.

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

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

IF(false, x, l) → LAST(head(l), tail(l))
LAST(x, l) → IF(empty(l), x, l)

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(x, l)) → false
head(cons(x, l)) → x
tail(nil) → nil
tail(cons(x, l)) → l
rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
last(x, l) → if(empty(l), x, l)
if(true, x, l) → x
if(false, x, l) → last(head(l), tail(l))
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

The set Q consists of the following terms:

empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
rev(nil)
rev(cons(x0, x1))
last(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
rev2(x0, nil)
rev2(x0, cons(x1, x2))

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

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

IF(false, x, l) → LAST(head(l), tail(l))
LAST(x, l) → IF(empty(l), x, l)

The TRS R consists of the following rules:

head(cons(x, l)) → x
tail(nil) → nil
tail(cons(x, l)) → l
empty(nil) → true
empty(cons(x, l)) → false

The set Q consists of the following terms:

empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
rev(nil)
rev(cons(x0, x1))
last(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
rev2(x0, nil)
rev2(x0, cons(x1, x2))

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.

rev(nil)
rev(cons(x0, x1))
last(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
rev2(x0, nil)
rev2(x0, cons(x1, x2))



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

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

IF(false, x, l) → LAST(head(l), tail(l))
LAST(x, l) → IF(empty(l), x, l)

The TRS R consists of the following rules:

head(cons(x, l)) → x
tail(nil) → nil
tail(cons(x, l)) → l
empty(nil) → true
empty(cons(x, l)) → false

The set Q consists of the following terms:

empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule LAST(x, l) → IF(empty(l), x, l) at position [0] we obtained the following new rules:

LAST(y0, cons(x0, x1)) → IF(false, y0, cons(x0, x1))
LAST(y0, nil) → IF(true, y0, nil)



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

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

LAST(y0, cons(x0, x1)) → IF(false, y0, cons(x0, x1))
IF(false, x, l) → LAST(head(l), tail(l))
LAST(y0, nil) → IF(true, y0, nil)

The TRS R consists of the following rules:

head(cons(x, l)) → x
tail(nil) → nil
tail(cons(x, l)) → l
empty(nil) → true
empty(cons(x, l)) → false

The set Q consists of the following terms:

empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))

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

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

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

LAST(y0, cons(x0, x1)) → IF(false, y0, cons(x0, x1))
IF(false, x, l) → LAST(head(l), tail(l))

The TRS R consists of the following rules:

head(cons(x, l)) → x
tail(nil) → nil
tail(cons(x, l)) → l
empty(nil) → true
empty(cons(x, l)) → false

The set Q consists of the following terms:

empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ UsableRulesProof
QDP
                                    ↳ QReductionProof
              ↳ QDP

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

LAST(y0, cons(x0, x1)) → IF(false, y0, cons(x0, x1))
IF(false, x, l) → LAST(head(l), tail(l))

The TRS R consists of the following rules:

head(cons(x, l)) → x
tail(nil) → nil
tail(cons(x, l)) → l

The set Q consists of the following terms:

empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(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.

empty(nil)
empty(cons(x0, x1))



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

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

LAST(y0, cons(x0, x1)) → IF(false, y0, cons(x0, x1))
IF(false, x, l) → LAST(head(l), tail(l))

The TRS R consists of the following rules:

head(cons(x, l)) → x
tail(nil) → nil
tail(cons(x, l)) → l

The set Q consists of the following terms:

head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule IF(false, x, l) → LAST(head(l), tail(l)) at position [1] we obtained the following new rules:

IF(false, y0, cons(x0, x1)) → LAST(head(cons(x0, x1)), x1)
IF(false, y0, nil) → LAST(head(nil), nil)



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

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

LAST(y0, cons(x0, x1)) → IF(false, y0, cons(x0, x1))
IF(false, y0, nil) → LAST(head(nil), nil)
IF(false, y0, cons(x0, x1)) → LAST(head(cons(x0, x1)), x1)

The TRS R consists of the following rules:

head(cons(x, l)) → x
tail(nil) → nil
tail(cons(x, l)) → l

The set Q consists of the following terms:

head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))

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

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

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

LAST(y0, cons(x0, x1)) → IF(false, y0, cons(x0, x1))
IF(false, y0, cons(x0, x1)) → LAST(head(cons(x0, x1)), x1)

The TRS R consists of the following rules:

head(cons(x, l)) → x
tail(nil) → nil
tail(cons(x, l)) → l

The set Q consists of the following terms:

head(cons(x0, x1))
tail(nil)
tail(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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ UsableRulesProof
                                  ↳ QDP
                                    ↳ QReductionProof
                                      ↳ QDP
                                        ↳ Narrowing
                                          ↳ QDP
                                            ↳ DependencyGraphProof
                                              ↳ QDP
                                                ↳ UsableRulesProof
QDP
                                                    ↳ QReductionProof
              ↳ QDP

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

LAST(y0, cons(x0, x1)) → IF(false, y0, cons(x0, x1))
IF(false, y0, cons(x0, x1)) → LAST(head(cons(x0, x1)), x1)

The TRS R consists of the following rules:

head(cons(x, l)) → x

The set Q consists of the following terms:

head(cons(x0, x1))
tail(nil)
tail(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.

tail(nil)
tail(cons(x0, x1))



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

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

LAST(y0, cons(x0, x1)) → IF(false, y0, cons(x0, x1))
IF(false, y0, cons(x0, x1)) → LAST(head(cons(x0, x1)), x1)

The TRS R consists of the following rules:

head(cons(x, l)) → x

The set Q consists of the following terms:

head(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule IF(false, y0, cons(x0, x1)) → LAST(head(cons(x0, x1)), x1) at position [0] we obtained the following new rules:

IF(false, y0, cons(x0, x1)) → LAST(x0, x1)



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

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

LAST(y0, cons(x0, x1)) → IF(false, y0, cons(x0, x1))
IF(false, y0, cons(x0, x1)) → LAST(x0, x1)

The TRS R consists of the following rules:

head(cons(x, l)) → x

The set Q consists of the following terms:

head(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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ UsableRulesProof
                                  ↳ QDP
                                    ↳ QReductionProof
                                      ↳ QDP
                                        ↳ Narrowing
                                          ↳ QDP
                                            ↳ DependencyGraphProof
                                              ↳ QDP
                                                ↳ UsableRulesProof
                                                  ↳ QDP
                                                    ↳ QReductionProof
                                                      ↳ QDP
                                                        ↳ Rewriting
                                                          ↳ QDP
                                                            ↳ UsableRulesProof
QDP
                                                                ↳ QReductionProof
              ↳ QDP

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

LAST(y0, cons(x0, x1)) → IF(false, y0, cons(x0, x1))
IF(false, y0, cons(x0, x1)) → LAST(x0, x1)

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

head(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.

head(cons(x0, x1))



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

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

LAST(y0, cons(x0, x1)) → IF(false, y0, cons(x0, x1))
IF(false, y0, cons(x0, x1)) → LAST(x0, x1)

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

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

REV(cons(x, l)) → REV2(x, l)
REV2(x, cons(y, l)) → REV2(y, l)
REV2(x, cons(y, l)) → REV(cons(x, rev2(y, l)))

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(x, l)) → false
head(cons(x, l)) → x
tail(nil) → nil
tail(cons(x, l)) → l
rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
last(x, l) → if(empty(l), x, l)
if(true, x, l) → x
if(false, x, l) → last(head(l), tail(l))
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

The set Q consists of the following terms:

empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
rev(nil)
rev(cons(x0, x1))
last(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
rev2(x0, nil)
rev2(x0, cons(x1, x2))

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

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

REV(cons(x, l)) → REV2(x, l)
REV2(x, cons(y, l)) → REV2(y, l)
REV2(x, cons(y, l)) → REV(cons(x, rev2(y, l)))

The TRS R consists of the following rules:

rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))

The set Q consists of the following terms:

empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
rev(nil)
rev(cons(x0, x1))
last(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
rev2(x0, nil)
rev2(x0, cons(x1, x2))

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.

empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
last(x0, x1)
if(true, x0, x1)
if(false, x0, x1)



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

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

REV(cons(x, l)) → REV2(x, l)
REV2(x, cons(y, l)) → REV2(y, l)
REV2(x, cons(y, l)) → REV(cons(x, rev2(y, l)))

The TRS R consists of the following rules:

rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))

The set Q consists of the following terms:

rev(nil)
rev(cons(x0, x1))
rev2(x0, nil)
rev2(x0, cons(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


REV2(x, cons(y, l)) → REV2(y, l)
REV2(x, cons(y, l)) → REV(cons(x, rev2(y, l)))
The remaining pairs can at least be oriented weakly.

REV(cons(x, l)) → REV2(x, l)
Used ordering: Polynomial interpretation [25]:

POL(REV(x1)) = x1   
POL(REV2(x1, x2)) = 1 + x2   
POL(cons(x1, x2)) = 1 + x2   
POL(nil) = 0   
POL(rev(x1)) = x1   
POL(rev1(x1, x2)) = 0   
POL(rev2(x1, x2)) = x2   

The following usable rules [17] were oriented:

rev2(x, nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))



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

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

REV(cons(x, l)) → REV2(x, l)

The TRS R consists of the following rules:

rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))

The set Q consists of the following terms:

rev(nil)
rev(cons(x0, x1))
rev2(x0, nil)
rev2(x0, cons(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.