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:

is_empty(nil) → true
is_empty(cons(x, l)) → false
hd(cons(x, l)) → x
tl(cons(x, l)) → l
append(l1, l2) → ifappend(l1, l2, is_empty(l1))
ifappend(l1, l2, true) → l2
ifappend(l1, l2, false) → cons(hd(l1), append(tl(l1), l2))

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

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

is_empty(nil) → true
is_empty(cons(x, l)) → false
hd(cons(x, l)) → x
tl(cons(x, l)) → l
append(l1, l2) → ifappend(l1, l2, is_empty(l1))
ifappend(l1, l2, true) → l2
ifappend(l1, l2, false) → cons(hd(l1), append(tl(l1), l2))

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:

is_empty(nil) → true
is_empty(cons(x, l)) → false
hd(cons(x, l)) → x
tl(cons(x, l)) → l
append(l1, l2) → ifappend(l1, l2, is_empty(l1))
ifappend(l1, l2, true) → l2
ifappend(l1, l2, false) → cons(hd(l1), append(tl(l1), l2))

The set Q consists of the following terms:

is_empty(nil)
is_empty(cons(x0, x1))
hd(cons(x0, x1))
tl(cons(x0, x1))
append(x0, x1)
ifappend(x0, x1, true)
ifappend(x0, x1, false)


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

APPEND(l1, l2) → IFAPPEND(l1, l2, is_empty(l1))
IFAPPEND(l1, l2, false) → TL(l1)
IFAPPEND(l1, l2, false) → APPEND(tl(l1), l2)
APPEND(l1, l2) → IS_EMPTY(l1)
IFAPPEND(l1, l2, false) → HD(l1)

The TRS R consists of the following rules:

is_empty(nil) → true
is_empty(cons(x, l)) → false
hd(cons(x, l)) → x
tl(cons(x, l)) → l
append(l1, l2) → ifappend(l1, l2, is_empty(l1))
ifappend(l1, l2, true) → l2
ifappend(l1, l2, false) → cons(hd(l1), append(tl(l1), l2))

The set Q consists of the following terms:

is_empty(nil)
is_empty(cons(x0, x1))
hd(cons(x0, x1))
tl(cons(x0, x1))
append(x0, x1)
ifappend(x0, x1, true)
ifappend(x0, x1, false)

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:

APPEND(l1, l2) → IFAPPEND(l1, l2, is_empty(l1))
IFAPPEND(l1, l2, false) → TL(l1)
IFAPPEND(l1, l2, false) → APPEND(tl(l1), l2)
APPEND(l1, l2) → IS_EMPTY(l1)
IFAPPEND(l1, l2, false) → HD(l1)

The TRS R consists of the following rules:

is_empty(nil) → true
is_empty(cons(x, l)) → false
hd(cons(x, l)) → x
tl(cons(x, l)) → l
append(l1, l2) → ifappend(l1, l2, is_empty(l1))
ifappend(l1, l2, true) → l2
ifappend(l1, l2, false) → cons(hd(l1), append(tl(l1), l2))

The set Q consists of the following terms:

is_empty(nil)
is_empty(cons(x0, x1))
hd(cons(x0, x1))
tl(cons(x0, x1))
append(x0, x1)
ifappend(x0, x1, true)
ifappend(x0, x1, false)

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

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

APPEND(l1, l2) → IFAPPEND(l1, l2, is_empty(l1))
IFAPPEND(l1, l2, false) → APPEND(tl(l1), l2)

The TRS R consists of the following rules:

is_empty(nil) → true
is_empty(cons(x, l)) → false
hd(cons(x, l)) → x
tl(cons(x, l)) → l
append(l1, l2) → ifappend(l1, l2, is_empty(l1))
ifappend(l1, l2, true) → l2
ifappend(l1, l2, false) → cons(hd(l1), append(tl(l1), l2))

The set Q consists of the following terms:

is_empty(nil)
is_empty(cons(x0, x1))
hd(cons(x0, x1))
tl(cons(x0, x1))
append(x0, x1)
ifappend(x0, x1, true)
ifappend(x0, x1, false)

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
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof

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

APPEND(l1, l2) → IFAPPEND(l1, l2, is_empty(l1))
IFAPPEND(l1, l2, false) → APPEND(tl(l1), l2)

The TRS R consists of the following rules:

is_empty(nil) → true
is_empty(cons(x, l)) → false
tl(cons(x, l)) → l

The set Q consists of the following terms:

is_empty(nil)
is_empty(cons(x0, x1))
hd(cons(x0, x1))
tl(cons(x0, x1))
append(x0, x1)
ifappend(x0, x1, true)
ifappend(x0, x1, false)

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.

hd(cons(x0, x1))
append(x0, x1)
ifappend(x0, x1, true)
ifappend(x0, x1, false)



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

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

APPEND(l1, l2) → IFAPPEND(l1, l2, is_empty(l1))
IFAPPEND(l1, l2, false) → APPEND(tl(l1), l2)

The TRS R consists of the following rules:

is_empty(nil) → true
is_empty(cons(x, l)) → false
tl(cons(x, l)) → l

The set Q consists of the following terms:

is_empty(nil)
is_empty(cons(x0, x1))
tl(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule APPEND(l1, l2) → IFAPPEND(l1, l2, is_empty(l1)) at position [2] we obtained the following new rules:

APPEND(cons(x0, x1), y1) → IFAPPEND(cons(x0, x1), y1, false)
APPEND(nil, y1) → IFAPPEND(nil, y1, true)



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

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

APPEND(cons(x0, x1), y1) → IFAPPEND(cons(x0, x1), y1, false)
IFAPPEND(l1, l2, false) → APPEND(tl(l1), l2)
APPEND(nil, y1) → IFAPPEND(nil, y1, true)

The TRS R consists of the following rules:

is_empty(nil) → true
is_empty(cons(x, l)) → false
tl(cons(x, l)) → l

The set Q consists of the following terms:

is_empty(nil)
is_empty(cons(x0, x1))
tl(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
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof
                    ↳ QDP
                      ↳ Narrowing
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ QDP
                              ↳ UsableRulesProof

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

APPEND(cons(x0, x1), y1) → IFAPPEND(cons(x0, x1), y1, false)
IFAPPEND(l1, l2, false) → APPEND(tl(l1), l2)

The TRS R consists of the following rules:

is_empty(nil) → true
is_empty(cons(x, l)) → false
tl(cons(x, l)) → l

The set Q consists of the following terms:

is_empty(nil)
is_empty(cons(x0, x1))
tl(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
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof
                    ↳ QDP
                      ↳ Narrowing
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ QDP
                              ↳ UsableRulesProof
                                ↳ QDP
                                  ↳ QReductionProof

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

APPEND(cons(x0, x1), y1) → IFAPPEND(cons(x0, x1), y1, false)
IFAPPEND(l1, l2, false) → APPEND(tl(l1), l2)

The TRS R consists of the following rules:

tl(cons(x, l)) → l

The set Q consists of the following terms:

is_empty(nil)
is_empty(cons(x0, x1))
tl(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.

is_empty(nil)
is_empty(cons(x0, x1))



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

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

APPEND(cons(x0, x1), y1) → IFAPPEND(cons(x0, x1), y1, false)
IFAPPEND(l1, l2, false) → APPEND(tl(l1), l2)

The TRS R consists of the following rules:

tl(cons(x, l)) → l

The set Q consists of the following terms:

tl(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule IFAPPEND(l1, l2, false) → APPEND(tl(l1), l2) at position [0] we obtained the following new rules:

IFAPPEND(cons(x0, x1), y1, false) → APPEND(x1, y1)



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

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

APPEND(cons(x0, x1), y1) → IFAPPEND(cons(x0, x1), y1, false)
IFAPPEND(cons(x0, x1), y1, false) → APPEND(x1, y1)

The TRS R consists of the following rules:

tl(cons(x, l)) → l

The set Q consists of the following terms:

tl(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
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof
                    ↳ QDP
                      ↳ Narrowing
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ QDP
                              ↳ UsableRulesProof
                                ↳ QDP
                                  ↳ QReductionProof
                                    ↳ QDP
                                      ↳ Narrowing
                                        ↳ QDP
                                          ↳ UsableRulesProof
                                            ↳ QDP
                                              ↳ QReductionProof

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

APPEND(cons(x0, x1), y1) → IFAPPEND(cons(x0, x1), y1, false)
IFAPPEND(cons(x0, x1), y1, false) → APPEND(x1, y1)

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

tl(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.

tl(cons(x0, x1))



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

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

APPEND(cons(x0, x1), y1) → IFAPPEND(cons(x0, x1), y1, false)
IFAPPEND(cons(x0, x1), y1, false) → APPEND(x1, y1)

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: