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:

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



↳ 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)) → cons(x, 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(cons(x0, x1))
hd(cons(x0, x1))
is_empty(nil)
tl(cons(x0, x1))
ifappend(x0, x1, true)
ifappend(x0, x1, false)
append(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:

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

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)) → cons(x, 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(cons(x0, x1))
hd(cons(x0, x1))
is_empty(nil)
tl(cons(x0, x1))
ifappend(x0, x1, true)
ifappend(x0, x1, false)
append(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 3 less nodes.

↳ 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)) → cons(x, 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(cons(x0, x1))
hd(cons(x0, x1))
is_empty(nil)
tl(cons(x0, x1))
ifappend(x0, x1, true)
ifappend(x0, x1, false)
append(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
        ↳ 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)) → cons(x, l)

The set Q consists of the following terms:

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

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



↳ 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)) → cons(x, l)

The set Q consists of the following terms:

is_empty(cons(x0, x1))
is_empty(nil)
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
  ↳ 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)) → cons(x, l)

The set Q consists of the following terms:

is_empty(cons(x0, x1))
is_empty(nil)
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
  ↳ 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)) → cons(x, l)

The set Q consists of the following terms:

is_empty(cons(x0, x1))
is_empty(nil)
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
  ↳ 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)) → cons(x, l)

The set Q consists of the following terms:

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



↳ 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)) → cons(x, 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(cons(x0, x1), y1)



↳ 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:

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

The TRS R consists of the following rules:

tl(cons(x, l)) → cons(x, 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
  ↳ 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:

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

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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ Narrowing
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ UsableRulesProof
                            ↳ QDP
                              ↳ QReductionProof
                                ↳ QDP
                                  ↳ Narrowing
                                    ↳ QDP
                                      ↳ UsableRulesProof
                                        ↳ QDP
                                          ↳ QReductionProof
                                            ↳ QDP
                                              ↳ NonTerminationProof

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

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

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 narrowing to the left:

The TRS P consists of the following rules:

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

The TRS R consists of the following rules:none


s = APPEND(cons(x0', x1'), y1') evaluates to t =APPEND(cons(x0', x1'), y1')

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



Rewriting sequence

APPEND(cons(x0', x1'), y1') → IFAPPEND(cons(x0', x1'), y1', false)
with rule APPEND(cons(x0'', x1''), y1'') → IFAPPEND(cons(x0'', x1''), y1'', false) at position [] and matcher [x1'' / x1', y1'' / y1', x0'' / x0']

IFAPPEND(cons(x0', x1'), y1', false) → APPEND(cons(x0', x1'), y1')
with rule IFAPPEND(cons(x0, x1), y1, false) → APPEND(cons(x0, x1), y1)

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.