Termination w.r.t. Q of the given QTRS could be disproven:

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 QDP
↳7 UsableRulesProof (⇔)
↳8 QDP
↳9 QReductionProof (⇔)
↳10 QDP
↳11 Narrowing (⇔)
↳12 QDP
↳13 DependencyGraphProof (⇔)
↳14 QDP
↳15 UsableRulesProof (⇔)
↳16 QDP
↳17 QReductionProof (⇔)
↳18 QDP
↳19 Narrowing (⇔)
↳20 QDP
↳21 UsableRulesProof (⇔)
↳22 QDP
↳23 QReductionProof (⇔)
↳24 QDP
↳25 NonTerminationProof (⇔)
↳26 FALSE

(0) Obligation:

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))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

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

(2) Obligation:

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(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)

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.

(6) Obligation:

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

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

(7) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(8) Obligation:

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

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

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(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.

(9) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

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

(10) Obligation:

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

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

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

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

(11) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule APPEND(l1, l2) → IFAPPEND(l1, l2, is_empty(l1)) at position [2] we obtained the following new rules [LPAR04]:

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

(12) Obligation:

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

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

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

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

(13) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(14) Obligation:

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

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

(15) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(16) Obligation:

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

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

(17) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

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

(18) Obligation:

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.

(19) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule IFAPPEND(l1, l2, false) → APPEND(tl(l1), l2) at position [0] we obtained the following new rules [LPAR04]:

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

(20) Obligation:

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(cons(x0, x1), y1)

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.

(21) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(22) Obligation:

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(cons(x0, 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.

(23) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

tl(cons(x0, x1))

(24) Obligation:

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(cons(x0, x1), y1)

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

(25) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

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

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Semiunifier: [ ]
  • Matcher: [ ]



Rewriting sequence

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

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

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.



(26) FALSE