YES Termination w.r.t. Q proof of Secret_06_TRS_reverse.ari

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

0 QTRS
↳1 Overlay + Local Confluence (⇔, 8 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 0 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 AND
    ↳7 QDP
        ↳8 UsableRulesProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 QReductionProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔, 0 ms)
        ↳13 YES
    ↳14 QDP
        ↳15 UsableRulesProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 QReductionProof (⇔, 0 ms)
        ↳18 QDP
        ↳19 QDPSizeChangeProof (⇔, 0 ms)
        ↳20 YES
    ↳21 QDP
        ↳22 UsableRulesProof (⇔, 0 ms)
        ↳23 QDP
        ↳24 QReductionProof (⇔, 0 ms)
        ↳25 QDP
        ↳26 QDPSizeChangeProof (⇔, 0 ms)
        ↳27 YES
    ↳28 QDP
        ↳29 UsableRulesProof (⇔, 0 ms)
        ↳30 QDP
        ↳31 QReductionProof (⇔, 0 ms)
        ↳32 QDP
        ↳33 TransformationProof (⇔, 0 ms)
        ↳34 QDP
        ↳35 DependencyGraphProof (⇔, 0 ms)
        ↳36 QDP
        ↳37 UsableRulesProof (⇔, 0 ms)
        ↳38 QDP
        ↳39 QReductionProof (⇔, 0 ms)
        ↳40 QDP
        ↳41 QDPOrderProof (⇔, 20 ms)
        ↳42 QDP
        ↳43 PisEmptyProof (⇔, 0 ms)
        ↳44 YES

(0) Obligation:

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

isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
reverse(xs) → rev(xs, nil)
rev(xs, ys) → if(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
if(true, xs, ys, zs) → zs
if(false, xs, ys, zs) → rev(xs, ys)

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:

isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
reverse(xs) → rev(xs, nil)
rev(xs, ys) → if(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
if(true, xs, ys, zs) → zs
if(false, xs, ys, zs) → rev(xs, ys)

The set Q consists of the following terms:

isEmpty(nil)
isEmpty(cons(x0, x1))
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
append(nil, x0)
append(cons(x0, x1), x2)
reverse(x0)
rev(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

(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:

LAST(cons(x, cons(y, ys))) → LAST(cons(y, ys))
DROPLAST(cons(x, cons(y, ys))) → DROPLAST(cons(y, ys))
APPEND(cons(x, xs), ys) → APPEND(xs, ys)
REVERSE(xs) → REV(xs, nil)
REV(xs, ys) → IF(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
REV(xs, ys) → ISEMPTY(xs)
REV(xs, ys) → DROPLAST(xs)
REV(xs, ys) → APPEND(ys, last(xs))
REV(xs, ys) → LAST(xs)
IF(false, xs, ys, zs) → REV(xs, ys)

The TRS R consists of the following rules:

isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
reverse(xs) → rev(xs, nil)
rev(xs, ys) → if(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
if(true, xs, ys, zs) → zs
if(false, xs, ys, zs) → rev(xs, ys)

The set Q consists of the following terms:

isEmpty(nil)
isEmpty(cons(x0, x1))
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
append(nil, x0)
append(cons(x0, x1), x2)
reverse(x0)
rev(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 5 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

APPEND(cons(x, xs), ys) → APPEND(xs, ys)

The TRS R consists of the following rules:

isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
reverse(xs) → rev(xs, nil)
rev(xs, ys) → if(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
if(true, xs, ys, zs) → zs
if(false, xs, ys, zs) → rev(xs, ys)

The set Q consists of the following terms:

isEmpty(nil)
isEmpty(cons(x0, x1))
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
append(nil, x0)
append(cons(x0, x1), x2)
reverse(x0)
rev(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

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

(8) 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.

(9) Obligation:

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

APPEND(cons(x, xs), ys) → APPEND(xs, ys)

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

isEmpty(nil)
isEmpty(cons(x0, x1))
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
append(nil, x0)
append(cons(x0, x1), x2)
reverse(x0)
rev(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

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

(10) 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].

isEmpty(nil)
isEmpty(cons(x0, x1))
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
append(nil, x0)
append(cons(x0, x1), x2)
reverse(x0)
rev(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

(11) Obligation:

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

APPEND(cons(x, xs), ys) → APPEND(xs, ys)

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

(12) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

  • APPEND(cons(x, xs), ys) → APPEND(xs, ys)
    The graph contains the following edges 1 > 1, 2 >= 2

(13) YES

(14) Obligation:

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

DROPLAST(cons(x, cons(y, ys))) → DROPLAST(cons(y, ys))

The TRS R consists of the following rules:

isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
reverse(xs) → rev(xs, nil)
rev(xs, ys) → if(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
if(true, xs, ys, zs) → zs
if(false, xs, ys, zs) → rev(xs, ys)

The set Q consists of the following terms:

isEmpty(nil)
isEmpty(cons(x0, x1))
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
append(nil, x0)
append(cons(x0, x1), x2)
reverse(x0)
rev(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

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:

DROPLAST(cons(x, cons(y, ys))) → DROPLAST(cons(y, ys))

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

isEmpty(nil)
isEmpty(cons(x0, x1))
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
append(nil, x0)
append(cons(x0, x1), x2)
reverse(x0)
rev(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

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].

isEmpty(nil)
isEmpty(cons(x0, x1))
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
append(nil, x0)
append(cons(x0, x1), x2)
reverse(x0)
rev(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

(18) Obligation:

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

DROPLAST(cons(x, cons(y, ys))) → DROPLAST(cons(y, ys))

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

(19) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

  • DROPLAST(cons(x, cons(y, ys))) → DROPLAST(cons(y, ys))
    The graph contains the following edges 1 > 1

(20) YES

(21) Obligation:

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

LAST(cons(x, cons(y, ys))) → LAST(cons(y, ys))

The TRS R consists of the following rules:

isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
reverse(xs) → rev(xs, nil)
rev(xs, ys) → if(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
if(true, xs, ys, zs) → zs
if(false, xs, ys, zs) → rev(xs, ys)

The set Q consists of the following terms:

isEmpty(nil)
isEmpty(cons(x0, x1))
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
append(nil, x0)
append(cons(x0, x1), x2)
reverse(x0)
rev(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

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

(22) 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.

(23) Obligation:

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

LAST(cons(x, cons(y, ys))) → LAST(cons(y, ys))

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

isEmpty(nil)
isEmpty(cons(x0, x1))
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
append(nil, x0)
append(cons(x0, x1), x2)
reverse(x0)
rev(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

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

(24) 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].

isEmpty(nil)
isEmpty(cons(x0, x1))
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
append(nil, x0)
append(cons(x0, x1), x2)
reverse(x0)
rev(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

(25) Obligation:

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

LAST(cons(x, cons(y, ys))) → LAST(cons(y, ys))

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

(26) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

  • LAST(cons(x, cons(y, ys))) → LAST(cons(y, ys))
    The graph contains the following edges 1 > 1

(27) YES

(28) Obligation:

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

REV(xs, ys) → IF(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
IF(false, xs, ys, zs) → REV(xs, ys)

The TRS R consists of the following rules:

isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
reverse(xs) → rev(xs, nil)
rev(xs, ys) → if(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
if(true, xs, ys, zs) → zs
if(false, xs, ys, zs) → rev(xs, ys)

The set Q consists of the following terms:

isEmpty(nil)
isEmpty(cons(x0, x1))
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
append(nil, x0)
append(cons(x0, x1), x2)
reverse(x0)
rev(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

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

(29) 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.

(30) Obligation:

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

REV(xs, ys) → IF(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
IF(false, xs, ys, zs) → REV(xs, ys)

The TRS R consists of the following rules:

isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))

The set Q consists of the following terms:

isEmpty(nil)
isEmpty(cons(x0, x1))
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
append(nil, x0)
append(cons(x0, x1), x2)
reverse(x0)
rev(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

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

(31) 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].

reverse(x0)
rev(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

(32) Obligation:

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

REV(xs, ys) → IF(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
IF(false, xs, ys, zs) → REV(xs, ys)

The TRS R consists of the following rules:

isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))

The set Q consists of the following terms:

isEmpty(nil)
isEmpty(cons(x0, x1))
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
append(nil, x0)
append(cons(x0, x1), x2)

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

(33) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule REV(xs, ys) → IF(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys) at position [0] we obtained the following new rules [LPAR04]:

REV(nil, y1) → IF(true, dropLast(nil), append(y1, last(nil)), y1) → REV(nil, y1) → IF(true, dropLast(nil), append(y1, last(nil)), y1)
REV(cons(x0, x1), y1) → IF(false, dropLast(cons(x0, x1)), append(y1, last(cons(x0, x1))), y1) → REV(cons(x0, x1), y1) → IF(false, dropLast(cons(x0, x1)), append(y1, last(cons(x0, x1))), y1)

(34) Obligation:

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

IF(false, xs, ys, zs) → REV(xs, ys)
REV(nil, y1) → IF(true, dropLast(nil), append(y1, last(nil)), y1)
REV(cons(x0, x1), y1) → IF(false, dropLast(cons(x0, x1)), append(y1, last(cons(x0, x1))), y1)

The TRS R consists of the following rules:

isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))

The set Q consists of the following terms:

isEmpty(nil)
isEmpty(cons(x0, x1))
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
append(nil, x0)
append(cons(x0, x1), x2)

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

(35) DependencyGraphProof (EQUIVALENT transformation)

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

(36) Obligation:

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

REV(cons(x0, x1), y1) → IF(false, dropLast(cons(x0, x1)), append(y1, last(cons(x0, x1))), y1)
IF(false, xs, ys, zs) → REV(xs, ys)

The TRS R consists of the following rules:

isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))

The set Q consists of the following terms:

isEmpty(nil)
isEmpty(cons(x0, x1))
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
append(nil, x0)
append(cons(x0, x1), x2)

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

(37) 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.

(38) Obligation:

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

REV(cons(x0, x1), y1) → IF(false, dropLast(cons(x0, x1)), append(y1, last(cons(x0, x1))), y1)
IF(false, xs, ys, zs) → REV(xs, ys)

The TRS R consists of the following rules:

dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))

The set Q consists of the following terms:

isEmpty(nil)
isEmpty(cons(x0, x1))
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
append(nil, x0)
append(cons(x0, x1), x2)

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

(39) 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].

isEmpty(nil)
isEmpty(cons(x0, x1))

(40) Obligation:

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

REV(cons(x0, x1), y1) → IF(false, dropLast(cons(x0, x1)), append(y1, last(cons(x0, x1))), y1)
IF(false, xs, ys, zs) → REV(xs, ys)

The TRS R consists of the following rules:

dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))

The set Q consists of the following terms:

last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
append(nil, x0)
append(cons(x0, x1), x2)

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

(41) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


REV(cons(x0, x1), y1) → IF(false, dropLast(cons(x0, x1)), append(y1, last(cons(x0, x1))), y1)
IF(false, xs, ys, zs) → REV(xs, ys)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( IF(x1, ..., x4) ) = 2x2 + 2

POL( dropLast(x1) ) = max{0, x1 - 2}

POL( cons(x1, x2) ) = 2x2 + 2

POL( nil ) = 2

POL( append(x1, x2) ) = max{0, -2}

POL( last(x1) ) = max{0, -2}

POL( REV(x1, x2) ) = 2x1 + 1

POL( false ) = 1


The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))

(42) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))

The set Q consists of the following terms:

last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
append(nil, x0)
append(cons(x0, x1), x2)

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

(43) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(44) YES