Termination w.r.t. Q proof of /tmp/tmpOXv8QA/reverse.trs

Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS

↳1 Overlay + Local Confluence (⇔)

↳2 QTRS

↳3 DependencyPairsProof (⇔)

↳4 QDP

↳5 DependencyGraphProof (⇔)

↳6 AND

  ↳7 QDP

    ↳8 QDPOrderProof (⇔)

    ↳9 QDP

    ↳10 PisEmptyProof (⇔)

    ↳11 TRUE

  ↳12 QDP

    ↳13 QDPOrderProof (⇔)

    ↳14 QDP

    ↳15 PisEmptyProof (⇔)

    ↳16 TRUE

  ↳17 QDP

    ↳18 QDPOrderProof (⇔)

    ↳19 QDP

    ↳20 PisEmptyProof (⇔)

    ↳21 TRUE

  ↳22 QDP

(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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
APPEND(x1, x2) = APPEND(x1, x2)
cons(x1, x2) = cons(x2)

Lexicographic path order with status [LPO].
Quasi-Precedence:

[APPEND₂, cons₁]

Status:

APPEND₂: [2,1]
cons₁: [1]

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

Q DP problem:
P is empty.
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.

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) 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.

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
DROPLAST(x1) = x1
cons(x1, x2) = cons(x2)

Lexicographic path order with status [LPO].
Quasi-Precedence:

trivial

Status:

cons₁: [1]

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

Q DP problem:
P is empty.
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) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE

(17) 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.

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
LAST(x1) = x1
cons(x1, x2) = cons(x2)

Lexicographic path order with status [LPO].
Quasi-Precedence:

trivial

Status:

cons₁: [1]

The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

Q DP problem:
P is empty.
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.

(20) PisEmptyProof (EQUIVALENT transformation)

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

(21) TRUE

(22) 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.