Termination w.r.t. Q proof of AProVE_07

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

0 QTRS

↳1 Overlay + Local Confluence (⇔, 0 ms)

↳2 QTRS

↳3 DependencyPairsProof (⇔, 5 ms)

↳4 QDP

↳5 DependencyGraphProof (⇔, 0 ms)

↳6 AND

    ↳7 QDP

        ↳8 UsableRulesProof (⇔, 0 ms)

        ↳9 QDP

        ↳10 QReductionProof (⇔, 0 ms)

        ↳11 QDP

        ↳12 QDPOrderProof (⇔, 17 ms)

        ↳13 QDP

        ↳14 DependencyGraphProof (⇔, 0 ms)

        ↳15 TRUE

    ↳16 QDP

        ↳17 UsableRulesProof (⇔, 0 ms)

        ↳18 QDP

        ↳19 QReductionProof (⇔, 0 ms)

        ↳20 QDP

        ↳21 QDPOrderProof (⇔, 0 ms)

        ↳22 QDP

        ↳23 DependencyGraphProof (⇔, 0 ms)

        ↳24 TRUE

(0) Obligation:

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

empty(nil) → true
empty(cons(x, l)) → false
head(cons(x, l)) → x
tail(nil) → nil
tail(cons(x, l)) → l
rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
last(x, l) → if(empty(l), x, l)
if(true, x, l) → x
if(false, x, l) → last(head(l), tail(l))
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

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:

empty(nil) → true
empty(cons(x, l)) → false
head(cons(x, l)) → x
tail(nil) → nil
tail(cons(x, l)) → l
rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
last(x, l) → if(empty(l), x, l)
if(true, x, l) → x
if(false, x, l) → last(head(l), tail(l))
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

The set Q consists of the following terms:

empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
rev(nil)
rev(cons(x0, x1))
last(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
rev2(x0, nil)
rev2(x0, cons(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:

REV(cons(x, l)) → REV2(x, l)
LAST(x, l) → IF(empty(l), x, l)
LAST(x, l) → EMPTY(l)
IF(false, x, l) → LAST(head(l), tail(l))
IF(false, x, l) → HEAD(l)
IF(false, x, l) → TAIL(l)
REV2(x, cons(y, l)) → REV(cons(x, rev2(y, l)))
REV2(x, cons(y, l)) → REV2(y, l)

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(x, l)) → false
head(cons(x, l)) → x
tail(nil) → nil
tail(cons(x, l)) → l
rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
last(x, l) → if(empty(l), x, l)
if(true, x, l) → x
if(false, x, l) → last(head(l), tail(l))
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

The set Q consists of the following terms:

empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
rev(nil)
rev(cons(x0, x1))
last(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
rev2(x0, nil)
rev2(x0, cons(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 2 SCCs with 3 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

IF(false, x, l) → LAST(head(l), tail(l))
LAST(x, l) → IF(empty(l), x, l)

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(x, l)) → false
head(cons(x, l)) → x
tail(nil) → nil
tail(cons(x, l)) → l
rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
last(x, l) → if(empty(l), x, l)
if(true, x, l) → x
if(false, x, l) → last(head(l), tail(l))
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

The set Q consists of the following terms:

empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
rev(nil)
rev(cons(x0, x1))
last(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
rev2(x0, nil)
rev2(x0, cons(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:

IF(false, x, l) → LAST(head(l), tail(l))
LAST(x, l) → IF(empty(l), x, l)

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(x, l)) → false
head(cons(x, l)) → x
tail(nil) → nil
tail(cons(x, l)) → l

The set Q consists of the following terms:

empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
rev(nil)
rev(cons(x0, x1))
last(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
rev2(x0, nil)
rev2(x0, cons(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].

rev(nil)
rev(cons(x0, x1))
last(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
rev2(x0, nil)
rev2(x0, cons(x1, x2))

(11) Obligation:

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

IF(false, x, l) → LAST(head(l), tail(l))
LAST(x, l) → IF(empty(l), x, l)

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(x, l)) → false
head(cons(x, l)) → x
tail(nil) → nil
tail(cons(x, l)) → l

The set Q consists of the following terms:

empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))

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

(12) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

LAST(x, l) → IF(empty(l), x, l)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO,RATPOLO]:

POL(IF(x₁, x₂, x₃)) = [1/4]x₁ + x₃
POL(LAST(x₁, x₂)) = [1/2] + [2]x₂
POL(cons(x₁, x₂)) = [4] + [4]x₂
POL(empty(x₁)) = [1/4] + [4]x₁
POL(false) = [2]
POL(head(x₁)) = 0
POL(nil) = 0
POL(tail(x₁)) = [1/2]x₁
POL(true) = [1/4]

The value of delta used in the strict ordering is 7/16.
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

tail(nil) → nil
tail(cons(x, l)) → l
empty(nil) → true
empty(cons(x, l)) → false

(13) Obligation:

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

IF(false, x, l) → LAST(head(l), tail(l))

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(x, l)) → false
head(cons(x, l)) → x
tail(nil) → nil
tail(cons(x, l)) → l

The set Q consists of the following terms:

empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))

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

(14) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(15) TRUE

(16) Obligation:

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

REV2(x, cons(y, l)) → REV(cons(x, rev2(y, l)))
REV(cons(x, l)) → REV2(x, l)
REV2(x, cons(y, l)) → REV2(y, l)

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(x, l)) → false
head(cons(x, l)) → x
tail(nil) → nil
tail(cons(x, l)) → l
rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
last(x, l) → if(empty(l), x, l)
if(true, x, l) → x
if(false, x, l) → last(head(l), tail(l))
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

The set Q consists of the following terms:

empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
rev(nil)
rev(cons(x0, x1))
last(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
rev2(x0, nil)
rev2(x0, cons(x1, x2))

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

(17) 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.

(18) Obligation:

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

REV2(x, cons(y, l)) → REV(cons(x, rev2(y, l)))
REV(cons(x, l)) → REV2(x, l)
REV2(x, cons(y, l)) → REV2(y, l)

The TRS R consists of the following rules:

rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))

The set Q consists of the following terms:

empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
rev(nil)
rev(cons(x0, x1))
last(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
rev2(x0, nil)
rev2(x0, cons(x1, x2))

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

(19) 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].

empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
last(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

(20) Obligation:

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

REV2(x, cons(y, l)) → REV(cons(x, rev2(y, l)))
REV(cons(x, l)) → REV2(x, l)
REV2(x, cons(y, l)) → REV2(y, l)

The TRS R consists of the following rules:

rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))

The set Q consists of the following terms:

rev(nil)
rev(cons(x0, x1))
rev2(x0, nil)
rev2(x0, cons(x1, x2))

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

(21) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

REV2(x, cons(y, l)) → REV(cons(x, rev2(y, l)))
REV2(x, cons(y, l)) → REV2(y, l)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( REV(x₁) ) = x₁ + 1

POL( cons(x₁, x₂) ) = x₁ + x₂ + 1

POL( rev2(x₁, x₂) ) = x₁ + x₂

POL( nil ) = 0

POL( rev(x₁) ) = x₁

POL( rev1(x₁, x₂) ) = 0

POL( REV2(x₁, x₂) ) = x₁ + x₂ + 2

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

rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))

(22) Obligation:

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

REV(cons(x, l)) → REV2(x, l)

The TRS R consists of the following rules:

rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))

The set Q consists of the following terms:

rev(nil)
rev(cons(x0, x1))
rev2(x0, nil)
rev2(x0, cons(x1, x2))

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

(23) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(0) Obligation:

(1) Overlay + Local Confluence (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyPairsProof (EQUIVALENT transformation)

(4) Obligation:

(5) DependencyGraphProof (EQUIVALENT transformation)

(6) Complex Obligation (AND)

(7) Obligation:

(8) UsableRulesProof (EQUIVALENT transformation)

(9) Obligation:

(10) QReductionProof (EQUIVALENT transformation)

(11) Obligation:

(12) QDPOrderProof (EQUIVALENT transformation)

(13) Obligation:

(14) DependencyGraphProof (EQUIVALENT transformation)

(15) TRUE

(16) Obligation:

(17) UsableRulesProof (EQUIVALENT transformation)

(18) Obligation:

(19) QReductionProof (EQUIVALENT transformation)

(20) Obligation:

(21) QDPOrderProof (EQUIVALENT transformation)

(22) Obligation:

(23) DependencyGraphProof (EQUIVALENT transformation)

(24) TRUE