Termination w.r.t. Q proof of AG01

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

0 QTRS

↳1 Overlay + Local Confluence (⇔, 6 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 QDPOrderProof (⇔, 27 ms)

        ↳18 QDP

        ↳19 PisEmptyProof (⇔, 0 ms)

        ↳20 YES

(0) Obligation:

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

rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0, nil) → 0
rev1(s(x), nil) → s(x)
rev1(x, cons(y, l)) → rev1(y, 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:

rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0, nil) → 0
rev1(s(x), nil) → s(x)
rev1(x, cons(y, l)) → rev1(y, l)
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

The set Q consists of the following terms:

rev(nil)
rev(cons(x0, x1))
rev1(0, nil)
rev1(s(x0), nil)
rev1(x0, cons(x1, x2))
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)) → REV1(x, l)
REV(cons(x, l)) → REV2(x, l)
REV1(x, cons(y, l)) → REV1(y, 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:

rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0, nil) → 0
rev1(s(x), nil) → s(x)
rev1(x, cons(y, l)) → rev1(y, l)
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

The set Q consists of the following terms:

rev(nil)
rev(cons(x0, x1))
rev1(0, nil)
rev1(s(x0), nil)
rev1(x0, cons(x1, x2))
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 1 less node.

(6) Complex Obligation (AND)

(7) Obligation:

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

REV1(x, cons(y, l)) → REV1(y, l)

The TRS R consists of the following rules:

rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0, nil) → 0
rev1(s(x), nil) → s(x)
rev1(x, cons(y, l)) → rev1(y, l)
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

The set Q consists of the following terms:

rev(nil)
rev(cons(x0, x1))
rev1(0, nil)
rev1(s(x0), nil)
rev1(x0, cons(x1, x2))
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:

REV1(x, cons(y, l)) → REV1(y, l)

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

rev(nil)
rev(cons(x0, x1))
rev1(0, nil)
rev1(s(x0), nil)
rev1(x0, cons(x1, x2))
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))
rev1(0, nil)
rev1(s(x0), nil)
rev1(x0, cons(x1, x2))
rev2(x0, nil)
rev2(x0, cons(x1, x2))

(11) Obligation:

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

REV1(x, cons(y, l)) → REV1(y, l)

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:

REV1(x, cons(y, l)) → REV1(y, l)
The graph contains the following edges 2 > 1, 2 > 2

(13) YES

(14) Obligation:

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

REV(cons(x, l)) → REV2(x, 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:

rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0, nil) → 0
rev1(s(x), nil) → s(x)
rev1(x, cons(y, l)) → rev1(y, l)
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

The set Q consists of the following terms:

rev(nil)
rev(cons(x0, x1))
rev1(0, nil)
rev1(s(x0), nil)
rev1(x0, cons(x1, x2))
rev2(x0, nil)
rev2(x0, cons(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:

REV(cons(x, l)) → REV2(x, 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:

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))
rev1(0, nil) → 0
rev1(s(x), nil) → s(x)
rev1(x, cons(y, l)) → rev1(y, l)

The set Q consists of the following terms:

rev(nil)
rev(cons(x0, x1))
rev1(0, nil)
rev1(s(x0), nil)
rev1(x0, cons(x1, x2))
rev2(x0, nil)
rev2(x0, cons(x1, x2))

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

(17) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

REV(cons(x, l)) → REV2(x, l)
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) ) = 2x₂ + 2

POL( rev2(x₁, x₂) ) = x₂

POL( nil ) = 0

POL( rev(x₁) ) = x₁

POL( rev1(x₁, x₂) ) = x₁ + 2

POL( 0 ) = 2

POL( s(x₁) ) = 2

POL( REV2(x₁, x₂) ) = 2x₂ + 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))

(18) Obligation:

Q DP problem:
P is empty.
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))
rev1(0, nil) → 0
rev1(s(x), nil) → s(x)
rev1(x, cons(y, l)) → rev1(y, l)

The set Q consists of the following terms:

rev(nil)
rev(cons(x0, x1))
rev1(0, nil)
rev1(s(x0), nil)
rev1(x0, cons(x1, x2))
rev2(x0, nil)
rev2(x0, cons(x1, x2))

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

(19) PisEmptyProof (EQUIVALENT transformation)

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

(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) QDPSizeChangeProof (EQUIVALENT transformation)

(13) YES

(14) Obligation:

(15) UsableRulesProof (EQUIVALENT transformation)

(16) Obligation:

(17) QDPOrderProof (EQUIVALENT transformation)

(18) Obligation:

(19) PisEmptyProof (EQUIVALENT transformation)

(20) YES