YES Termination w.r.t. Q proof of Transformed_CSR_04_ExIntrod_GM01_GM.ari

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

0 QTRS
↳1 QTRSRRRProof (⇔, 77 ms)
↳2 QTRS
↳3 QTRSRRRProof (⇔, 0 ms)
↳4 QTRS
↳5 QTRSRRRProof (⇔, 0 ms)
↳6 QTRS
↳7 DependencyPairsProof (⇔, 0 ms)
↳8 QDP
↳9 DependencyGraphProof (⇔, 0 ms)
↳10 QDP
↳11 MRRProof (⇔, 2 ms)
↳12 QDP
↳13 DependencyGraphProof (⇔, 0 ms)
↳14 QDP
↳15 MRRProof (⇔, 0 ms)
↳16 QDP
↳17 QDPOrderProof (⇔, 7 ms)
↳18 QDP
↳19 DependencyGraphProof (⇔, 0 ms)
↳20 QDP
↳21 UsableRulesProof (⇔, 0 ms)
↳22 QDP
↳23 QDPSizeChangeProof (⇔, 0 ms)
↳24 YES

(0) Obligation:

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

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(nil) → nil
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__head(cons(X, L)) → mark(X)
a__tail(cons(X, L)) → mark(L)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(X) → tail(X)

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(a__adx(x1)) = x1   
POL(a__head(x1)) = 2 + 2·x1   
POL(a__incr(x1)) = x1   
POL(a__nats) = 2   
POL(a__tail(x1)) = 2 + x1   
POL(a__zeros) = 2   
POL(adx(x1)) = x1   
POL(cons(x1, x2)) = x1 + x2   
POL(head(x1)) = 2 + 2·x1   
POL(incr(x1)) = x1   
POL(mark(x1)) = x1   
POL(nats) = 2   
POL(nil) = 0   
POL(s(x1)) = x1   
POL(tail(x1)) = 2 + x1   
POL(zeros) = 2   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

a__head(cons(X, L)) → mark(X)
a__tail(cons(X, L)) → mark(L)


(2) Obligation:

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

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(nil) → nil
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(X) → tail(X)

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(a__adx(x1)) = 2 + 2·x1   
POL(a__head(x1)) = 2 + 2·x1   
POL(a__incr(x1)) = x1   
POL(a__nats) = 2   
POL(a__tail(x1)) = 2 + x1   
POL(a__zeros) = 0   
POL(adx(x1)) = 2 + 2·x1   
POL(cons(x1, x2)) = x1 + x2   
POL(head(x1)) = 2 + 2·x1   
POL(incr(x1)) = x1   
POL(mark(x1)) = x1   
POL(nats) = 2   
POL(nil) = 2   
POL(s(x1)) = x1   
POL(tail(x1)) = 2 + x1   
POL(zeros) = 0   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

a__adx(nil) → nil


(4) Obligation:

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

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(X) → tail(X)

Q is empty.

(5) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(a__adx(x1)) = x1   
POL(a__head(x1)) = 1 + x1   
POL(a__incr(x1)) = x1   
POL(a__nats) = 2   
POL(a__tail(x1)) = 1 + 2·x1   
POL(a__zeros) = 0   
POL(adx(x1)) = x1   
POL(cons(x1, x2)) = x1 + x2   
POL(head(x1)) = 1 + x1   
POL(incr(x1)) = x1   
POL(mark(x1)) = x1   
POL(nats) = 2   
POL(nil) = 2   
POL(s(x1)) = x1   
POL(tail(x1)) = 1 + 2·x1   
POL(zeros) = 0   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

a__nats → a__adx(a__zeros)


(6) Obligation:

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

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__zeros → cons(0, zeros)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(X) → tail(X)

Q is empty.

(7) DependencyPairsProof (EQUIVALENT transformation)

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

(8) Obligation:

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

A__INCR(cons(X, L)) → MARK(X)
A__ADX(cons(X, L)) → A__INCR(cons(mark(X), adx(L)))
A__ADX(cons(X, L)) → MARK(X)
MARK(incr(X)) → A__INCR(mark(X))
MARK(incr(X)) → MARK(X)
MARK(adx(X)) → A__ADX(mark(X))
MARK(adx(X)) → MARK(X)
MARK(nats) → A__NATS
MARK(zeros) → A__ZEROS
MARK(head(X)) → A__HEAD(mark(X))
MARK(head(X)) → MARK(X)
MARK(tail(X)) → A__TAIL(mark(X))
MARK(tail(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__zeros → cons(0, zeros)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(X) → tail(X)

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

(9) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes.

(10) Obligation:

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

MARK(incr(X)) → A__INCR(mark(X))
A__INCR(cons(X, L)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(adx(X)) → A__ADX(mark(X))
A__ADX(cons(X, L)) → A__INCR(cons(mark(X), adx(L)))
A__ADX(cons(X, L)) → MARK(X)
MARK(adx(X)) → MARK(X)
MARK(head(X)) → MARK(X)
MARK(tail(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__zeros → cons(0, zeros)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(X) → tail(X)

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

(11) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

A__ADX(cons(X, L)) → A__INCR(cons(mark(X), adx(L)))
A__ADX(cons(X, L)) → MARK(X)
MARK(adx(X)) → MARK(X)
MARK(head(X)) → MARK(X)


Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(A__ADX(x1)) = 2 + 2·x1   
POL(A__INCR(x1)) = x1   
POL(MARK(x1)) = 2·x1   
POL(a__adx(x1)) = 1 + 2·x1   
POL(a__head(x1)) = 2 + 2·x1   
POL(a__incr(x1)) = x1   
POL(a__nats) = 1   
POL(a__tail(x1)) = x1   
POL(a__zeros) = 2   
POL(adx(x1)) = 1 + 2·x1   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(head(x1)) = 2 + 2·x1   
POL(incr(x1)) = x1   
POL(mark(x1)) = x1   
POL(nats) = 1   
POL(nil) = 0   
POL(s(x1)) = x1   
POL(tail(x1)) = x1   
POL(zeros) = 2   

(12) Obligation:

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

MARK(incr(X)) → A__INCR(mark(X))
A__INCR(cons(X, L)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(adx(X)) → A__ADX(mark(X))
MARK(tail(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__zeros → cons(0, zeros)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(X) → tail(X)

Q is empty.
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:

A__INCR(cons(X, L)) → MARK(X)
MARK(incr(X)) → A__INCR(mark(X))
MARK(incr(X)) → MARK(X)
MARK(tail(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__zeros → cons(0, zeros)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(X) → tail(X)

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

(15) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

MARK(tail(X)) → MARK(X)


Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(A__INCR(x1)) = 2·x1   
POL(MARK(x1)) = 2·x1   
POL(a__adx(x1)) = x1   
POL(a__head(x1)) = 2 + x1   
POL(a__incr(x1)) = x1   
POL(a__nats) = 0   
POL(a__tail(x1)) = 1 + x1   
POL(a__zeros) = 0   
POL(adx(x1)) = x1   
POL(cons(x1, x2)) = x1 + 2·x2   
POL(head(x1)) = 2 + x1   
POL(incr(x1)) = x1   
POL(mark(x1)) = x1   
POL(nats) = 0   
POL(nil) = 0   
POL(s(x1)) = x1   
POL(tail(x1)) = 1 + x1   
POL(zeros) = 0   

(16) Obligation:

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

A__INCR(cons(X, L)) → MARK(X)
MARK(incr(X)) → A__INCR(mark(X))
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__zeros → cons(0, zeros)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(X) → tail(X)

Q is empty.
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.


A__INCR(cons(X, L)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( A__INCR(x1) ) = x1 + 1

POL( mark(x1) ) = x1

POL( incr(x1) ) = x1

POL( a__incr(x1) ) = x1

POL( adx(x1) ) = x1 + 1

POL( a__adx(x1) ) = x1 + 1

POL( nats ) = 0

POL( a__nats ) = 0

POL( zeros ) = 1

POL( a__zeros ) = 1

POL( head(x1) ) = 2

POL( a__head(x1) ) = 2

POL( tail(x1) ) = 0

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

POL( nil ) = 2

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

POL( s(x1) ) = x1

POL( 0 ) = 0

POL( MARK(x1) ) = x1 + 1


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

mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__adx(X) → adx(X)
a__incr(nil) → nil
a__incr(X) → incr(X)
a__head(X) → head(X)
a__tail(X) → tail(X)
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__zeros → cons(0, zeros)
a__zeros → zeros
a__nats → nats

(18) Obligation:

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

MARK(incr(X)) → A__INCR(mark(X))
MARK(incr(X)) → MARK(X)
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__zeros → cons(0, zeros)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(X) → tail(X)

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

(19) DependencyGraphProof (EQUIVALENT transformation)

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

(20) Obligation:

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

MARK(s(X)) → MARK(X)
MARK(incr(X)) → MARK(X)

The TRS R consists of the following rules:

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__zeros → cons(0, zeros)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(X) → tail(X)

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

(21) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(22) Obligation:

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

MARK(s(X)) → MARK(X)
MARK(incr(X)) → MARK(X)

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

(23) 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:

  • MARK(s(X)) → MARK(X)
    The graph contains the following edges 1 > 1

  • MARK(incr(X)) → MARK(X)
    The graph contains the following edges 1 > 1

(24) YES