Termination w.r.t. Q proof of Transformed_CSR_04_Ex26_Luc03b

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

0 QTRS

↳1 DependencyPairsProof (⇔, 0 ms)

↳2 QDP

↳3 DependencyGraphProof (⇔, 0 ms)

↳4 QDP

↳5 QDPOrderProof (⇔, 87 ms)

↳6 QDP

↳7 QDPOrderProof (⇔, 29 ms)

↳8 QDP

↳9 DependencyGraphProof (⇔, 0 ms)

↳10 AND

    ↳11 QDP

        ↳12 UsableRulesProof (⇔, 0 ms)

        ↳13 QDP

        ↳14 QDPSizeChangeProof (⇔, 0 ms)

        ↳15 YES

    ↳16 QDP

        ↳17 QDPOrderProof (⇔, 49 ms)

        ↳18 QDP

        ↳19 PisEmptyProof (⇔, 0 ms)

        ↳20 YES

(0) Obligation:

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

terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(n__add(sqr(activate(X)), dbl(activate(X))))
dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(activate(X), activate(Z)))
terms(X) → n__terms(X)
add(X1, X2) → n__add(X1, X2)
s(X) → n__s(X)
dbl(X) → n__dbl(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

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

(2) Obligation:

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

TERMS(N) → SQR(N)
TERMS(N) → S(N)
SQR(s(X)) → S(n__add(sqr(activate(X)), dbl(activate(X))))
SQR(s(X)) → SQR(activate(X))
SQR(s(X)) → ACTIVATE(X)
SQR(s(X)) → DBL(activate(X))
DBL(s(X)) → S(n__s(n__dbl(activate(X))))
DBL(s(X)) → ACTIVATE(X)
ADD(s(X), Y) → S(n__add(activate(X), Y))
ADD(s(X), Y) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__terms(X)) → TERMS(X)
ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)
ACTIVATE(n__s(X)) → S(X)
ACTIVATE(n__dbl(X)) → DBL(X)
ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(n__add(sqr(activate(X)), dbl(activate(X))))
dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(activate(X), activate(Z)))
terms(X) → n__terms(X)
add(X1, X2) → n__add(X1, X2)
s(X) → n__s(X)
dbl(X) → n__dbl(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 5 less nodes.

(4) Obligation:

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

SQR(s(X)) → SQR(activate(X))
SQR(s(X)) → ACTIVATE(X)
ACTIVATE(n__terms(X)) → TERMS(X)
TERMS(N) → SQR(N)
SQR(s(X)) → DBL(activate(X))
DBL(s(X)) → ACTIVATE(X)
ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)
ADD(s(X), Y) → ACTIVATE(X)
ACTIVATE(n__dbl(X)) → DBL(X)
ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(n__add(sqr(activate(X)), dbl(activate(X))))
dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(activate(X), activate(Z)))
terms(X) → n__terms(X)
add(X1, X2) → n__add(X1, X2)
s(X) → n__s(X)
dbl(X) → n__dbl(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X

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

(5) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

ACTIVATE(n__dbl(X)) → DBL(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(SQR(x₁)) = 4A + 3A · x₁

POL(s(x₁)) = -I + 0A · x₁

POL(activate(x₁)) = 0A + 0A · x₁

POL(ACTIVATE(x₁)) = 4A + 1A · x₁

POL(n__terms(x₁)) = -I + 5A · x₁

POL(TERMS(x₁)) = 4A + 5A · x₁

POL(DBL(x₁)) = 4A + 3A · x₁

POL(n__add(x₁, x₂)) = 4A + 0A · x₁ + 0A · x₂

POL(ADD(x₁, x₂)) = 5A + 1A · x₁ + 1A · x₂

POL(n__dbl(x₁)) = 5A + 5A · x₁

POL(n__first(x₁, x₂)) = 2A + 2A · x₁ + 2A · x₂

POL(FIRST(x₁, x₂)) = 4A + 3A · x₁ + 1A · x₂

POL(cons(x₁, x₂)) = -I + -I · x₁ + 0A · x₂

POL(terms(x₁)) = -I + 5A · x₁

POL(add(x₁, x₂)) = 4A + 0A · x₁ + 0A · x₂

POL(n__s(x₁)) = -I + 0A · x₁

POL(dbl(x₁)) = 5A + 5A · x₁

POL(first(x₁, x₂)) = 2A + 2A · x₁ + 2A · x₂

POL(sqr(x₁)) = 5A + 5A · x₁

POL(0) = 5A

POL(recip(x₁)) = 1A + 0A · x₁

POL(nil) = 4A

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

activate(n__terms(X)) → terms(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X
sqr(s(X)) → s(n__add(sqr(activate(X)), dbl(activate(X))))
sqr(0) → 0
dbl(0) → 0
dbl(X) → n__dbl(X)
s(X) → n__s(X)
terms(X) → n__terms(X)
terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
add(0, X) → X
add(X1, X2) → n__add(X1, X2)
add(s(X), Y) → s(n__add(activate(X), Y))
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
first(0, X) → nil
first(X1, X2) → n__first(X1, X2)
first(s(X), cons(Y, Z)) → cons(Y, n__first(activate(X), activate(Z)))

(6) Obligation:

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

SQR(s(X)) → SQR(activate(X))
SQR(s(X)) → ACTIVATE(X)
ACTIVATE(n__terms(X)) → TERMS(X)
TERMS(N) → SQR(N)
SQR(s(X)) → DBL(activate(X))
DBL(s(X)) → ACTIVATE(X)
ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)
ADD(s(X), Y) → ACTIVATE(X)
ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(n__add(sqr(activate(X)), dbl(activate(X))))
dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(activate(X), activate(Z)))
terms(X) → n__terms(X)
add(X1, X2) → n__add(X1, X2)
s(X) → n__s(X)
dbl(X) → n__dbl(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X

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

(7) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

ACTIVATE(n__terms(X)) → TERMS(X)
ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)

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

POL( DBL(x₁) ) = x₁ + 2

POL( SQR(x₁) ) = 2x₁ + 2

POL( activate(x₁) ) = x₁

POL( n__terms(x₁) ) = 2x₁ + 2

POL( terms(x₁) ) = 2x₁ + 2

POL( n__add(x₁, x₂) ) = x₁ + x₂

POL( add(x₁, x₂) ) = x₁ + x₂

POL( n__s(x₁) ) = x₁

POL( s(x₁) ) = x₁

POL( n__dbl(x₁) ) = max{0, -2}

POL( dbl(x₁) ) = 0

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

POL( first(x₁, x₂) ) = x₁ + 2x₂ + 2

POL( sqr(x₁) ) = 2x₁ + 1

POL( cons(x₁, x₂) ) = x₁ + x₂

POL( recip(x₁) ) = max{0, -2}

POL( 0 ) = 0

POL( nil ) = 2

POL( ACTIVATE(x₁) ) = x₁ + 2

POL( TERMS(x₁) ) = 2x₁ + 2

POL( ADD(x₁, x₂) ) = x₁ + 2

POL( FIRST(x₁, x₂) ) = x₁ + 2x₂ + 2

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

activate(n__terms(X)) → terms(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X
sqr(s(X)) → s(n__add(sqr(activate(X)), dbl(activate(X))))
sqr(0) → 0
dbl(0) → 0
dbl(X) → n__dbl(X)
s(X) → n__s(X)
terms(X) → n__terms(X)
terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
add(0, X) → X
add(X1, X2) → n__add(X1, X2)
add(s(X), Y) → s(n__add(activate(X), Y))
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
first(0, X) → nil
first(X1, X2) → n__first(X1, X2)
first(s(X), cons(Y, Z)) → cons(Y, n__first(activate(X), activate(Z)))

(8) Obligation:

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

SQR(s(X)) → SQR(activate(X))
SQR(s(X)) → ACTIVATE(X)
TERMS(N) → SQR(N)
SQR(s(X)) → DBL(activate(X))
DBL(s(X)) → ACTIVATE(X)
ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)
ADD(s(X), Y) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(n__add(sqr(activate(X)), dbl(activate(X))))
dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(activate(X), activate(Z)))
terms(X) → n__terms(X)
add(X1, X2) → n__add(X1, X2)
s(X) → n__s(X)
dbl(X) → n__dbl(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → 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 2 SCCs with 6 less nodes.

(10) Complex Obligation (AND)

(11) Obligation:

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

ADD(s(X), Y) → ACTIVATE(X)
ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(n__add(sqr(activate(X)), dbl(activate(X))))
dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(activate(X), activate(Z)))
terms(X) → n__terms(X)
add(X1, X2) → n__add(X1, X2)
s(X) → n__s(X)
dbl(X) → n__dbl(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X

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

(12) 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.

(13) Obligation:

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

ADD(s(X), Y) → ACTIVATE(X)
ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)

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

(14) 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:

ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)
The graph contains the following edges 1 > 1, 1 > 2
ADD(s(X), Y) → ACTIVATE(X)
The graph contains the following edges 1 > 1

(15) YES

(16) Obligation:

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

SQR(s(X)) → SQR(activate(X))

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(n__add(sqr(activate(X)), dbl(activate(X))))
dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(activate(X), activate(Z)))
terms(X) → n__terms(X)
add(X1, X2) → n__add(X1, X2)
s(X) → n__s(X)
dbl(X) → n__dbl(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → 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.

SQR(s(X)) → SQR(activate(X))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SQR(x1) = SQR(x1)
s(x1) = s(x1)
activate(x1) = x1
n__terms(x1) = n__terms
terms(x1) = terms
n__add(x1, x2) = n__add(x1, x2)
add(x1, x2) = add(x1, x2)
n__s(x1) = n__s(x1)
n__dbl(x1) = n__dbl(x1)
dbl(x1) = dbl(x1)
n__first(x1, x2) = n__first(x1)
first(x1, x2) = first(x1)
sqr(x1) = sqr(x1)
0 = 0
cons(x1, x2) = x2
recip(x1) = recip
nil = nil

Recursive path order with status [RPO].
Quasi-Precedence:

SQR₁ > [s₁, n_{_s₁}, recip]
[n_{_terms}, terms] > [s₁, n_{_s₁}, recip]
[n_{_dbl₁}, dbl₁, sqr₁] > [n_{_add₂}, add₂] > [s₁, n_{_s₁}, recip]
[n_{_dbl₁}, dbl₁, sqr₁] > 0 > nil > [s₁, n_{_s₁}, recip]
[n_{_first₁}, first₁] > [s₁, n_{_s₁}, recip]

Status:

SQR₁: [1]
s₁: multiset
n_{_terms}: []
terms: []
n_{_add₂}: multiset
add₂: multiset
n_{_s₁}: multiset
n_{_dbl₁}: [1]
dbl₁: [1]
n_{_first₁}: [1]
first₁: [1]
sqr₁: [1]
0: multiset
recip: multiset
nil: multiset

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

activate(n__terms(X)) → terms(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X
sqr(s(X)) → s(n__add(sqr(activate(X)), dbl(activate(X))))
sqr(0) → 0
dbl(0) → 0
dbl(X) → n__dbl(X)
s(X) → n__s(X)
terms(X) → n__terms(X)
terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
add(0, X) → X
add(X1, X2) → n__add(X1, X2)
add(s(X), Y) → s(n__add(activate(X), Y))
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
first(0, X) → nil
first(X1, X2) → n__first(X1, X2)
first(s(X), cons(Y, Z)) → cons(Y, n__first(activate(X), activate(Z)))

(18) Obligation:

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

terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(n__add(sqr(activate(X)), dbl(activate(X))))
dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(activate(X), activate(Z)))
terms(X) → n__terms(X)
add(X1, X2) → n__add(X1, X2)
s(X) → n__s(X)
dbl(X) → n__dbl(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X

Q is empty.
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) DependencyPairsProof (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyGraphProof (EQUIVALENT transformation)

(4) Obligation:

(5) QDPOrderProof (EQUIVALENT transformation)

(6) Obligation:

(7) QDPOrderProof (EQUIVALENT transformation)

(8) Obligation:

(9) DependencyGraphProof (EQUIVALENT transformation)

(10) Complex Obligation (AND)

(11) Obligation:

(12) UsableRulesProof (EQUIVALENT transformation)

(13) Obligation:

(14) QDPSizeChangeProof (EQUIVALENT transformation)

(15) YES

(16) Obligation:

(17) QDPOrderProof (EQUIVALENT transformation)

(18) Obligation:

(19) PisEmptyProof (EQUIVALENT transformation)

(20) YES