(0) Obligation:

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

a__fst(0, Z) → nil
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__len(nil) → 0
a__len(cons(X, Z)) → s(len(Z))
mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(len(X)) → a__len(mark(X))
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__fst(X1, X2) → fst(X1, X2)
a__from(X) → from(X)
a__add(X1, X2) → add(X1, X2)
a__len(X) → len(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:

A__FST(s(X), cons(Y, Z)) → MARK(Y)
A__FROM(X) → MARK(X)
A__ADD(0, X) → MARK(X)
MARK(fst(X1, X2)) → A__FST(mark(X1), mark(X2))
MARK(fst(X1, X2)) → MARK(X1)
MARK(fst(X1, X2)) → MARK(X2)
MARK(from(X)) → A__FROM(mark(X))
MARK(from(X)) → MARK(X)
MARK(add(X1, X2)) → A__ADD(mark(X1), mark(X2))
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
MARK(len(X)) → A__LEN(mark(X))
MARK(len(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__fst(0, Z) → nil
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__len(nil) → 0
a__len(cons(X, Z)) → s(len(Z))
mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(len(X)) → a__len(mark(X))
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__fst(X1, X2) → fst(X1, X2)
a__from(X) → from(X)
a__add(X1, X2) → add(X1, X2)
a__len(X) → len(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 1 less node.

(4) Obligation:

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

MARK(fst(X1, X2)) → A__FST(mark(X1), mark(X2))
A__FST(s(X), cons(Y, Z)) → MARK(Y)
MARK(fst(X1, X2)) → MARK(X1)
MARK(fst(X1, X2)) → MARK(X2)
MARK(from(X)) → A__FROM(mark(X))
A__FROM(X) → MARK(X)
MARK(from(X)) → MARK(X)
MARK(add(X1, X2)) → A__ADD(mark(X1), mark(X2))
A__ADD(0, X) → MARK(X)
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
MARK(len(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__fst(0, Z) → nil
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__len(nil) → 0
a__len(cons(X, Z)) → s(len(Z))
mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(len(X)) → a__len(mark(X))
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__fst(X1, X2) → fst(X1, X2)
a__from(X) → from(X)
a__add(X1, X2) → add(X1, X2)
a__len(X) → len(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].


The following pairs can be oriented strictly and are deleted.


MARK(add(X1, X2)) → A__ADD(mark(X1), mark(X2))
A__ADD(0, X) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
MARK(x0, x1)  =  MARK(x1)
A__FST(x0, x1, x2)  =  A__FST(x2)
A__FROM(x0, x1)  =  A__FROM(x1)
A__ADD(x0, x1, x2)  =  A__ADD(x0, x2)

Tags:
MARK has argument tags [1,2] and root tag 0
A__FST has argument tags [0,9,2] and root tag 0
A__FROM has argument tags [9,2] and root tag 0
A__ADD has argument tags [1,11,2] and root tag 2

Comparison: MIN
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(0) = 1   
POL(A__ADD(x1, x2)) = x1 + x2   
POL(A__FROM(x1)) = 1   
POL(A__FST(x1, x2)) = 1 + x1   
POL(MARK(x1)) = 1   
POL(a__add(x1, x2)) = x1 + x2   
POL(a__from(x1)) = x1   
POL(a__fst(x1, x2)) = x1 + x2   
POL(a__len(x1)) = x1   
POL(add(x1, x2)) = x1 + x2   
POL(cons(x1, x2)) = x1   
POL(from(x1)) = x1   
POL(fst(x1, x2)) = x1 + x2   
POL(len(x1)) = x1   
POL(mark(x1)) = x1   
POL(nil) = 1   
POL(s(x1)) = 0   

The following usable rules [FROCOS05] were oriented:

mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
a__add(0, X) → mark(X)
mark(len(X)) → a__len(mark(X))
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__from(X) → cons(mark(X), from(s(X)))
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))
a__from(X) → from(X)
a__fst(0, Z) → nil
a__fst(X1, X2) → fst(X1, X2)
a__add(s(X), Y) → s(add(X, Y))
a__add(X1, X2) → add(X1, X2)
a__len(nil) → 0
a__len(cons(X, Z)) → s(len(Z))
a__len(X) → len(X)

(6) Obligation:

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

MARK(fst(X1, X2)) → A__FST(mark(X1), mark(X2))
A__FST(s(X), cons(Y, Z)) → MARK(Y)
MARK(fst(X1, X2)) → MARK(X1)
MARK(fst(X1, X2)) → MARK(X2)
MARK(from(X)) → A__FROM(mark(X))
A__FROM(X) → MARK(X)
MARK(from(X)) → MARK(X)
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
MARK(len(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__fst(0, Z) → nil
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__len(nil) → 0
a__len(cons(X, Z)) → s(len(Z))
mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(len(X)) → a__len(mark(X))
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__fst(X1, X2) → fst(X1, X2)
a__from(X) → from(X)
a__add(X1, X2) → add(X1, X2)
a__len(X) → len(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].


The following pairs can be oriented strictly and are deleted.


MARK(len(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
MARK(x0, x1)  =  MARK(x1)
A__FST(x0, x1, x2)  =  A__FST(x0, x2)
A__FROM(x0, x1)  =  A__FROM(x0, x1)

Tags:
MARK has argument tags [5,2] and root tag 0
A__FST has argument tags [2,5,2] and root tag 0
A__FROM has argument tags [2,1] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(A__FROM(x1)) = x1   
POL(A__FST(x1, x2)) = 0   
POL(MARK(x1)) = 0   
POL(a__add(x1, x2)) = x1 + x2   
POL(a__from(x1)) = x1   
POL(a__fst(x1, x2)) = x1 + x2   
POL(a__len(x1)) = 1 + x1   
POL(add(x1, x2)) = x1 + x2   
POL(cons(x1, x2)) = x1   
POL(from(x1)) = x1   
POL(fst(x1, x2)) = x1 + x2   
POL(len(x1)) = 1 + x1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(s(x1)) = 0   

The following usable rules [FROCOS05] were oriented:

mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
a__add(0, X) → mark(X)
mark(len(X)) → a__len(mark(X))
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__from(X) → cons(mark(X), from(s(X)))
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))
a__from(X) → from(X)
a__fst(0, Z) → nil
a__fst(X1, X2) → fst(X1, X2)
a__add(s(X), Y) → s(add(X, Y))
a__add(X1, X2) → add(X1, X2)
a__len(nil) → 0
a__len(cons(X, Z)) → s(len(Z))
a__len(X) → len(X)

(8) Obligation:

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

MARK(fst(X1, X2)) → A__FST(mark(X1), mark(X2))
A__FST(s(X), cons(Y, Z)) → MARK(Y)
MARK(fst(X1, X2)) → MARK(X1)
MARK(fst(X1, X2)) → MARK(X2)
MARK(from(X)) → A__FROM(mark(X))
A__FROM(X) → MARK(X)
MARK(from(X)) → MARK(X)
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__fst(0, Z) → nil
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__len(nil) → 0
a__len(cons(X, Z)) → s(len(Z))
mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(len(X)) → a__len(mark(X))
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__fst(X1, X2) → fst(X1, X2)
a__from(X) → from(X)
a__add(X1, X2) → add(X1, X2)
a__len(X) → len(X)

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

(9) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
MARK(x0, x1)  =  MARK(x0, x1)
A__FST(x0, x1, x2)  =  A__FST(x0)
A__FROM(x0, x1)  =  A__FROM(x0, x1)

Tags:
MARK has argument tags [0,0] and root tag 0
A__FST has argument tags [0,2,4] and root tag 0
A__FROM has argument tags [0,4] and root tag 0

Comparison: MIN
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(A__FROM(x1)) = x1   
POL(A__FST(x1, x2)) = x1 + x2   
POL(MARK(x1)) = x1   
POL(a__add(x1, x2)) = 1 + x1 + x2   
POL(a__from(x1)) = x1   
POL(a__fst(x1, x2)) = x1 + x2   
POL(a__len(x1)) = 0   
POL(add(x1, x2)) = 1 + x1 + x2   
POL(cons(x1, x2)) = x1   
POL(from(x1)) = x1   
POL(fst(x1, x2)) = x1 + x2   
POL(len(x1)) = 0   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(s(x1)) = x1   

The following usable rules [FROCOS05] were oriented:

mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
a__add(0, X) → mark(X)
mark(len(X)) → a__len(mark(X))
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__from(X) → cons(mark(X), from(s(X)))
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))
a__from(X) → from(X)
a__fst(0, Z) → nil
a__fst(X1, X2) → fst(X1, X2)
a__add(s(X), Y) → s(add(X, Y))
a__add(X1, X2) → add(X1, X2)
a__len(nil) → 0
a__len(cons(X, Z)) → s(len(Z))
a__len(X) → len(X)

(10) Obligation:

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

MARK(fst(X1, X2)) → A__FST(mark(X1), mark(X2))
A__FST(s(X), cons(Y, Z)) → MARK(Y)
MARK(fst(X1, X2)) → MARK(X1)
MARK(fst(X1, X2)) → MARK(X2)
MARK(from(X)) → A__FROM(mark(X))
A__FROM(X) → MARK(X)
MARK(from(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__fst(0, Z) → nil
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__len(nil) → 0
a__len(cons(X, Z)) → s(len(Z))
mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(len(X)) → a__len(mark(X))
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__fst(X1, X2) → fst(X1, X2)
a__from(X) → from(X)
a__add(X1, X2) → add(X1, X2)
a__len(X) → len(X)

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(from(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
MARK(x0, x1)  =  MARK(x0, x1)
A__FST(x0, x1, x2)  =  A__FST(x1, x2)
A__FROM(x0, x1)  =  A__FROM(x0, x1)

Tags:
MARK has argument tags [2,4] and root tag 0
A__FST has argument tags [4,4,4] and root tag 0
A__FROM has argument tags [4,4] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(A__FROM(x1)) = 1   
POL(A__FST(x1, x2)) = 0   
POL(MARK(x1)) = 1   
POL(a__add(x1, x2)) = x1 + x2   
POL(a__from(x1)) = 1 + x1   
POL(a__fst(x1, x2)) = x1 + x2   
POL(a__len(x1)) = 1   
POL(add(x1, x2)) = x1 + x2   
POL(cons(x1, x2)) = x1   
POL(from(x1)) = 1 + x1   
POL(fst(x1, x2)) = x1 + x2   
POL(len(x1)) = 1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(s(x1)) = 1   

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

MARK(fst(X1, X2)) → A__FST(mark(X1), mark(X2))
A__FST(s(X), cons(Y, Z)) → MARK(Y)
MARK(fst(X1, X2)) → MARK(X1)
MARK(fst(X1, X2)) → MARK(X2)
MARK(from(X)) → A__FROM(mark(X))
A__FROM(X) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__fst(0, Z) → nil
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__len(nil) → 0
a__len(cons(X, Z)) → s(len(Z))
mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(len(X)) → a__len(mark(X))
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__fst(X1, X2) → fst(X1, X2)
a__from(X) → from(X)
a__add(X1, X2) → add(X1, X2)
a__len(X) → len(X)

Q is empty.
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.


MARK(fst(X1, X2)) → A__FST(mark(X1), mark(X2))
A__FST(s(X), cons(Y, Z)) → MARK(Y)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
MARK(x0, x1)  =  MARK(x1)
A__FST(x0, x1, x2)  =  A__FST(x0)
A__FROM(x0, x1)  =  A__FROM(x0, x1)

Tags:
MARK has argument tags [7,0] and root tag 2
A__FST has argument tags [0,6,0] and root tag 1
A__FROM has argument tags [0,2] and root tag 2

Comparison: MIN
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(0) = 1   
POL(A__FROM(x1)) = x1   
POL(A__FST(x1, x2)) = x1 + x2   
POL(MARK(x1)) = 0   
POL(a__add(x1, x2)) = 1 + x2   
POL(a__from(x1)) = x1   
POL(a__fst(x1, x2)) = x1 + x2   
POL(a__len(x1)) = 1   
POL(add(x1, x2)) = 1 + x2   
POL(cons(x1, x2)) = x1   
POL(from(x1)) = x1   
POL(fst(x1, x2)) = x1 + x2   
POL(len(x1)) = 1   
POL(mark(x1)) = x1   
POL(nil) = 1   
POL(s(x1)) = 1   

The following usable rules [FROCOS05] were oriented:

mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
a__add(0, X) → mark(X)
mark(len(X)) → a__len(mark(X))
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__from(X) → cons(mark(X), from(s(X)))
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))
a__from(X) → from(X)
a__fst(0, Z) → nil
a__fst(X1, X2) → fst(X1, X2)
a__add(s(X), Y) → s(add(X, Y))
a__add(X1, X2) → add(X1, X2)
a__len(nil) → 0
a__len(cons(X, Z)) → s(len(Z))
a__len(X) → len(X)

(14) Obligation:

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

MARK(fst(X1, X2)) → MARK(X1)
MARK(fst(X1, X2)) → MARK(X2)
MARK(from(X)) → A__FROM(mark(X))
A__FROM(X) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__fst(0, Z) → nil
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__len(nil) → 0
a__len(cons(X, Z)) → s(len(Z))
mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(len(X)) → a__len(mark(X))
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__fst(X1, X2) → fst(X1, X2)
a__from(X) → from(X)
a__add(X1, X2) → add(X1, X2)
a__len(X) → len(X)

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

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


A__FROM(X) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
MARK(x0, x1)  =  MARK(x1)
A__FROM(x0, x1)  =  A__FROM(x0)

Tags:
MARK has argument tags [1,1] and root tag 0
A__FROM has argument tags [1,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(A__FROM(x1)) = 1 + x1   
POL(MARK(x1)) = 0   
POL(a__add(x1, x2)) = x2   
POL(a__from(x1)) = 1 + x1   
POL(a__fst(x1, x2)) = x1 + x2   
POL(a__len(x1)) = 0   
POL(add(x1, x2)) = x2   
POL(cons(x1, x2)) = x1   
POL(from(x1)) = 1 + x1   
POL(fst(x1, x2)) = x1 + x2   
POL(len(x1)) = 0   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(s(x1)) = 0   

The following usable rules [FROCOS05] were oriented:

mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
a__add(0, X) → mark(X)
mark(len(X)) → a__len(mark(X))
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__from(X) → cons(mark(X), from(s(X)))
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))
a__from(X) → from(X)
a__fst(0, Z) → nil
a__fst(X1, X2) → fst(X1, X2)
a__add(s(X), Y) → s(add(X, Y))
a__add(X1, X2) → add(X1, X2)
a__len(nil) → 0
a__len(cons(X, Z)) → s(len(Z))
a__len(X) → len(X)

(16) Obligation:

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

MARK(fst(X1, X2)) → MARK(X1)
MARK(fst(X1, X2)) → MARK(X2)
MARK(from(X)) → A__FROM(mark(X))
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__fst(0, Z) → nil
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__len(nil) → 0
a__len(cons(X, Z)) → s(len(Z))
mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(len(X)) → a__len(mark(X))
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__fst(X1, X2) → fst(X1, X2)
a__from(X) → from(X)
a__add(X1, X2) → add(X1, X2)
a__len(X) → len(X)

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

(17) DependencyGraphProof (EQUIVALENT transformation)

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

(18) Obligation:

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

MARK(fst(X1, X2)) → MARK(X2)
MARK(fst(X1, X2)) → MARK(X1)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__fst(0, Z) → nil
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__len(nil) → 0
a__len(cons(X, Z)) → s(len(Z))
mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(len(X)) → a__len(mark(X))
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__fst(X1, X2) → fst(X1, X2)
a__from(X) → from(X)
a__add(X1, X2) → add(X1, X2)
a__len(X) → len(X)

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

(19) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(cons(X1, X2)) → MARK(X1)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
MARK(x0, x1)  =  MARK(x0)

Tags:
MARK has argument tags [0,1] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(MARK(x1)) = x1   
POL(cons(x1, x2)) = 1 + x1   
POL(fst(x1, x2)) = x1 + x2   

The following usable rules [FROCOS05] were oriented: none

(20) Obligation:

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

MARK(fst(X1, X2)) → MARK(X2)
MARK(fst(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__fst(0, Z) → nil
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__len(nil) → 0
a__len(cons(X, Z)) → s(len(Z))
mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(len(X)) → a__len(mark(X))
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__fst(X1, X2) → fst(X1, X2)
a__from(X) → from(X)
a__add(X1, X2) → add(X1, X2)
a__len(X) → len(X)

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

(21) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(fst(X1, X2)) → MARK(X2)
MARK(fst(X1, X2)) → MARK(X1)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
MARK(x0, x1)  =  MARK(x1)

Tags:
MARK has argument tags [1,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(MARK(x1)) = 1   
POL(fst(x1, x2)) = 1 + x1 + x2   

The following usable rules [FROCOS05] were oriented: none

(22) Obligation:

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

a__fst(0, Z) → nil
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__len(nil) → 0
a__len(cons(X, Z)) → s(len(Z))
mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(len(X)) → a__len(mark(X))
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__fst(X1, X2) → fst(X1, X2)
a__from(X) → from(X)
a__add(X1, X2) → add(X1, X2)
a__len(X) → len(X)

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

(23) PisEmptyProof (EQUIVALENT transformation)

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

(24) TRUE