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

(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) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Polynomial interpretation [POLO]:

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

From the DPs we obtained the following set of size-change graphs:

• A__FST(s(X), cons(Y, Z)) → MARK(Y) (allowed arguments on rhs = {1})
  The graph contains the following edges 2 > 1

• MARK(fst(X1, X2)) → A__FST(mark(X1), mark(X2)) (allowed arguments on rhs = {2})
  The graph contains the following edges 1 > 2

• A__FROM(X) → MARK(X) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 >= 1

• A__ADD(0, X) → MARK(X) (allowed arguments on rhs = {1})
  The graph contains the following edges 2 >= 1

• MARK(from(X)) → A__FROM(mark(X)) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

• MARK(add(X1, X2)) → A__ADD(mark(X1), mark(X2)) (allowed arguments on rhs = {2})
  The graph contains the following edges 1 > 2

• MARK(fst(X1, X2)) → MARK(X1) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

• MARK(fst(X1, X2)) → MARK(X2) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

• MARK(from(X)) → MARK(X) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

• MARK(add(X1, X2)) → MARK(X1) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

• MARK(add(X1, X2)) → MARK(X2) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

• MARK(len(X)) → MARK(X) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

• MARK(cons(X1, X2)) → MARK(X1) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

We oriented the following set of usable rules [AAECC05,FROCOS05].

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

(6) TRUE