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__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(0) → 0
mark(nil) → nil
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__first(X1, X2) → first(X1, X2)
a__from(X) → from(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__FIRST(s(X), cons(Y, Z)) → MARK(Y)
A__FROM(X) → MARK(X)
MARK(first(X1, X2)) → A__FIRST(mark(X1), mark(X2))
MARK(first(X1, X2)) → MARK(X1)
MARK(first(X1, X2)) → MARK(X2)
MARK(from(X)) → A__FROM(mark(X))
MARK(from(X)) → MARK(X)
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(0) → 0
mark(nil) → nil
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__first(X1, X2) → first(X1, X2)
a__from(X) → from(X)

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

(3) 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) = 0   
POL(a__first(x₁, x₂)) = 1 + x₁ + x₂   
POL(a__from(x₁)) = 1 + x₁   
POL(cons(x₁, x₂)) = 1 + x₁   
POL(first(x₁, x₂)) = 1 + x₁ + x₂   
POL(from(x₁)) = 1 + x₁   
POL(mark(x₁)) = x₁   
POL(nil) = 0   
POL(s(x₁)) = 1 + x₁   

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

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

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

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

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

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

• MARK(first(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(s(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(mark(X))
mark(nil) → nil
mark(from(X)) → a__from(mark(X))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
a__from(X) → from(X)
a__from(X) → cons(mark(X), from(s(X)))
a__first(X1, X2) → first(X1, X2)
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
a__first(0, X) → nil

(4) TRUE