Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar14/TRCSRSLASHEx2_Luc03b

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

fst₂(0, Z) -> nil
fst₂(s, cons₁(Y)) -> cons₁(Y)
from₁(X) -> cons₁(X)
add₂(0, X) -> X
add₂(s, Y) -> s
len₁(nil) -> 0
len₁(cons₁(X)) -> s

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

fst₂(0, Z) -> nil
fst₂(s, cons₁(Y)) -> cons₁(Y)
from₁(X) -> cons₁(X)
add₂(0, X) -> X
add₂(s, Y) -> s
len₁(nil) -> 0
len₁(cons₁(X)) -> s

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

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

fst₂(0, Z) -> nil
fst₂(s, cons₁(Y)) -> cons₁(Y)
from₁(X) -> cons₁(X)
add₂(0, X) -> X
add₂(s, Y) -> s
len₁(nil) -> 0
len₁(cons₁(X)) -> s

The set Q consists of the following terms:

fst₂(0, x0)
fst₂(s, cons₁(x0))
from₁(x0)
add₂(0, x0)
add₂(s, x0)
len₁(nil)
len₁(cons₁(x0))

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

fst₂(0, Z) -> nil
fst₂(s, cons₁(Y)) -> cons₁(Y)
from₁(X) -> cons₁(X)
add₂(0, X) -> X
add₂(s, Y) -> s
len₁(nil) -> 0
len₁(cons₁(X)) -> s

The set Q consists of the following terms:

fst₂(0, x0)
fst₂(s, cons₁(x0))
from₁(x0)
add₂(0, x0)
add₂(s, x0)
len₁(nil)
len₁(cons₁(x0))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ PisEmptyProof

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

fst₂(0, Z) -> nil
fst₂(s, cons₁(Y)) -> cons₁(Y)
from₁(X) -> cons₁(X)
add₂(0, X) -> X
add₂(s, Y) -> s
len₁(nil) -> 0
len₁(cons₁(X)) -> s

The set Q consists of the following terms:

fst₂(0, x0)
fst₂(s, cons₁(x0))
from₁(x0)
add₂(0, x0)
add₂(s, x0)
len₁(nil)
len₁(cons₁(x0))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.