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))


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.