Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

cons2(x, cons2(y, z)) -> big
inf1(x) -> cons2(x, inf1(s1(x)))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

cons2(x, cons2(y, z)) -> big
inf1(x) -> cons2(x, inf1(s1(x)))

Q is empty.

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

INF1(x) -> CONS2(x, inf1(s1(x)))
INF1(x) -> INF1(s1(x))

The TRS R consists of the following rules:

cons2(x, cons2(y, z)) -> big
inf1(x) -> cons2(x, inf1(s1(x)))

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ DependencyGraphProof

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

INF1(x) -> CONS2(x, inf1(s1(x)))
INF1(x) -> INF1(s1(x))

The TRS R consists of the following rules:

cons2(x, cons2(y, z)) -> big
inf1(x) -> cons2(x, inf1(s1(x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP

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

INF1(x) -> INF1(s1(x))

The TRS R consists of the following rules:

cons2(x, cons2(y, z)) -> big
inf1(x) -> cons2(x, inf1(s1(x)))

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