Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar3/nonterminSLASHcaribooSLASHex2.trs

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.