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:

s1(s1(s0(s0(x)))) → s0(s0(s0(s1(s1(s1(x))))))

Q is empty.


QTRS
  ↳ Overlay + Local Confluence

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

s1(s1(s0(s0(x)))) → s0(s0(s0(s1(s1(s1(x))))))

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
QTRS
      ↳ DependencyPairsProof

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

s1(s1(s0(s0(x)))) → s0(s0(s0(s1(s1(s1(x))))))

The set Q consists of the following terms:

s1(s1(s0(s0(x0))))


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

S1(s1(s0(s0(x)))) → S1(x)
S1(s1(s0(s0(x)))) → S1(s1(x))
S1(s1(s0(s0(x)))) → S1(s1(s1(x)))

The TRS R consists of the following rules:

s1(s1(s0(s0(x)))) → s0(s0(s0(s1(s1(s1(x))))))

The set Q consists of the following terms:

s1(s1(s0(s0(x0))))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
QDP

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

S1(s1(s0(s0(x)))) → S1(x)
S1(s1(s0(s0(x)))) → S1(s1(x))
S1(s1(s0(s0(x)))) → S1(s1(s1(x)))

The TRS R consists of the following rules:

s1(s1(s0(s0(x)))) → s0(s0(s0(s1(s1(s1(x))))))

The set Q consists of the following terms:

s1(s1(s0(s0(x0))))

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