Termination w.r.t. Q proof of TRS/Cimedpqs.trs

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:

f₁(f₁(x)) -> f₁(c₁(f₁(x)))
f₁(f₁(x)) -> f₁(d₁(f₁(x)))
g₁(c₁(x)) -> x
g₁(d₁(x)) -> x
g₁(c₁(0)) -> g₁(d₁(1))
g₁(c₁(1)) -> g₁(d₁(0))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₁(f₁(x)) -> f₁(c₁(f₁(x)))
f₁(f₁(x)) -> f₁(d₁(f₁(x)))
g₁(c₁(x)) -> x
g₁(d₁(x)) -> x
g₁(c₁(0)) -> g₁(d₁(1))
g₁(c₁(1)) -> g₁(d₁(0))

Q is empty.

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

F₁(f₁(x)) -> F₁(c₁(f₁(x)))
G₁(c₁(1)) -> G₁(d₁(0))
F₁(f₁(x)) -> F₁(d₁(f₁(x)))
G₁(c₁(0)) -> G₁(d₁(1))

The TRS R consists of the following rules:

f₁(f₁(x)) -> f₁(c₁(f₁(x)))
f₁(f₁(x)) -> f₁(d₁(f₁(x)))
g₁(c₁(x)) -> x
g₁(d₁(x)) -> x
g₁(c₁(0)) -> g₁(d₁(1))
g₁(c₁(1)) -> g₁(d₁(0))

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:

F₁(f₁(x)) -> F₁(c₁(f₁(x)))
G₁(c₁(1)) -> G₁(d₁(0))
F₁(f₁(x)) -> F₁(d₁(f₁(x)))
G₁(c₁(0)) -> G₁(d₁(1))

The TRS R consists of the following rules:

f₁(f₁(x)) -> f₁(c₁(f₁(x)))
f₁(f₁(x)) -> f₁(d₁(f₁(x)))
g₁(c₁(x)) -> x
g₁(d₁(x)) -> x
g₁(c₁(0)) -> g₁(d₁(1))
g₁(c₁(1)) -> g₁(d₁(0))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 4 less nodes.