Termination w.r.t. Q proof of TRS/secret06jambox7.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:

c₁(c₁(c₁(a₂(x, y)))) -> b₂(c₁(c₁(c₁(c₁(y)))), x)
c₁(c₁(b₂(c₁(y), 0))) -> a₂(0, c₁(c₁(a₂(y, 0))))
c₁(c₁(a₂(a₂(y, 0), x))) -> c₁(y)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

c₁(c₁(c₁(a₂(x, y)))) -> b₂(c₁(c₁(c₁(c₁(y)))), x)
c₁(c₁(b₂(c₁(y), 0))) -> a₂(0, c₁(c₁(a₂(y, 0))))
c₁(c₁(a₂(a₂(y, 0), x))) -> c₁(y)

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:

C₁(c₁(c₁(a₂(x, y)))) -> C₁(c₁(c₁(c₁(y))))
C₁(c₁(c₁(a₂(x, y)))) -> C₁(y)
C₁(c₁(b₂(c₁(y), 0))) -> C₁(a₂(y, 0))
C₁(c₁(c₁(a₂(x, y)))) -> C₁(c₁(c₁(y)))
C₁(c₁(a₂(a₂(y, 0), x))) -> C₁(y)
C₁(c₁(c₁(a₂(x, y)))) -> C₁(c₁(y))
C₁(c₁(b₂(c₁(y), 0))) -> C₁(c₁(a₂(y, 0)))

The TRS R consists of the following rules:

c₁(c₁(c₁(a₂(x, y)))) -> b₂(c₁(c₁(c₁(c₁(y)))), x)
c₁(c₁(b₂(c₁(y), 0))) -> a₂(0, c₁(c₁(a₂(y, 0))))
c₁(c₁(a₂(a₂(y, 0), x))) -> c₁(y)

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:

C₁(c₁(c₁(a₂(x, y)))) -> C₁(c₁(c₁(c₁(y))))
C₁(c₁(c₁(a₂(x, y)))) -> C₁(y)
C₁(c₁(b₂(c₁(y), 0))) -> C₁(a₂(y, 0))
C₁(c₁(c₁(a₂(x, y)))) -> C₁(c₁(c₁(y)))
C₁(c₁(a₂(a₂(y, 0), x))) -> C₁(y)
C₁(c₁(c₁(a₂(x, y)))) -> C₁(c₁(y))
C₁(c₁(b₂(c₁(y), 0))) -> C₁(c₁(a₂(y, 0)))

The TRS R consists of the following rules:

c₁(c₁(c₁(a₂(x, y)))) -> b₂(c₁(c₁(c₁(c₁(y)))), x)
c₁(c₁(b₂(c₁(y), 0))) -> a₂(0, c₁(c₁(a₂(y, 0))))
c₁(c₁(a₂(a₂(y, 0), x))) -> c₁(y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

C₁(c₁(c₁(a₂(x, y)))) -> C₁(c₁(c₁(c₁(y))))
C₁(c₁(c₁(a₂(x, y)))) -> C₁(y)
C₁(c₁(c₁(a₂(x, y)))) -> C₁(c₁(c₁(y)))
C₁(c₁(a₂(a₂(y, 0), x))) -> C₁(y)
C₁(c₁(c₁(a₂(x, y)))) -> C₁(c₁(y))
C₁(c₁(b₂(c₁(y), 0))) -> C₁(c₁(a₂(y, 0)))

The TRS R consists of the following rules:

c₁(c₁(c₁(a₂(x, y)))) -> b₂(c₁(c₁(c₁(c₁(y)))), x)
c₁(c₁(b₂(c₁(y), 0))) -> a₂(0, c₁(c₁(a₂(y, 0))))
c₁(c₁(a₂(a₂(y, 0), x))) -> c₁(y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

C₁(c₁(c₁(a₂(x, y)))) -> C₁(c₁(c₁(c₁(y))))
C₁(c₁(c₁(a₂(x, y)))) -> C₁(y)
C₁(c₁(c₁(a₂(x, y)))) -> C₁(c₁(c₁(y)))
C₁(c₁(a₂(a₂(y, 0), x))) -> C₁(y)
C₁(c₁(c₁(a₂(x, y)))) -> C₁(c₁(y))
C₁(c₁(b₂(c₁(y), 0))) -> C₁(c₁(a₂(y, 0)))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(C₁(x₁)) = 4 + x₁
POL(a₂(x₁, x₂)) = 1 + x₁ + 2·x₂
POL(b₂(x₁, x₂)) = 3 + x₁
POL(c₁(x₁)) = 2·x₁

The following usable rules [14] were oriented:

c₁(c₁(b₂(c₁(y), 0))) -> a₂(0, c₁(c₁(a₂(y, 0))))
c₁(c₁(a₂(a₂(y, 0), x))) -> c₁(y)
c₁(c₁(c₁(a₂(x, y)))) -> b₂(c₁(c₁(c₁(c₁(y)))), x)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

c₁(c₁(c₁(a₂(x, y)))) -> b₂(c₁(c₁(c₁(c₁(y)))), x)
c₁(c₁(b₂(c₁(y), 0))) -> a₂(0, c₁(c₁(a₂(y, 0))))
c₁(c₁(a₂(a₂(y, 0), x))) -> c₁(y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.