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 Order [17,21] with Interpretation:

POL( C₁(x₁) ) = max{0, x₁ - 3}

POL( c₁(x₁) ) = 2x₁

POL( a₂(x₁, x₂) ) = x₁ + 2x₂ + 1

POL( 0 ) = 0

POL( b₂(x₁, x₂) ) = x₁ + 3

The following usable rules [14] were oriented:

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

↳ 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.