Termination w.r.t. Q proof of Secret_06_TRS

Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS

↳1 DependencyPairsProof (⇔, 8 ms)

↳2 QDP

↳3 DependencyGraphProof (⇔, 0 ms)

↳4 QDP

↳5 TransformationProof (⇔, 0 ms)

↳6 QDP

↳7 DependencyGraphProof (⇔, 0 ms)

↳8 QDP

↳9 TransformationProof (⇔, 0 ms)

↳10 QDP

↳11 QDPOrderProof (⇔, 27 ms)

↳12 QDP

↳13 QDPOrderProof (⇔, 131 ms)

↳14 QDP

↳15 DependencyGraphProof (⇔, 0 ms)

↳16 TRUE

(0) Obligation:

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

c(c(c(y))) → c(c(a(y, 0)))
c(a(a(0, x), y)) → a(c(c(c(0))), y)
c(y) → y

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

C(c(c(y))) → C(c(a(y, 0)))
C(c(c(y))) → C(a(y, 0))
C(a(a(0, x), y)) → C(c(c(0)))
C(a(a(0, x), y)) → C(c(0))
C(a(a(0, x), y)) → C(0)

The TRS R consists of the following rules:

c(c(c(y))) → c(c(a(y, 0)))
c(a(a(0, x), y)) → a(c(c(c(0))), y)
c(y) → y

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(4) Obligation:

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

C(c(c(y))) → C(a(y, 0))
C(a(a(0, x), y)) → C(c(c(0)))
C(c(c(y))) → C(c(a(y, 0)))
C(a(a(0, x), y)) → C(c(0))

The TRS R consists of the following rules:

c(c(c(y))) → c(c(a(y, 0)))
c(a(a(0, x), y)) → a(c(c(c(0))), y)
c(y) → y

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

(5) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule C(a(a(0, x), y)) → C(c(0)) at position [0] we obtained the following new rules [LPAR04]:

C(a(a(0, y0), y1)) → C(0) → C(a(a(0, y0), y1)) → C(0)

(6) Obligation:

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

C(c(c(y))) → C(a(y, 0))
C(a(a(0, x), y)) → C(c(c(0)))
C(c(c(y))) → C(c(a(y, 0)))
C(a(a(0, y0), y1)) → C(0)

The TRS R consists of the following rules:

c(c(c(y))) → c(c(a(y, 0)))
c(a(a(0, x), y)) → a(c(c(c(0))), y)
c(y) → y

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(8) Obligation:

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

C(a(a(0, x), y)) → C(c(c(0)))
C(c(c(y))) → C(c(a(y, 0)))
C(c(c(y))) → C(a(y, 0))

The TRS R consists of the following rules:

c(c(c(y))) → c(c(a(y, 0)))
c(a(a(0, x), y)) → a(c(c(c(0))), y)
c(y) → y

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

(9) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule C(c(c(y))) → C(c(a(y, 0))) at position [0] we obtained the following new rules [LPAR04]:

C(c(c(a(0, x0)))) → C(a(c(c(c(0))), 0)) → C(c(c(a(0, x0)))) → C(a(c(c(c(0))), 0))
C(c(c(y0))) → C(a(y0, 0)) → C(c(c(y))) → C(a(y, 0))

(10) Obligation:

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

C(a(a(0, x), y)) → C(c(c(0)))
C(c(c(y))) → C(a(y, 0))
C(c(c(a(0, x0)))) → C(a(c(c(c(0))), 0))

The TRS R consists of the following rules:

c(c(c(y))) → c(c(a(y, 0)))
c(a(a(0, x), y)) → a(c(c(c(0))), y)
c(y) → y

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

C(c(c(a(0, x0)))) → C(a(c(c(c(0))), 0))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( C(x₁) ) = max{0, x₁ - 2}

POL( c(x₁) ) = x₁ + 1

POL( a(x₁, x₂) ) = 1

POL( 0 ) = 0

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

c(y) → y
c(c(c(y))) → c(c(a(y, 0)))
c(a(a(0, x), y)) → a(c(c(c(0))), y)

(12) Obligation:

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

C(a(a(0, x), y)) → C(c(c(0)))
C(c(c(y))) → C(a(y, 0))

The TRS R consists of the following rules:

c(c(c(y))) → c(c(a(y, 0)))
c(a(a(0, x), y)) → a(c(c(c(0))), y)
c(y) → y

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

C(a(a(0, x), y)) → C(c(c(0)))

The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(C(x₁)) = 0A +
[ -I, 0A, 0A ]

· x₁

POL(a(x₁, x₂)) =
/ -I \

| -I |

\ 0A /

+
/ -I 0A 1A \

| -I 0A -I |

\ 0A 0A 1A /

· x₁ +
/ 0A 3A -I \

| -I -I -I |

\ -I 2A -I /

· x₂

POL(0) =
/ 0A \

| -I |

\ -I /

POL(c(x₁)) =
/ -I \

| -I |

\ 0A /

+
/ 0A 0A 0A \

| -I 1A 0A |

\ 0A 0A 0A /

· x₁

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

c(y) → y
c(c(c(y))) → c(c(a(y, 0)))
c(a(a(0, x), y)) → a(c(c(c(0))), y)

(14) Obligation:

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

C(c(c(y))) → C(a(y, 0))

The TRS R consists of the following rules:

c(c(c(y))) → c(c(a(y, 0)))
c(a(a(0, x), y)) → a(c(c(c(0))), y)
c(y) → y

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

(15) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(0) Obligation:

(1) DependencyPairsProof (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyGraphProof (EQUIVALENT transformation)

(4) Obligation:

(5) TransformationProof (EQUIVALENT transformation)

(6) Obligation:

(7) DependencyGraphProof (EQUIVALENT transformation)

(8) Obligation:

(9) TransformationProof (EQUIVALENT transformation)

(10) Obligation:

(11) QDPOrderProof (EQUIVALENT transformation)

(12) Obligation:

(13) QDPOrderProof (EQUIVALENT transformation)

(14) Obligation:

(15) DependencyGraphProof (EQUIVALENT transformation)

(16) TRUE