Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Secret_06

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

0 QTRS

↳1 DependencyPairsProof (⇔, 24 ms)

↳2 QDP

↳3 QDPOrderProof (⇔, 49 ms)

↳4 QDP

↳5 MRRProof (⇔, 49 ms)

↳6 QDP

↳7 QDPOrderProof (⇔, 19 ms)

↳8 QDP

↳9 QDPOrderProof (⇔, 115 ms)

↳10 QDP

↳11 QDPOrderProof (⇔, 358 ms)

↳12 QDP

↳13 DependencyGraphProof (⇔, 0 ms)

↳14 TRUE

(0) Obligation:

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

c(b(a(a(x)))) → a(a(b(c(x))))
b(a(a(a(x)))) → a(a(a(b(x))))
a(b(c(x))) → c(b(a(x)))
c(c(b(b(x)))) → b(b(c(c(x))))

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(b(a(a(x)))) → A(a(b(c(x))))
C(b(a(a(x)))) → A(b(c(x)))
C(b(a(a(x)))) → B(c(x))
C(b(a(a(x)))) → C(x)
B(a(a(a(x)))) → A(a(a(b(x))))
B(a(a(a(x)))) → A(a(b(x)))
B(a(a(a(x)))) → A(b(x))
B(a(a(a(x)))) → B(x)
A(b(c(x))) → C(b(a(x)))
A(b(c(x))) → B(a(x))
A(b(c(x))) → A(x)
C(c(b(b(x)))) → B(b(c(c(x))))
C(c(b(b(x)))) → B(c(c(x)))
C(c(b(b(x)))) → C(c(x))
C(c(b(b(x)))) → C(x)

The TRS R consists of the following rules:

c(b(a(a(x)))) → a(a(b(c(x))))
b(a(a(a(x)))) → a(a(a(b(x))))
a(b(c(x))) → c(b(a(x)))
c(c(b(b(x)))) → b(b(c(c(x))))

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

(3) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

A(b(c(x))) → B(a(x))
A(b(c(x))) → A(x)
C(c(b(b(x)))) → C(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(A(x₁)) = x₁
POL(B(x₁)) = x₁
POL(C(x₁)) = 1 + x₁
POL(a(x₁)) = x₁
POL(b(x₁)) = x₁
POL(c(x₁)) = 1 + x₁

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

a(b(c(x))) → c(b(a(x)))
c(b(a(a(x)))) → a(a(b(c(x))))
c(c(b(b(x)))) → b(b(c(c(x))))
b(a(a(a(x)))) → a(a(a(b(x))))

(4) Obligation:

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

C(b(a(a(x)))) → A(a(b(c(x))))
C(b(a(a(x)))) → A(b(c(x)))
C(b(a(a(x)))) → B(c(x))
C(b(a(a(x)))) → C(x)
B(a(a(a(x)))) → A(a(a(b(x))))
B(a(a(a(x)))) → A(a(b(x)))
B(a(a(a(x)))) → A(b(x))
B(a(a(a(x)))) → B(x)
A(b(c(x))) → C(b(a(x)))
C(c(b(b(x)))) → B(b(c(c(x))))
C(c(b(b(x)))) → B(c(c(x)))
C(c(b(b(x)))) → C(c(x))

The TRS R consists of the following rules:

c(b(a(a(x)))) → a(a(b(c(x))))
b(a(a(a(x)))) → a(a(a(b(x))))
a(b(c(x))) → c(b(a(x)))
c(c(b(b(x)))) → b(b(c(c(x))))

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

(5) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

C(b(a(a(x)))) → C(x)
C(c(b(b(x)))) → B(c(c(x)))
C(c(b(b(x)))) → C(c(x))

Used ordering: Polynomial interpretation [POLO]:

POL(A(x₁)) = 1 + x₁
POL(B(x₁)) = 2 + 2·x₁
POL(C(x₁)) = 2 + 2·x₁
POL(a(x₁)) = x₁
POL(b(x₁)) = 1 + 2·x₁
POL(c(x₁)) = 1 + 2·x₁

(6) Obligation:

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

C(b(a(a(x)))) → A(a(b(c(x))))
C(b(a(a(x)))) → A(b(c(x)))
C(b(a(a(x)))) → B(c(x))
B(a(a(a(x)))) → A(a(a(b(x))))
B(a(a(a(x)))) → A(a(b(x)))
B(a(a(a(x)))) → A(b(x))
B(a(a(a(x)))) → B(x)
A(b(c(x))) → C(b(a(x)))
C(c(b(b(x)))) → B(b(c(c(x))))

The TRS R consists of the following rules:

c(b(a(a(x)))) → a(a(b(c(x))))
b(a(a(a(x)))) → a(a(a(b(x))))
a(b(c(x))) → c(b(a(x)))
c(c(b(b(x)))) → b(b(c(c(x))))

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

(7) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

C(b(a(a(x)))) → A(b(c(x)))
C(b(a(a(x)))) → B(c(x))
B(a(a(a(x)))) → A(a(b(x)))
B(a(a(a(x)))) → A(b(x))
B(a(a(a(x)))) → B(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(A(x₁)) = 1 + x₁
POL(B(x₁)) = x₁
POL(C(x₁)) = x₁
POL(a(x₁)) = 1 + x₁
POL(b(x₁)) = x₁
POL(c(x₁)) = x₁

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

a(b(c(x))) → c(b(a(x)))
c(b(a(a(x)))) → a(a(b(c(x))))
c(c(b(b(x)))) → b(b(c(c(x))))
b(a(a(a(x)))) → a(a(a(b(x))))

(8) Obligation:

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

C(b(a(a(x)))) → A(a(b(c(x))))
B(a(a(a(x)))) → A(a(a(b(x))))
A(b(c(x))) → C(b(a(x)))
C(c(b(b(x)))) → B(b(c(c(x))))

The TRS R consists of the following rules:

c(b(a(a(x)))) → a(a(b(c(x))))
b(a(a(a(x)))) → a(a(a(b(x))))
a(b(c(x))) → c(b(a(x)))
c(c(b(b(x)))) → b(b(c(c(x))))

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

(9) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

B(a(a(a(x)))) → A(a(a(b(x))))
C(c(b(b(x)))) → B(b(c(c(x))))

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

POL( A(x₁) ) = 0

POL( B(x₁) ) = 2

POL( C(x₁) ) = max{0, 2x₁ - 2}

POL( a(x₁) ) = 0

POL( b(x₁) ) = 2x₁ + 1

POL( c(x₁) ) = max{0, 2x₁ - 2}

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

a(b(c(x))) → c(b(a(x)))
c(b(a(a(x)))) → a(a(b(c(x))))
c(c(b(b(x)))) → b(b(c(c(x))))
b(a(a(a(x)))) → a(a(a(b(x))))

(10) Obligation:

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

C(b(a(a(x)))) → A(a(b(c(x))))
A(b(c(x))) → C(b(a(x)))

The TRS R consists of the following rules:

c(b(a(a(x)))) → a(a(b(c(x))))
b(a(a(a(x)))) → a(a(a(b(x))))
a(b(c(x))) → c(b(a(x)))
c(c(b(b(x)))) → b(b(c(c(x))))

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(b(a(a(x)))) → A(a(b(c(x))))

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

POL(C(x₁)) = 0A +
[ 0A, 0A, -I ]

· x₁

POL(b(x₁)) =
/ -I \

| -I |

\ -I /

+
/ 0A -I 0A \

| -I -I 0A |

\ -I 0A -I /

· x₁

POL(a(x₁)) =
/ 0A \

| 0A |

\ -I /

+
/ 1A 1A 0A \

| 0A 0A 1A |

\ 0A 0A 0A /

· x₁

POL(A(x₁)) = 0A +
[ -I, -I, 0A ]

· x₁

POL(c(x₁)) =
/ 0A \

| 0A |

\ -I /

+
/ 1A 1A 0A \

| 1A 1A 0A |

\ 0A 0A -I /

· x₁

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

a(b(c(x))) → c(b(a(x)))
c(b(a(a(x)))) → a(a(b(c(x))))
c(c(b(b(x)))) → b(b(c(c(x))))
b(a(a(a(x)))) → a(a(a(b(x))))

(12) Obligation:

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

A(b(c(x))) → C(b(a(x)))

The TRS R consists of the following rules:

c(b(a(a(x)))) → a(a(b(c(x))))
b(a(a(a(x)))) → a(a(a(b(x))))
a(b(c(x))) → c(b(a(x)))
c(c(b(b(x)))) → b(b(c(c(x))))

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

(13) 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) QDPOrderProof (EQUIVALENT transformation)

(4) Obligation:

(5) MRRProof (EQUIVALENT transformation)

(6) Obligation:

(7) QDPOrderProof (EQUIVALENT transformation)

(8) Obligation:

(9) QDPOrderProof (EQUIVALENT transformation)

(10) Obligation:

(11) QDPOrderProof (EQUIVALENT transformation)

(12) Obligation:

(13) DependencyGraphProof (EQUIVALENT transformation)

(14) TRUE