YES Termination w.r.t. Q proof of Secret_06_TRS_gen-28.ari

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

0 QTRS
↳1 DependencyPairsProof (⇔, 0 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 QDP
↳5 QDPOrderProof (⇔, 278 ms)
↳6 QDP
↳7 DependencyGraphProof (⇔, 0 ms)
↳8 QDP
↳9 TransformationProof (⇔, 0 ms)
↳10 QDP
↳11 DependencyGraphProof (⇔, 0 ms)
↳12 QDP
↳13 TransformationProof (⇔, 0 ms)
↳14 QDP
↳15 DependencyGraphProof (⇔, 0 ms)
↳16 TRUE

(0) Obligation:

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

b(y, z) → f(c(c(y, z, z), a, a))
b(b(z, y), a) → z
c(f(z), f(c(a, x, a)), y) → c(f(b(x, z)), c(z, y, a), a)

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:

B(y, z) → C(c(y, z, z), a, a)
B(y, z) → C(y, z, z)
C(f(z), f(c(a, x, a)), y) → C(f(b(x, z)), c(z, y, a), a)
C(f(z), f(c(a, x, a)), y) → B(x, z)
C(f(z), f(c(a, x, a)), y) → C(z, y, a)

The TRS R consists of the following rules:

b(y, z) → f(c(c(y, z, z), a, a))
b(b(z, y), a) → z
c(f(z), f(c(a, x, a)), y) → c(f(b(x, z)), c(z, y, a), a)

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:

B(y, z) → C(y, z, z)
C(f(z), f(c(a, x, a)), y) → C(f(b(x, z)), c(z, y, a), a)
C(f(z), f(c(a, x, a)), y) → B(x, z)
C(f(z), f(c(a, x, a)), y) → C(z, y, a)

The TRS R consists of the following rules:

b(y, z) → f(c(c(y, z, z), a, a))
b(b(z, y), a) → z
c(f(z), f(c(a, x, a)), y) → c(f(b(x, z)), c(z, y, a), a)

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

(5) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


B(y, z) → C(y, z, z)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(B(x1, x2)) = 3A +
[1A,3A,1A]
·x1 +
[0A,0A,2A]
·x2

POL(C(x1, x2, x3)) = 2A +
[0A,0A,0A]
·x1 +
[-I,-I,1A]
·x2 +
[-I,-I,1A]
·x3

POL(f(x1)) =
/0A\
|-I|
\-I/
 +
/-I0A2A\
|-I0A-I|
\0A-I1A/
·x1

POL(c(x1, x2, x3)) =
/0A\
|-I|
\1A/
 +
/-I-I1A\
|-I-I0A|
\-I-I-I/
·x1 +
/0A2A0A\
|0A1A-I|
\-I-I-I/
·x2 +
/2A3A0A\
|3A2A0A|
\-I-I-I/
·x3

POL(a) =
/0A\
|0A|
\3A/

POL(b(x1, x2)) =
/3A\
|3A|
\3A/
 +
/0A-I-I\
|1A0A0A|
\-I0A-I/
·x1 +
/-I-I0A\
|-I-I0A|
\-I-I-I/
·x2

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

b(y, z) → f(c(c(y, z, z), a, a))
b(b(z, y), a) → z
c(f(z), f(c(a, x, a)), y) → c(f(b(x, z)), c(z, y, a), a)

(6) Obligation:

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

C(f(z), f(c(a, x, a)), y) → C(f(b(x, z)), c(z, y, a), a)
C(f(z), f(c(a, x, a)), y) → B(x, z)
C(f(z), f(c(a, x, a)), y) → C(z, y, a)

The TRS R consists of the following rules:

b(y, z) → f(c(c(y, z, z), a, a))
b(b(z, y), a) → z
c(f(z), f(c(a, x, a)), y) → c(f(b(x, z)), c(z, y, a), a)

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(f(z), f(c(a, x, a)), y) → C(z, y, a)
C(f(z), f(c(a, x, a)), y) → C(f(b(x, z)), c(z, y, a), a)

The TRS R consists of the following rules:

b(y, z) → f(c(c(y, z, z), a, a))
b(b(z, y), a) → z
c(f(z), f(c(a, x, a)), y) → c(f(b(x, z)), c(z, y, a), a)

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

(9) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule C(f(z), f(c(a, x, a)), y) → C(z, y, a) we obtained the following new rules [LPAR04]:

C(f(x0), f(c(a, x1, a)), a) → C(x0, a, a) → C(f(x0), f(c(a, x1, a)), a) → C(x0, a, a)

(10) Obligation:

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

C(f(z), f(c(a, x, a)), y) → C(f(b(x, z)), c(z, y, a), a)
C(f(x0), f(c(a, x1, a)), a) → C(x0, a, a)

The TRS R consists of the following rules:

b(y, z) → f(c(c(y, z, z), a, a))
b(b(z, y), a) → z
c(f(z), f(c(a, x, a)), y) → c(f(b(x, z)), c(z, y, a), a)

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

(11) DependencyGraphProof (EQUIVALENT transformation)

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

(12) Obligation:

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

C(f(z), f(c(a, x, a)), y) → C(f(b(x, z)), c(z, y, a), a)

The TRS R consists of the following rules:

b(y, z) → f(c(c(y, z, z), a, a))
b(b(z, y), a) → z
c(f(z), f(c(a, x, a)), y) → c(f(b(x, z)), c(z, y, a), a)

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

(13) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule C(f(z), f(c(a, x, a)), y) → C(f(b(x, z)), c(z, y, a), a) we obtained the following new rules [LPAR04]:

C(f(y_2), f(c(a, x1, a)), a) → C(f(b(x1, y_2)), c(y_2, a, a), a) → C(f(y_2), f(c(a, x1, a)), a) → C(f(b(x1, y_2)), c(y_2, a, a), a)

(14) Obligation:

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

C(f(y_2), f(c(a, x1, a)), a) → C(f(b(x1, y_2)), c(y_2, a, a), a)

The TRS R consists of the following rules:

b(y, z) → f(c(c(y, z, z), a, a))
b(b(z, y), a) → z
c(f(z), f(c(a, x, a)), y) → c(f(b(x, z)), c(z, y, a), a)

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.

(16) TRUE