YES Termination w.r.t. Q proof of Secret_06_TRS_gen-14.ari

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

0 QTRS
↳1 DependencyPairsProof (⇔, 5 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 QDP
↳5 QDPOrderProof (⇔, 0 ms)
↳6 QDP
↳7 PisEmptyProof (⇔, 0 ms)
↳8 YES

(0) Obligation:

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

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

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

The TRS R consists of the following rules:

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

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 2 less nodes.

(4) Obligation:

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

B(f(b(x, z)), y) → B(y, z)
B(f(b(x, z)), y) → B(z, b(y, z))

The TRS R consists of the following rules:

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

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(f(b(x, z)), y) → B(y, z)
B(f(b(x, z)), y) → B(z, b(y, z))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] to (N^2, +, *, >=, >) :

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

POL(f(x1)) =
/1\
\1/
 +
/00\
\10/
·x1

POL(b(x1, x2)) =
/1\
\1/
 +
/01\
\01/
·x1 +
/11\
\10/
·x2

POL(c(x1, x2, x3)) =
/0\
\1/
 +
/00\
\00/
·x1 +
/00\
\00/
·x2 +
/00\
\11/
·x3

POL(a) =
/0\
\0/

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

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

(6) Obligation:

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

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

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

(7) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(8) YES