YES Termination w.r.t. Q proof of Secret_06_TRS_8.ari

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

0 QTRS
↳1 DependencyPairsProof (⇔, 1 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 QDP
↳5 MRRProof (⇔, 0 ms)
↳6 QDP
↳7 TransformationProof (⇔, 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(b(0, y), x) → y
c(c(c(y))) → c(c(a(a(c(b(0, y)), 0), 0)))
a(y, 0) → b(y, 0)

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(a(c(b(0, y)), 0), 0)))
C(c(c(y))) → C(a(a(c(b(0, y)), 0), 0))
C(c(c(y))) → A(a(c(b(0, y)), 0), 0)
C(c(c(y))) → A(c(b(0, y)), 0)
C(c(c(y))) → C(b(0, y))
C(c(c(y))) → B(0, y)
A(y, 0) → B(y, 0)

The TRS R consists of the following rules:

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

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 5 less nodes.

(4) Obligation:

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

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

The TRS R consists of the following rules:

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

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(c(c(y))) → C(a(a(c(b(0, y)), 0), 0))


Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(C(x1)) = 2·x1   
POL(a(x1, x2)) = x1 + 2·x2   
POL(b(x1, x2)) = x1 + x2   
POL(c(x1)) = 2 + 2·x1   

(6) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(7) TransformationProof (EQUIVALENT transformation)

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

C(c(c(y0))) → C(c(b(a(c(b(0, y0)), 0), 0))) → C(c(c(y0))) → C(c(b(a(c(b(0, y0)), 0), 0)))
C(c(c(y0))) → C(c(a(b(c(b(0, y0)), 0), 0))) → C(c(c(y0))) → C(c(a(b(c(b(0, y0)), 0), 0)))

(8) Obligation:

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

C(c(c(y0))) → C(c(b(a(c(b(0, y0)), 0), 0)))
C(c(c(y0))) → C(c(a(b(c(b(0, y0)), 0), 0)))

The TRS R consists of the following rules:

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

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(y0))) → C(c(b(a(c(b(0, y0)), 0), 0))) at position [0] we obtained the following new rules [LPAR04]:

C(c(c(y0))) → C(c(b(b(c(b(0, y0)), 0), 0))) → C(c(c(y0))) → C(c(b(b(c(b(0, y0)), 0), 0)))

(10) Obligation:

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

C(c(c(y0))) → C(c(a(b(c(b(0, y0)), 0), 0)))
C(c(c(y0))) → C(c(b(b(c(b(0, y0)), 0), 0)))

The TRS R consists of the following rules:

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

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(c(c(y0))) → C(c(a(b(c(b(0, y0)), 0), 0)))

The TRS R consists of the following rules:

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

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

(13) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule C(c(c(y0))) → C(c(a(b(c(b(0, y0)), 0), 0))) at position [0] we obtained the following new rules [LPAR04]:

C(c(c(y0))) → C(c(b(b(c(b(0, y0)), 0), 0))) → C(c(c(y0))) → C(c(b(b(c(b(0, y0)), 0), 0)))

(14) Obligation:

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

C(c(c(y0))) → C(c(b(b(c(b(0, y0)), 0), 0)))

The TRS R consists of the following rules:

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

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