YES Termination w.r.t. Q proof of Transformed_CSR_04_ExProp7_Luc06_Z.ari

(0) Obligation:

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

f(0) → cons(0, n__f(s(0)))
f(s(0)) → f(p(s(0)))
p(s(X)) → X
f(X) → n__f(X)
activate(n__f(X)) → f(X)
activate(X) → X

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Combined order from the following AFS and order.
f(x1)  =  f(x1)
0  =  0
cons(x1, x2)  =  cons(x1, x2)
n__f(x1)  =  x1
s(x1)  =  x1
p(x1)  =  x1
activate(x1)  =  activate(x1)

Recursive path order with status [RPO].
Quasi-Precedence:
0 > [f1, activate1] > cons2

Status:
f1: multiset
0: multiset
cons2: [2,1]
activate1: multiset

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

f(0) → cons(0, n__f(s(0)))
f(X) → n__f(X)
activate(X) → X


(2) Obligation:

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

f(s(0)) → f(p(s(0)))
p(s(X)) → X
activate(n__f(X)) → f(X)

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Combined order from the following AFS and order.
f(x1)  =  f(x1)
s(x1)  =  s(x1)
0  =  0
p(x1)  =  x1
activate(x1)  =  x1
n__f(x1)  =  n__f(x1)

Recursive path order with status [RPO].
Quasi-Precedence:
nf1 > f1 > s1 > 0

Status:
f1: multiset
s1: multiset
0: multiset
nf1: [1]

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

p(s(X)) → X
activate(n__f(X)) → f(X)


(4) Obligation:

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

f(s(0)) → f(p(s(0)))

Q is empty.

(5) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is none

The TRS R 2 is

f(s(0)) → f(p(s(0)))

The signature Sigma is {f}

(6) Obligation:

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

f(s(0)) → f(p(s(0)))

The set Q consists of the following terms:

f(s(0))

(7) DependencyPairsProof (EQUIVALENT transformation)

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

(8) Obligation:

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

F(s(0)) → F(p(s(0)))

The TRS R consists of the following rules:

f(s(0)) → f(p(s(0)))

The set Q consists of the following terms:

f(s(0))

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

(9) DependencyGraphProof (EQUIVALENT transformation)

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

(10) TRUE