(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