(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(0) → cons(0, n__f(n__s(n__0)))
f(s(0)) → f(p(s(0)))
p(s(X)) → X
f(X) → n__f(X)
s(X) → n__s(X)
0 → n__0
activate(n__f(X)) → f(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(X) → X
Q is empty.
(1) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(0) = 0
POL(activate(x1)) = 1 + 2·x1
POL(cons(x1, x2)) = x1 + x2
POL(f(x1)) = 1 + 2·x1
POL(n__0) = 0
POL(n__f(x1)) = 1 + 2·x1
POL(n__s(x1)) = x1
POL(p(x1)) = x1
POL(s(x1)) = x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
activate(n__0) → 0
activate(X) → X
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(0) → cons(0, n__f(n__s(n__0)))
f(s(0)) → f(p(s(0)))
p(s(X)) → X
f(X) → n__f(X)
s(X) → n__s(X)
0 → n__0
activate(n__f(X)) → f(activate(X))
activate(n__s(X)) → s(activate(X))
Q is empty.
(3) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(0) = 2
POL(activate(x1)) = 2·x1
POL(cons(x1, x2)) = x1 + x2
POL(f(x1)) = 2 + 2·x1
POL(n__0) = 0
POL(n__f(x1)) = 2 + 2·x1
POL(n__s(x1)) = 1 + x1
POL(p(x1)) = x1
POL(s(x1)) = 1 + x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
p(s(X)) → X
0 → n__0
activate(n__f(X)) → f(activate(X))
activate(n__s(X)) → s(activate(X))
(4) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(0) → cons(0, n__f(n__s(n__0)))
f(s(0)) → f(p(s(0)))
f(X) → n__f(X)
s(X) → n__s(X)
Q is empty.
(5) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(0) = 1
POL(cons(x1, x2)) = x1 + x2
POL(f(x1)) = 2 + 2·x1
POL(n__0) = 0
POL(n__f(x1)) = 2 + x1
POL(n__s(x1)) = 1 + x1
POL(p(x1)) = x1
POL(s(x1)) = 2 + x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
s(X) → n__s(X)
(6) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(0) → cons(0, n__f(n__s(n__0)))
f(s(0)) → f(p(s(0)))
f(X) → n__f(X)
Q is empty.
(7) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(0) = 2
POL(cons(x1, x2)) = 2·x1 + x2
POL(f(x1)) = 2 + 2·x1
POL(n__0) = 0
POL(n__f(x1)) = 1 + x1
POL(n__s(x1)) = 2·x1
POL(p(x1)) = x1
POL(s(x1)) = 2 + 2·x1
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(n__s(n__0)))
f(X) → n__f(X)
(8) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(s(0)) → f(p(s(0)))
Q is empty.
(9) 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}
(10) 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))
(11) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(12) 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.
(13) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
(14) TRUE