(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(X, Y)))
Q is empty.
(1) AAECC Innermost (EQUIVALENT transformation)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is
g(cons(s(X), Y)) → s(X)
g(cons(0, Y)) → g(Y)
The TRS R 2 is
f(s(X)) → f(X)
h(cons(X, Y)) → h(g(cons(X, Y)))
The signature Sigma is {
f,
h}
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(X, Y)))
The set Q consists of the following terms:
f(s(x0))
g(cons(0, x0))
g(cons(s(x0), x1))
h(cons(x0, x1))
(3) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(s(X)) → F(X)
G(cons(0, Y)) → G(Y)
H(cons(X, Y)) → H(g(cons(X, Y)))
H(cons(X, Y)) → G(cons(X, Y))
The TRS R consists of the following rules:
f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(X, Y)))
The set Q consists of the following terms:
f(s(x0))
g(cons(0, x0))
g(cons(s(x0), x1))
h(cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 2 less nodes.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
G(cons(0, Y)) → G(Y)
The TRS R consists of the following rules:
f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(X, Y)))
The set Q consists of the following terms:
f(s(x0))
g(cons(0, x0))
g(cons(s(x0), x1))
h(cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(8) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
G(cons(0, Y)) → G(Y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
G(
x1) =
G(
x1)
cons(
x1,
x2) =
cons(
x1,
x2)
0 =
0
f(
x1) =
f(
x1)
s(
x1) =
s(
x1)
g(
x1) =
x1
h(
x1) =
h
Recursive path order with status [RPO].
Precedence:
0 > G1
h > cons2
Status:
G1: [1]
cons2: multiset
0: multiset
f1: [1]
s1: multiset
h: multiset
The following usable rules [FROCOS05] were oriented:
f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(X, Y)))
(9) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(X, Y)))
The set Q consists of the following terms:
f(s(x0))
g(cons(0, x0))
g(cons(s(x0), x1))
h(cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(10) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(11) TRUE
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(s(X)) → F(X)
The TRS R consists of the following rules:
f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(X, Y)))
The set Q consists of the following terms:
f(s(x0))
g(cons(0, x0))
g(cons(s(x0), x1))
h(cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(13) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
F(s(X)) → F(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(
x1) =
F(
x1)
s(
x1) =
s(
x1)
f(
x1) =
f(
x1)
g(
x1) =
g(
x1)
cons(
x1,
x2) =
cons(
x1,
x2)
0 =
0
h(
x1) =
h
Recursive path order with status [RPO].
Precedence:
0 > g1 > s1
h > cons2
Status:
F1: [1]
s1: multiset
f1: [1]
g1: [1]
cons2: multiset
0: multiset
h: []
The following usable rules [FROCOS05] were oriented:
f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(X, Y)))
(14) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(X, Y)))
The set Q consists of the following terms:
f(s(x0))
g(cons(0, x0))
g(cons(s(x0), x1))
h(cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(15) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(16) TRUE