(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))
Q is empty.
(1) Overlay + Local Confluence (EQUIVALENT transformation)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))
The set Q consists of the following terms:
f(x0, empty)
f(x0, cons(x1, x2))
g(empty, x0)
g(cons(x0, x1), x2)
(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(a, empty) → G(a, empty)
F(a, cons(x, k)) → F(cons(x, a), k)
G(cons(x, k), d) → G(k, cons(x, d))
The TRS R consists of the following rules:
f(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))
The set Q consists of the following terms:
f(x0, empty)
f(x0, cons(x1, x2))
g(empty, x0)
g(cons(x0, x1), x2)
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 1 less node.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
G(cons(x, k), d) → G(k, cons(x, d))
The TRS R consists of the following rules:
f(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))
The set Q consists of the following terms:
f(x0, empty)
f(x0, cons(x1, x2))
g(empty, x0)
g(cons(x0, x1), x2)
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(x, k), d) → G(k, cons(x, d))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
G(
x1,
x2) =
x1
cons(
x1,
x2) =
cons(
x1,
x2)
Recursive Path Order [RPO].
Precedence:
trivial
The following usable rules [FROCOS05] were oriented:
none
(9) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))
The set Q consists of the following terms:
f(x0, empty)
f(x0, cons(x1, x2))
g(empty, x0)
g(cons(x0, x1), x2)
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(a, cons(x, k)) → F(cons(x, a), k)
The TRS R consists of the following rules:
f(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))
The set Q consists of the following terms:
f(x0, empty)
f(x0, cons(x1, x2))
g(empty, x0)
g(cons(x0, x1), x2)
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(a, cons(x, k)) → F(cons(x, a), k)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(
x1,
x2) =
x2
cons(
x1,
x2) =
cons(
x1,
x2)
Recursive Path Order [RPO].
Precedence:
trivial
The following usable rules [FROCOS05] were oriented:
none
(14) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))
The set Q consists of the following terms:
f(x0, empty)
f(x0, cons(x1, x2))
g(empty, x0)
g(cons(x0, x1), x2)
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