(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(a, f(b, x)) → f(b, f(a, x))
f(b, f(c, x)) → f(c, f(b, x))
f(c, f(a, x)) → f(a, f(c, x))
Q is empty.
(1) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(2) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(a, f(b, x)) → F(b, f(a, x))
F(a, f(b, x)) → F(a, x)
F(b, f(c, x)) → F(c, f(b, x))
F(b, f(c, x)) → F(b, x)
F(c, f(a, x)) → F(a, f(c, x))
F(c, f(a, x)) → F(c, x)
The TRS R consists of the following rules:
f(a, f(b, x)) → f(b, f(a, x))
f(b, f(c, x)) → f(c, f(b, x))
f(c, f(a, x)) → f(a, f(c, x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04]. Here, we combined the reduction pair processor with the A-transformation [FROCOS05] which results in the following intermediate Q-DP Problem.
The a-transformed P is
a1(b(x)) → b1(a(x))
a1(b(x)) → a1(x)
b1(c(x)) → c1(b(x))
b1(c(x)) → b1(x)
c1(a(x)) → a1(c(x))
c1(a(x)) → c1(x)
The a-transformed usable rules are
a(b(x)) → b(a(x))
c(a(x)) → a(c(x))
b(c(x)) → c(b(x))
The following pairs can be oriented strictly and are deleted.
F(a, f(b, x)) → F(b, f(a, x))
F(b, f(c, x)) → F(c, f(b, x))
F(c, f(a, x)) → F(a, f(c, x))
F(c, f(a, x)) → F(c, x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
a1(
x1) =
a1
b1(
x1) =
b1
c1(
x1) =
x1
b(
x1) =
b
a(
x1) =
a(
x1)
c(
x1) =
x1
Recursive path order with status [RPO].
Precedence:
a1 > a1 > b1 > b
Status:
trivial
The following usable rules [FROCOS05] were oriented:
f(a, f(b, x)) → f(b, f(a, x))
f(c, f(a, x)) → f(a, f(c, x))
f(b, f(c, x)) → f(c, f(b, x))
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(a, f(b, x)) → F(a, x)
F(b, f(c, x)) → F(b, x)
The TRS R consists of the following rules:
f(a, f(b, x)) → f(b, f(a, x))
f(b, f(c, x)) → f(c, f(b, x))
f(c, f(a, x)) → f(a, f(c, x))
Q is empty.
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.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(b, f(c, x)) → F(b, x)
The TRS R consists of the following rules:
f(a, f(b, x)) → f(b, f(a, x))
f(b, f(c, x)) → f(c, f(b, x))
f(c, f(a, x)) → f(a, f(c, x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04]. Here, we combined the reduction pair processor with the A-transformation [FROCOS05] which results in the following intermediate Q-DP Problem.
The a-transformed P is
b1(c(x)) → b1(x)
The a-transformed usable rules are
none
The following pairs can be oriented strictly and are deleted.
F(b, f(c, x)) → F(b, x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
b1(
x1) =
x1
c(
x1) =
c(
x1)
Recursive path order with status [RPO].
Precedence:
trivial
Status:
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, f(b, x)) → f(b, f(a, x))
f(b, f(c, x)) → f(c, f(b, x))
f(c, f(a, x)) → f(a, f(c, x))
Q is empty.
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, f(b, x)) → F(a, x)
The TRS R consists of the following rules:
f(a, f(b, x)) → f(b, f(a, x))
f(b, f(c, x)) → f(c, f(b, x))
f(c, f(a, x)) → f(a, f(c, x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04]. Here, we combined the reduction pair processor with the A-transformation [FROCOS05] which results in the following intermediate Q-DP Problem.
The a-transformed P is
a1(b(x)) → a1(x)
The a-transformed usable rules are
none
The following pairs can be oriented strictly and are deleted.
F(a, f(b, x)) → F(a, x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
a1(
x1) =
x1
b(
x1) =
b(
x1)
Recursive path order with status [RPO].
Precedence:
trivial
Status:
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, f(b, x)) → f(b, f(a, x))
f(b, f(c, x)) → f(c, f(b, x))
f(c, f(a, x)) → f(a, f(c, x))
Q is empty.
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