(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(a, x) → f(b, f(c, x))
f(a, f(b, x)) → f(b, f(a, x))
f(d, f(c, x)) → f(d, f(a, x))
f(a, f(c, x)) → f(c, f(a, 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, x) → F(b, f(c, x))
F(a, x) → F(c, x)
F(a, f(b, x)) → F(b, f(a, x))
F(a, f(b, x)) → F(a, x)
F(d, f(c, x)) → F(d, f(a, x))
F(d, f(c, x)) → F(a, x)
F(a, f(c, x)) → F(c, f(a, x))
F(a, f(c, x)) → F(a, x)
The TRS R consists of the following rules:
f(a, x) → f(b, f(c, x))
f(a, f(b, x)) → f(b, f(a, x))
f(d, f(c, x)) → f(d, f(a, x))
f(a, f(c, x)) → f(c, f(a, x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 5 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(a, f(c, x)) → F(a, x)
F(a, f(b, x)) → F(a, x)
The TRS R consists of the following rules:
f(a, x) → f(b, f(c, x))
f(a, f(b, x)) → f(b, f(a, x))
f(d, f(c, x)) → f(d, f(a, x))
f(a, f(c, x)) → f(c, f(a, x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(a, f(c, x)) → F(a, x)
F(a, f(b, x)) → F(a, x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) ATransformationProof (EQUIVALENT transformation)
We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem.
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
a(c(x)) → a(x)
a(b(x)) → a(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(10) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- a(c(x)) → a(x)
The graph contains the following edges 1 > 1
- a(b(x)) → a(x)
The graph contains the following edges 1 > 1
(11) YES
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(d, f(c, x)) → F(d, f(a, x))
The TRS R consists of the following rules:
f(a, x) → f(b, f(c, x))
f(a, f(b, x)) → f(b, f(a, x))
f(d, f(c, x)) → f(d, f(a, x))
f(a, f(c, x)) → f(c, f(a, x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(d, f(c, x)) → F(d, f(a, x))
The TRS R consists of the following rules:
f(a, x) → f(b, f(c, x))
f(a, f(b, x)) → f(b, f(a, x))
f(a, f(c, x)) → f(c, f(a, x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(15) UsableRulesReductionPairsProof (EQUIVALENT transformation)
First, we A-transformed [FROCOS05] the QDP-Problem.
Then we obtain the following A-transformed DP problem.
The pairs P are:
d(c(x)) → d(a(x))
and the Q and R are:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → b(c(x))
a(b(x)) → b(a(x))
a(c(x)) → c(a(x))
Q is empty.
By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.
No dependency pairs are removed.
The following rules are removed from R:
f(a, x) → f(b, f(c, x))
f(a, f(b, x)) → f(b, f(a, x))
f(a, f(c, x)) → f(c, f(a, x))
Used ordering: POLO with Polynomial interpretation [POLO]:
POL(a(x1)) = x1
POL(b(x1)) = x1
POL(c(x1)) = x1
POL(d(x1)) = x1
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
d(c(x)) → d(a(x))
The TRS R consists of the following rules:
a(x) → b(c(x))
a(b(x)) → b(a(x))
a(c(x)) → c(a(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(17) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
d(c(x)) → d(a(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
d(
x1) =
x1
c(
x1) =
c(
x1)
a(
x1) =
x1
b(
x1) =
b
Knuth-Bendix order [KBO] with precedence:
trivial
and weight map:
b=1
c_1=1
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
a(x) → b(c(x))
a(b(x)) → b(a(x))
a(c(x)) → c(a(x))
(18) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
a(x) → b(c(x))
a(b(x)) → b(a(x))
a(c(x)) → c(a(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(19) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(20) YES