(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → b(c(x))
a(b(x)) → b(a(x))
d(c(x)) → d(a(x))
a(c(x)) → c(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:
A(b(x)) → A(x)
D(c(x)) → D(a(x))
D(c(x)) → A(x)
A(c(x)) → A(x)
The TRS R consists of the following rules:
a(x) → b(c(x))
a(b(x)) → b(a(x))
d(c(x)) → d(a(x))
a(c(x)) → c(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 1 less node.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(c(x)) → A(x)
A(b(x)) → A(x)
The TRS R consists of the following rules:
a(x) → b(c(x))
a(b(x)) → b(a(x))
d(c(x)) → d(a(x))
a(c(x)) → c(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:
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.
(8) 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
(9) YES
(10) 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))
d(c(x)) → d(a(x))
a(c(x)) → c(a(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(11) 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.
(12) 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.
(13) 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: Polynomial interpretation [POLO]:
POL(D(x1)) = x1
POL(a(x1)) = x1
POL(b(x1)) = 0
POL(c(x1)) = 1 + x1
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))
(14) 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.
(15) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(16) YES