(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(x)) → b(c(a(x)))
b(c(x)) → c(b(b(x)))
c(a(x)) → a(c(x))
Q is empty.
(1) QTRS Reverse (EQUIVALENT transformation)
We applied the QTRS Reverse Processor [REVERSE].
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a(x)) → a(c(b(x)))
c(b(x)) → b(b(c(x)))
a(c(x)) → c(a(x))
Q is empty.
(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:
B(a(x)) → A(c(b(x)))
B(a(x)) → C(b(x))
B(a(x)) → B(x)
C(b(x)) → B(b(c(x)))
C(b(x)) → B(c(x))
C(b(x)) → C(x)
A(c(x)) → C(a(x))
A(c(x)) → A(x)
The TRS R consists of the following rules:
b(a(x)) → a(c(b(x)))
c(b(x)) → b(b(c(x)))
a(c(x)) → c(a(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(5) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
B(a(x)) → C(b(x))
B(a(x)) → B(x)
A(c(x)) → C(a(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(A(x1)) = 1
POL(B(x1)) = x1
POL(C(x1)) = 0
POL(a(x1)) = 1 + x1
POL(b(x1)) = x1
POL(c(x1)) = 0
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
a(c(x)) → c(a(x))
c(b(x)) → b(b(c(x)))
b(a(x)) → a(c(b(x)))
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(a(x)) → A(c(b(x)))
C(b(x)) → B(b(c(x)))
C(b(x)) → B(c(x))
C(b(x)) → C(x)
A(c(x)) → A(x)
The TRS R consists of the following rules:
b(a(x)) → a(c(b(x)))
c(b(x)) → b(b(c(x)))
a(c(x)) → c(a(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(7) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes.
(8) Complex Obligation (AND)
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(c(x)) → A(x)
The TRS R consists of the following rules:
b(a(x)) → a(c(b(x)))
c(b(x)) → b(b(c(x)))
a(c(x)) → c(a(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(10) 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.
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(c(x)) → A(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(12) 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
(13) YES
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
C(b(x)) → C(x)
The TRS R consists of the following rules:
b(a(x)) → a(c(b(x)))
c(b(x)) → b(b(c(x)))
a(c(x)) → c(a(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(15) 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.
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
C(b(x)) → C(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(17) 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:
- C(b(x)) → C(x)
The graph contains the following edges 1 > 1
(18) YES