(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(x)) → b(b(a(x)))
b(c(x)) → c(b(b(x)))
c(a(x)) → a(c(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(b(b(x)))
c(b(x)) → b(b(c(x)))
a(c(x)) → c(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(b(b(x)))
B(a(x)) → B(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(c(a(x)))
A(c(x)) → C(a(x))
A(c(x)) → A(x)
The TRS R consists of the following rules:
b(a(x)) → a(b(b(x)))
c(b(x)) → b(b(c(x)))
a(c(x)) → c(c(a(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(5) MRRProof (EQUIVALENT transformation)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
B(a(x)) → B(b(x))
B(a(x)) → B(x)
Used ordering: Polynomial interpretation [POLO]:
POL(A(x1)) = 2 + x1
POL(B(x1)) = x1
POL(C(x1)) = x1
POL(a(x1)) = 2 + x1
POL(b(x1)) = x1
POL(c(x1)) = x1
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(a(x)) → A(b(b(x)))
C(b(x)) → B(b(c(x)))
C(b(x)) → B(c(x))
C(b(x)) → C(x)
A(c(x)) → C(c(a(x)))
A(c(x)) → C(a(x))
A(c(x)) → A(x)
The TRS R consists of the following rules:
b(a(x)) → a(b(b(x)))
c(b(x)) → b(b(c(x)))
a(c(x)) → c(c(a(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(7) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
B(a(x)) → A(b(b(x)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(A(x1)) = 0
POL(B(x1)) = x1
POL(C(x1)) = 0
POL(a(x1)) = 1
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(c(a(x)))
c(b(x)) → b(b(c(x)))
b(a(x)) → a(b(b(x)))
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
C(b(x)) → B(b(c(x)))
C(b(x)) → B(c(x))
C(b(x)) → C(x)
A(c(x)) → C(c(a(x)))
A(c(x)) → C(a(x))
A(c(x)) → A(x)
The TRS R consists of the following rules:
b(a(x)) → a(b(b(x)))
c(b(x)) → b(b(c(x)))
a(c(x)) → c(c(a(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 4 less nodes.
(10) Complex Obligation (AND)
(11) 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(b(b(x)))
c(b(x)) → b(b(c(x)))
a(c(x)) → c(c(a(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(12) 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.
(13) 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.
(14) 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
(15) YES
(16) 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(b(b(x)))
c(b(x)) → b(b(c(x)))
a(c(x)) → c(c(a(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(17) 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.
(18) 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.
(19) 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
(20) YES