(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(b(x))) → a(x)
a(a(x)) → a(b(a(x)))
b(c(x)) → c(a(a(x)))
a(c(x)) → c(b(b(x)))
a(a(a(x))) → b(a(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:
B(b(b(x))) → A(x)
A(a(x)) → A(b(a(x)))
A(a(x)) → B(a(x))
B(c(x)) → A(a(x))
B(c(x)) → A(x)
A(c(x)) → B(b(x))
A(c(x)) → B(x)
A(a(a(x))) → B(a(a(x)))
The TRS R consists of the following rules:
b(b(b(x))) → a(x)
a(a(x)) → a(b(a(x)))
b(c(x)) → c(a(a(x)))
a(c(x)) → c(b(b(x)))
a(a(a(x))) → b(a(a(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) 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(c(x)) → A(a(x))
B(c(x)) → A(x)
A(c(x)) → B(b(x))
A(c(x)) → B(x)
Used ordering: Polynomial interpretation [POLO]:
POL(A(x1)) = x1
POL(B(x1)) = x1
POL(a(x1)) = x1
POL(b(x1)) = x1
POL(c(x1)) = 2 + x1
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(b(b(x))) → A(x)
A(a(x)) → A(b(a(x)))
A(a(x)) → B(a(x))
A(a(a(x))) → B(a(a(x)))
The TRS R consists of the following rules:
b(b(b(x))) → a(x)
a(a(x)) → a(b(a(x)))
b(c(x)) → c(a(a(x)))
a(c(x)) → c(b(b(x)))
a(a(a(x))) → b(a(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(b(b(x))) → A(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
| POL(b(x1)) = | | + | | / | 0A | 0A | 0A | \ |
| | | 1A | 0A | -I | | |
| \ | 0A | 0A | 0A | / |
| · | x1 |
| POL(a(x1)) = | | + | | / | -I | 0A | -I | \ |
| | | -I | 1A | 0A | | |
| \ | 0A | 0A | 0A | / |
| · | x1 |
| POL(c(x1)) = | | + | | / | -I | -I | -I | \ |
| | | -I | -I | -I | | |
| \ | -I | -I | -I | / |
| · | x1 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
a(a(x)) → a(b(a(x)))
a(a(a(x))) → b(a(a(x)))
b(b(b(x))) → a(x)
a(c(x)) → c(b(b(x)))
b(c(x)) → c(a(a(x)))
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(a(x)) → A(b(a(x)))
A(a(x)) → B(a(x))
A(a(a(x))) → B(a(a(x)))
The TRS R consists of the following rules:
b(b(b(x))) → a(x)
a(a(x)) → a(b(a(x)))
b(c(x)) → c(a(a(x)))
a(c(x)) → c(b(b(x)))
a(a(a(x))) → b(a(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 1 SCC with 2 less nodes.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(a(x)) → A(b(a(x)))
The TRS R consists of the following rules:
b(b(b(x))) → a(x)
a(a(x)) → a(b(a(x)))
b(c(x)) → c(a(a(x)))
a(c(x)) → c(b(b(x)))
a(a(a(x))) → b(a(a(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
A(a(x)) → A(b(a(x)))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
| POL(a(x1)) = | | + | | / | 1A | 0A | 1A | \ |
| | | 0A | 0A | 0A | | |
| \ | 0A | -I | 0A | / |
| · | x1 |
| POL(b(x1)) = | | + | | / | -I | -I | 0A | \ |
| | | -I | -I | 0A | | |
| \ | 0A | 0A | 1A | / |
| · | x1 |
| POL(c(x1)) = | | + | | / | -I | -I | -I | \ |
| | | -I | -I | -I | | |
| \ | -I | -I | -I | / |
| · | x1 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
a(a(x)) → a(b(a(x)))
a(a(a(x))) → b(a(a(x)))
b(b(b(x))) → a(x)
a(c(x)) → c(b(b(x)))
b(c(x)) → c(a(a(x)))
(10) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
b(b(b(x))) → a(x)
a(a(x)) → a(b(a(x)))
b(c(x)) → c(a(a(x)))
a(c(x)) → c(b(b(x)))
a(a(a(x))) → b(a(a(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(11) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(12) YES