(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → x
a(b(x)) → c(b(c(b(c(a(x))))))
c(c(x)) → a(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:
a(x) → x
b(a(x)) → a(c(b(c(b(c(x))))))
c(c(x)) → 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(c(b(c(x))))))
B(a(x)) → C(b(c(b(c(x)))))
B(a(x)) → B(c(b(c(x))))
B(a(x)) → C(b(c(x)))
B(a(x)) → B(c(x))
B(a(x)) → C(x)
C(c(x)) → A(x)
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → a(c(b(c(b(c(x))))))
c(c(x)) → a(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 5 less nodes.
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(a(x)) → B(c(x))
B(a(x)) → B(c(b(c(x))))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → a(c(b(c(b(c(x))))))
c(c(x)) → 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)) → B(c(x))
B(a(x)) → B(c(b(c(x))))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
POL(a(x1)) = | | + | / | 0A | 0A | -I | \ |
| | 1A | 0A | -I | | |
\ | -I | -I | 0A | / |
| · | x1 |
POL(c(x1)) = | | + | / | 1A | 0A | -I | \ |
| | 0A | -I | -I | | |
\ | 0A | 0A | 0A | / |
| · | x1 |
POL(b(x1)) = | | + | / | -I | 0A | -I | \ |
| | 0A | 1A | -I | | |
\ | 0A | 0A | 0A | / |
| · | x1 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
c(c(x)) → a(x)
b(a(x)) → a(c(b(c(b(c(x))))))
a(x) → x
(8) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → a(c(b(c(b(c(x))))))
c(c(x)) → a(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(10) YES