(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → b(x)
a(b(x)) → c(a(c(a(x))))
c(b(c(x))) → b(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)) → C(a(c(a(x))))
A(b(x)) → A(c(a(x)))
A(b(x)) → C(a(x))
A(b(x)) → A(x)
The TRS R consists of the following rules:
a(x) → b(x)
a(b(x)) → c(a(c(a(x))))
c(b(c(x))) → b(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 1 SCC with 2 less nodes.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(b(x)) → A(x)
A(b(x)) → A(c(a(x)))
The TRS R consists of the following rules:
a(x) → b(x)
a(b(x)) → c(a(c(a(x))))
c(b(c(x))) → b(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:
A(b(x)) → A(x)
Used ordering: Polynomial interpretation [POLO]:
POL(A(x1)) = 2·x1
POL(a(x1)) = 2 + x1
POL(b(x1)) = 2 + x1
POL(c(x1)) = x1
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(b(x)) → A(c(a(x)))
The TRS R consists of the following rules:
a(x) → b(x)
a(b(x)) → c(a(c(a(x))))
c(b(c(x))) → b(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.
A(b(x)) → A(c(a(x)))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic integers [ARCTIC,STERNAGEL_THIEMANN_RTA14]:
| POL(b(x1)) = | | + | | / | -I | -I | -1A | \ |
| | | -I | -I | -1A | | |
| \ | -1A | 2A | -I | / |
| · | x1 |
| POL(c(x1)) = | | + | | / | -1A | -I | -I | \ |
| | | -I | -1A | -I | | |
| \ | -I | -I | 1A | / |
| · | x1 |
| POL(a(x1)) = | | + | | / | -I | -I | -1A | \ |
| | | -I | -I | -1A | | |
| \ | -1A | 2A | -I | / |
| · | x1 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
a(x) → b(x)
a(b(x)) → c(a(c(a(x))))
c(b(c(x))) → b(x)
(8) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
a(x) → b(x)
a(b(x)) → c(a(c(a(x))))
c(b(c(x))) → b(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