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