(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(x, a(b(y))) → f(a(b(x)), y)
f(x, b(c(y))) → f(b(c(x)), y)
f(x, c(a(y))) → f(c(a(x)), y)
f(a(x), y) → f(x, a(y))
f(b(x), y) → f(x, b(y))
f(c(x), y) → f(x, c(y))
Q is empty.
(1) AAECC Innermost (EQUIVALENT transformation)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is none
The TRS R 2 is
f(x, a(b(y))) → f(a(b(x)), y)
f(x, b(c(y))) → f(b(c(x)), y)
f(x, c(a(y))) → f(c(a(x)), y)
f(a(x), y) → f(x, a(y))
f(b(x), y) → f(x, b(y))
f(c(x), y) → f(x, c(y))
The signature Sigma is {
f}
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(x, a(b(y))) → f(a(b(x)), y)
f(x, b(c(y))) → f(b(c(x)), y)
f(x, c(a(y))) → f(c(a(x)), y)
f(a(x), y) → f(x, a(y))
f(b(x), y) → f(x, b(y))
f(c(x), y) → f(x, c(y))
The set Q consists of the following terms:
f(x0, a(b(x1)))
f(x0, b(c(x1)))
f(x0, c(a(x1)))
f(a(x0), x1)
f(b(x0), x1)
f(c(x0), x1)
(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:
F(x, a(b(y))) → F(a(b(x)), y)
F(x, b(c(y))) → F(b(c(x)), y)
F(x, c(a(y))) → F(c(a(x)), y)
F(a(x), y) → F(x, a(y))
F(b(x), y) → F(x, b(y))
F(c(x), y) → F(x, c(y))
The TRS R consists of the following rules:
f(x, a(b(y))) → f(a(b(x)), y)
f(x, b(c(y))) → f(b(c(x)), y)
f(x, c(a(y))) → f(c(a(x)), y)
f(a(x), y) → f(x, a(y))
f(b(x), y) → f(x, b(y))
f(c(x), y) → f(x, c(y))
The set Q consists of the following terms:
f(x0, a(b(x1)))
f(x0, b(c(x1)))
f(x0, c(a(x1)))
f(a(x0), x1)
f(b(x0), x1)
f(c(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
(5) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(x, a(b(y))) → F(a(b(x)), y)
F(x, b(c(y))) → F(b(c(x)), y)
F(x, c(a(y))) → F(c(a(x)), y)
F(a(x), y) → F(x, a(y))
F(b(x), y) → F(x, b(y))
F(c(x), y) → F(x, c(y))
R is empty.
The set Q consists of the following terms:
f(x0, a(b(x1)))
f(x0, b(c(x1)))
f(x0, c(a(x1)))
f(a(x0), x1)
f(b(x0), x1)
f(c(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
(7) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
f(x0, a(b(x1)))
f(x0, b(c(x1)))
f(x0, c(a(x1)))
f(a(x0), x1)
f(b(x0), x1)
f(c(x0), x1)
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(x, a(b(y))) → F(a(b(x)), y)
F(x, b(c(y))) → F(b(c(x)), y)
F(x, c(a(y))) → F(c(a(x)), y)
F(a(x), y) → F(x, a(y))
F(b(x), y) → F(x, b(y))
F(c(x), y) → F(x, c(y))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) QDPToSRSProof (SOUND transformation)
The finiteness of this DP problem is implied by strong termination of a SRS due to [UNKNOWN].
(10) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(x)) → b(a(x))
b(c(x)) → c(b(x))
c(a(x)) → a(c(x))
Q is empty.
(11) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(b(x)) → B(a(x))
A(b(x)) → A(x)
B(c(x)) → C(b(x))
B(c(x)) → B(x)
C(a(x)) → A(c(x))
C(a(x)) → C(x)
The TRS R consists of the following rules:
a(b(x)) → b(a(x))
b(c(x)) → c(b(x))
c(a(x)) → a(c(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) 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)
C(a(x)) → C(x)
Used ordering: Polynomial interpretation [POLO]:
POL(A(x1)) = 1 + 2·x1
POL(B(x1)) = 1 + 2·x1
POL(C(x1)) = x1
POL(a(x1)) = 1 + 2·x1
POL(b(x1)) = 1 + 2·x1
POL(c(x1)) = x1
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(b(x)) → B(a(x))
B(c(x)) → C(b(x))
B(c(x)) → B(x)
C(a(x)) → A(c(x))
The TRS R consists of the following rules:
a(b(x)) → b(a(x))
b(c(x)) → c(b(x))
c(a(x)) → a(c(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(15) 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)) → B(x)
C(a(x)) → A(c(x))
Strictly oriented rules of the TRS R:
b(c(x)) → c(b(x))
Used ordering: Polynomial interpretation [POLO]:
POL(A(x1)) = x1
POL(B(x1)) = 2·x1
POL(C(x1)) = 2 + x1
POL(a(x1)) = x1
POL(b(x1)) = 2·x1
POL(c(x1)) = 1 + x1
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(b(x)) → B(a(x))
B(c(x)) → C(b(x))
The TRS R consists of the following rules:
a(b(x)) → b(a(x))
c(a(x)) → a(c(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(17) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.
(18) TRUE