(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(a, f(b, x)) → f(b, f(a, x))
f(b, f(c, x)) → f(c, f(b, x))
f(c, f(a, x)) → f(a, f(c, 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:
F(a, f(b, x)) → F(b, f(a, x))
F(a, f(b, x)) → F(a, x)
F(b, f(c, x)) → F(c, f(b, x))
F(b, f(c, x)) → F(b, x)
F(c, f(a, x)) → F(a, f(c, x))
F(c, f(a, x)) → F(c, x)
The TRS R consists of the following rules:
f(a, f(b, x)) → f(b, f(a, x))
f(b, f(c, x)) → f(c, f(b, x))
f(c, f(a, x)) → f(a, f(c, x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) UsableRulesReductionPairsProof (EQUIVALENT transformation)
First, we A-transformed [FROCOS05] the QDP-Problem.
Then we obtain the following A-transformed DP problem.
The pairs P are:
a1(b(x)) → b1(a(x))
a1(b(x)) → a1(x)
b1(c(x)) → c1(b(x))
b1(c(x)) → b1(x)
c1(a(x)) → a1(c(x))
c1(a(x)) → c1(x)
and the Q and R are:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(x)) → b(a(x))
c(a(x)) → a(c(x))
b(c(x)) → c(b(x))
Q is empty.
By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.
No dependency pairs are removed.
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [POLO]:
POL(a(x1)) = 2·x1
POL(a1(x1)) = 1 + 2·x1
POL(b(x1)) = x1
POL(b1(x1)) = 1 + x1
POL(c(x1)) = x1
POL(c1(x1)) = 1 + x1
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
a1(b(x)) → b1(a(x))
a1(b(x)) → a1(x)
b1(c(x)) → c1(b(x))
b1(c(x)) → b1(x)
c1(a(x)) → a1(c(x))
c1(a(x)) → c1(x)
The TRS R consists of the following rules:
a(b(x)) → b(a(x))
c(a(x)) → a(c(x))
b(c(x)) → c(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:
a1(b(x)) → a1(x)
b1(c(x)) → b1(x)
c1(a(x)) → c1(x)
Used ordering: Polynomial interpretation [POLO]:
POL(a(x1)) = 1 + x1
POL(a1(x1)) = x1
POL(b(x1)) = 1 + x1
POL(b1(x1)) = x1
POL(c(x1)) = 1 + x1
POL(c1(x1)) = x1
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
a1(b(x)) → b1(a(x))
b1(c(x)) → c1(b(x))
c1(a(x)) → a1(c(x))
The TRS R consists of the following rules:
a(b(x)) → b(a(x))
c(a(x)) → a(c(x))
b(c(x)) → c(b(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(7) 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:
c1(a(x)) → a1(c(x))
Strictly oriented rules of the TRS R:
a(b(x)) → b(a(x))
Used ordering: Polynomial interpretation [POLO]:
POL(a(x1)) = 2·x1
POL(a1(x1)) = 2·x1
POL(b(x1)) = 1 + 2·x1
POL(b1(x1)) = 2 + 2·x1
POL(c(x1)) = x1
POL(c1(x1)) = 1 + x1
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
a1(b(x)) → b1(a(x))
b1(c(x)) → c1(b(x))
The TRS R consists of the following rules:
c(a(x)) → a(c(x))
b(c(x)) → c(b(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