(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a, b(c(z, x, y), a)) → b(b(z, c(y, z, a)), x)
f(c(a, b(b(z, a), y), x)) → f(c(x, b(z, x), y))
c(f(c(a, y, a)), x, z) → f(b(b(z, z), f(b(y, b(x, a)))))
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:
B(a, b(c(z, x, y), a)) → B(b(z, c(y, z, a)), x)
B(a, b(c(z, x, y), a)) → B(z, c(y, z, a))
B(a, b(c(z, x, y), a)) → C(y, z, a)
F(c(a, b(b(z, a), y), x)) → F(c(x, b(z, x), y))
F(c(a, b(b(z, a), y), x)) → C(x, b(z, x), y)
F(c(a, b(b(z, a), y), x)) → B(z, x)
C(f(c(a, y, a)), x, z) → F(b(b(z, z), f(b(y, b(x, a)))))
C(f(c(a, y, a)), x, z) → B(b(z, z), f(b(y, b(x, a))))
C(f(c(a, y, a)), x, z) → B(z, z)
C(f(c(a, y, a)), x, z) → F(b(y, b(x, a)))
C(f(c(a, y, a)), x, z) → B(y, b(x, a))
C(f(c(a, y, a)), x, z) → B(x, a)
The TRS R consists of the following rules:
b(a, b(c(z, x, y), a)) → b(b(z, c(y, z, a)), x)
f(c(a, b(b(z, a), y), x)) → f(c(x, b(z, x), y))
c(f(c(a, y, a)), x, z) → f(b(b(z, z), f(b(y, b(x, a)))))
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 2 SCCs with 9 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(a, b(c(z, x, y), a)) → C(y, z, a)
C(f(c(a, y, a)), x, z) → B(y, b(x, a))
The TRS R consists of the following rules:
b(a, b(c(z, x, y), a)) → b(b(z, c(y, z, a)), x)
f(c(a, b(b(z, a), y), x)) → f(c(x, b(z, x), y))
c(f(c(a, y, a)), x, z) → f(b(b(z, z), f(b(y, b(x, a)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(a, b(c(z, x, y), a)) → C(y, z, a)
C(f(c(a, y, a)), x, z) → B(y, b(x, a))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) UsableRulesReductionPairsProof (EQUIVALENT transformation)
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.
The following dependency pairs can be deleted:
B(a, b(c(z, x, y), a)) → C(y, z, a)
C(f(c(a, y, a)), x, z) → B(y, b(x, a))
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [POLO]:
POL(B(x1, x2)) = x1 + 2·x2
POL(C(x1, x2, x3)) = 2 + x1 + 2·x2 + x3
POL(a) = 1
POL(b(x1, x2)) = x1 + x2
POL(c(x1, x2, x3)) = 2 + x1 + x2 + 2·x3
POL(f(x1)) = 2 + x1
(9) Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(10) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(11) YES
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(c(a, b(b(z, a), y), x)) → F(c(x, b(z, x), y))
The TRS R consists of the following rules:
b(a, b(c(z, x, y), a)) → b(b(z, c(y, z, a)), x)
f(c(a, b(b(z, a), y), x)) → f(c(x, b(z, x), y))
c(f(c(a, y, a)), x, z) → f(b(b(z, z), f(b(y, b(x, a)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
F(c(a, b(b(z, a), y), x)) → F(c(x, b(z, x), y))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] to (N^2, +, *, >=, >) :
| POL(c(x1, x2, x3)) = | | + | | · | x1 | + | | · | x2 | + | | · | x3 |
| POL(b(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
b(a, b(c(z, x, y), a)) → b(b(z, c(y, z, a)), x)
c(f(c(a, y, a)), x, z) → f(b(b(z, z), f(b(y, b(x, a)))))
f(c(a, b(b(z, a), y), x)) → f(c(x, b(z, x), y))
(14) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
b(a, b(c(z, x, y), a)) → b(b(z, c(y, z, a)), x)
f(c(a, b(b(z, a), y), x)) → f(c(x, b(z, x), y))
c(f(c(a, y, a)), x, z) → f(b(b(z, z), f(b(y, b(x, a)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(15) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(16) YES