(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
c(c(z, y, a), a, a) → b(z, y)
f(c(x, y, z)) → c(z, f(b(y, z)), a)
b(z, b(c(a, y, a), f(f(x)))) → c(c(y, a, z), z, 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:
C(c(z, y, a), a, a) → B(z, y)
F(c(x, y, z)) → C(z, f(b(y, z)), a)
F(c(x, y, z)) → F(b(y, z))
F(c(x, y, z)) → B(y, z)
B(z, b(c(a, y, a), f(f(x)))) → C(c(y, a, z), z, x)
B(z, b(c(a, y, a), f(f(x)))) → C(y, a, z)
The TRS R consists of the following rules:
c(c(z, y, a), a, a) → b(z, y)
f(c(x, y, z)) → c(z, f(b(y, z)), a)
b(z, b(c(a, y, a), f(f(x)))) → c(c(y, a, z), z, 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 2 SCCs with 2 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(z, b(c(a, y, a), f(f(x)))) → C(c(y, a, z), z, x)
C(c(z, y, a), a, a) → B(z, y)
B(z, b(c(a, y, a), f(f(x)))) → C(y, a, z)
The TRS R consists of the following rules:
c(c(z, y, a), a, a) → b(z, y)
f(c(x, y, z)) → c(z, f(b(y, z)), a)
b(z, b(c(a, y, a), f(f(x)))) → c(c(y, a, z), z, x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
B(z, b(c(a, y, a), f(f(x)))) → C(c(y, a, z), z, x)
C(c(z, y, a), a, a) → B(z, y)
B(z, b(c(a, y, a), f(f(x)))) → C(y, a, z)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
B(
x1,
x2) =
x2
b(
x1,
x2) =
b(
x1,
x2)
c(
x1,
x2,
x3) =
c(
x1,
x2)
a =
a
f(
x1) =
f
C(
x1,
x2,
x3) =
x1
Knuth-Bendix order [KBO] with precedence:
trivial
and weight map:
a=4
b_2=11
c_2=8
f=7
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
b(z, b(c(a, y, a), f(f(x)))) → c(c(y, a, z), z, x)
c(c(z, y, a), a, a) → b(z, y)
(7) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
c(c(z, y, a), a, a) → b(z, y)
f(c(x, y, z)) → c(z, f(b(y, z)), a)
b(z, b(c(a, y, a), f(f(x)))) → c(c(y, a, z), z, x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(9) YES
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(c(x, y, z)) → F(b(y, z))
The TRS R consists of the following rules:
c(c(z, y, a), a, a) → b(z, y)
f(c(x, y, z)) → c(z, f(b(y, z)), a)
b(z, b(c(a, y, a), f(f(x)))) → c(c(y, a, z), z, x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(11) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
F(c(x, y, z)) → F(b(y, z))
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(z, b(c(a, y, a), f(f(x)))) → c(c(y, a, z), z, x)
c(c(z, y, a), a, a) → b(z, y)
(12) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
c(c(z, y, a), a, a) → b(z, y)
f(c(x, y, z)) → c(z, f(b(y, z)), a)
b(z, b(c(a, y, a), f(f(x)))) → c(c(y, a, z), z, x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(14) YES