(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))
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:
A(0, b(0, x)) → B(0, a(0, x))
A(0, b(0, x)) → A(0, x)
A(0, x) → B(0, b(0, x))
A(0, x) → B(0, x)
A(0, a(1, a(x, y))) → A(1, a(0, a(x, y)))
A(0, a(1, a(x, y))) → A(0, a(x, y))
B(0, a(1, a(x, y))) → B(1, a(0, a(x, y)))
B(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, a(x, y)) → A(1, a(1, a(x, y)))
A(0, a(x, y)) → A(1, a(x, y))
The TRS R consists of the following rules:
a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))
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 1 SCC with 4 less nodes.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, b(0, x)) → B(0, a(0, x))
A(0, b(0, x)) → A(0, x)
A(0, x) → B(0, b(0, x))
A(0, x) → B(0, x)
A(0, a(1, a(x, y))) → A(0, a(x, y))
The TRS R consists of the following rules:
a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.