(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
bin(x, 0) → s(0)
bin(0, s(y)) → 0
bin(s(x), s(y)) → +(bin(x, s(y)), bin(x, y))
Q is empty.
(1) Overlay + Local Confluence (EQUIVALENT transformation)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
bin(x, 0) → s(0)
bin(0, s(y)) → 0
bin(s(x), s(y)) → +(bin(x, s(y)), bin(x, y))
The set Q consists of the following terms:
bin(x0, 0)
bin(0, s(x0))
bin(s(x0), s(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:
BIN(s(x), s(y)) → BIN(x, s(y))
BIN(s(x), s(y)) → BIN(x, y)
The TRS R consists of the following rules:
bin(x, 0) → s(0)
bin(0, s(y)) → 0
bin(s(x), s(y)) → +(bin(x, s(y)), bin(x, y))
The set Q consists of the following terms:
bin(x0, 0)
bin(0, s(x0))
bin(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(5) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
BIN(s(x), s(y)) → BIN(x, s(y))
BIN(s(x), s(y)) → BIN(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
BIN(
x1,
x2) =
x1
s(
x1) =
s(
x1)
bin(
x1,
x2) =
bin(
x1,
x2)
0 =
0
+(
x1,
x2) =
x2
Recursive Path Order [RPO].
Precedence:
bin2 > s1
0 > s1
The following usable rules [FROCOS05] were oriented:
bin(x, 0) → s(0)
bin(0, s(y)) → 0
bin(s(x), s(y)) → +(bin(x, s(y)), bin(x, y))
(6) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
bin(x, 0) → s(0)
bin(0, s(y)) → 0
bin(s(x), s(y)) → +(bin(x, s(y)), bin(x, y))
The set Q consists of the following terms:
bin(x0, 0)
bin(0, s(x0))
bin(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(7) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(8) TRUE