(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) 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:

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))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(3) 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: Lexicographic path order with status [LPO].
Quasi-Precedence:
[BIN2, s1]

Status:
BIN2: [2,1]
s1: [1]


The following usable rules [FROCOS05] were oriented: none

(4) 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))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(5) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(6) TRUE