(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