Termination w.r.t. Q of the given QTRS could be proven:

(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, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
BIN(x1, x2)  =  x2
s(x1)  =  s(x1)
bin(x1, x2)  =  bin(x1, x2)
0  =  0
+(x1, x2)  =  x2

Lexicographic path order with status [LPO].
Precedence:

bin₂ > s₁
0 > s₁

Status:

s₁: [1]
bin₂: [1,2]
0: []

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:
The TRS P consists of the following rules:

BIN(s(x), s(y)) → BIN(x, s(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.

(7) 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))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
BIN(x1, x2)  =  BIN(x1, x2)
s(x1)  =  s(x1)
bin(x1, x2)  =  bin(x1, x2)
0  =  0
+(x1, x2)  =  x2

Lexicographic path order with status [LPO].
Precedence:

BIN₂ > s₁
bin₂ > s₁
0 > s₁

Status:

BIN₂: [1,2]
s₁: [1]
bin₂: [1,2]
0: []

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))

(8) 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.

(9) PisEmptyProof (EQUIVALENT transformation)

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

(10) TRUE