Termination w.r.t. Q of the following Term Rewriting System could be proven:

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.


↳ QTRS
  ↳ Overlay + Local Confluence

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.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof

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))


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

BIN(s(x), s(y)) → BIN(x, y)
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.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

BIN(s(x), s(y)) → BIN(x, y)
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.
We use the reduction pair processor [13].


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.

BIN(s(x), s(y)) → BIN(x, s(y))
Used ordering: Combined order from the following AFS and order.
BIN(x1, x2)  =  x2
s(x1)  =  s(x1)

Recursive Path Order [2].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


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.
none
Used ordering: Combined order from the following AFS and order.
BIN(x1, x2)  =  BIN(x1, x2)
s(x1)  =  s(x1)

Recursive Path Order [2].
Precedence:
BIN2 > s1

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ PisEmptyProof

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