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:

times(x, plus(y, 1)) → plus(times(x, plus(y, times(1, 0))), x)
times(x, 1) → x
plus(x, 0) → x
times(x, 0) → 0

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

Q restricted rewrite system:
The TRS R consists of the following rules:

times(x, plus(y, 1)) → plus(times(x, plus(y, times(1, 0))), x)
times(x, 1) → x
plus(x, 0) → x
times(x, 0) → 0

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:

times(x, plus(y, 1)) → plus(times(x, plus(y, times(1, 0))), x)
times(x, 1) → x
plus(x, 0) → x
times(x, 0) → 0

The set Q consists of the following terms:

times(x0, plus(x1, 1))
times(x0, 1)
plus(x0, 0)
times(x0, 0)


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

TIMES(x, plus(y, 1)) → TIMES(x, plus(y, times(1, 0)))
TIMES(x, plus(y, 1)) → TIMES(1, 0)
TIMES(x, plus(y, 1)) → PLUS(times(x, plus(y, times(1, 0))), x)
TIMES(x, plus(y, 1)) → PLUS(y, times(1, 0))

The TRS R consists of the following rules:

times(x, plus(y, 1)) → plus(times(x, plus(y, times(1, 0))), x)
times(x, 1) → x
plus(x, 0) → x
times(x, 0) → 0

The set Q consists of the following terms:

times(x0, plus(x1, 1))
times(x0, 1)
plus(x0, 0)
times(x0, 0)

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

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

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

TIMES(x, plus(y, 1)) → TIMES(x, plus(y, times(1, 0)))
TIMES(x, plus(y, 1)) → TIMES(1, 0)
TIMES(x, plus(y, 1)) → PLUS(times(x, plus(y, times(1, 0))), x)
TIMES(x, plus(y, 1)) → PLUS(y, times(1, 0))

The TRS R consists of the following rules:

times(x, plus(y, 1)) → plus(times(x, plus(y, times(1, 0))), x)
times(x, 1) → x
plus(x, 0) → x
times(x, 0) → 0

The set Q consists of the following terms:

times(x0, plus(x1, 1))
times(x0, 1)
plus(x0, 0)
times(x0, 0)

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof

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

TIMES(x, plus(y, 1)) → TIMES(x, plus(y, times(1, 0)))
TIMES(x, plus(y, 1)) → TIMES(1, 0)
TIMES(x, plus(y, 1)) → PLUS(times(x, plus(y, times(1, 0))), x)
TIMES(x, plus(y, 1)) → PLUS(y, times(1, 0))

The TRS R consists of the following rules:

times(x, plus(y, 1)) → plus(times(x, plus(y, times(1, 0))), x)
times(x, 1) → x
plus(x, 0) → x
times(x, 0) → 0

The set Q consists of the following terms:

times(x0, plus(x1, 1))
times(x0, 1)
plus(x0, 0)
times(x0, 0)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 3 less nodes.

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

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

TIMES(x, plus(y, 1)) → TIMES(x, plus(y, times(1, 0)))

The TRS R consists of the following rules:

times(x, plus(y, 1)) → plus(times(x, plus(y, times(1, 0))), x)
times(x, 1) → x
plus(x, 0) → x
times(x, 0) → 0

The set Q consists of the following terms:

times(x0, plus(x1, 1))
times(x0, 1)
plus(x0, 0)
times(x0, 0)

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.


TIMES(x, plus(y, 1)) → TIMES(x, plus(y, times(1, 0)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
TIMES(x1, x2)  =  x2
plus(x1, x2)  =  plus(x1, x2)
1  =  1
times(x1, x2)  =  times
0  =  0

Recursive path order with status [2].
Precedence:
1 > times > 0

Status:
plus2: [1,2]

The following usable rules [14] were oriented:

plus(x, 0) → x
times(x, 0) → 0



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

Q DP problem:
P is empty.
The TRS R consists of the following rules:

times(x, plus(y, 1)) → plus(times(x, plus(y, times(1, 0))), x)
times(x, 1) → x
plus(x, 0) → x
times(x, 0) → 0

The set Q consists of the following terms:

times(x0, plus(x1, 1))
times(x0, 1)
plus(x0, 0)
times(x0, 0)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.