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
  ↳ 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

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ 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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


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: Polynomial interpretation [25,35]:

POL(plus(x1, x2)) = (2)x_1 + (4)x_2   
POL(TIMES(x1, x2)) = (4)x_2   
POL(times(x1, x2)) = 0   
POL(1) = 1/2   
POL(0) = 0   
The value of delta used in the strict ordering is 8.
The following usable rules [17] were oriented:

plus(x, 0) → x
times(x, 0) → 0



↳ QTRS
  ↳ DependencyPairsProof
    ↳ 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

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