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 [19] 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 [1,15] 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

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
          ↳ 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

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 [15,17,22] contains 1 SCC with 3 less nodes.

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

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.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ RuleRemovalProof

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, 0) → 0
plus(x, 0) → x

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.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

plus(x, 0) → x

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(1) = 0   
POL(TIMES(x1, x2)) = x1 + 2·x2   
POL(plus(x1, x2)) = 1 + x1 + 2·x2   
POL(times(x1, x2)) = 2·x1 + 2·x2   



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ RuleRemovalProof
                    ↳ 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)))

The TRS R consists of the following rules:

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 [15,17,22] contains 0 SCCs with 1 less node.