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:

p1(s1(x)) -> x
fact1(0) -> s1(0)
fact1(s1(x)) -> *2(s1(x), fact1(p1(s1(x))))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

p1(s1(x)) -> x
fact1(0) -> s1(0)
fact1(s1(x)) -> *2(s1(x), fact1(p1(s1(x))))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))

Q is empty.

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

*12(s1(x), y) -> +12(*2(x, y), y)
+12(x, s1(y)) -> +12(x, y)
*12(s1(x), y) -> *12(x, y)
FACT1(s1(x)) -> P1(s1(x))
FACT1(s1(x)) -> *12(s1(x), fact1(p1(s1(x))))
FACT1(s1(x)) -> FACT1(p1(s1(x)))

The TRS R consists of the following rules:

p1(s1(x)) -> x
fact1(0) -> s1(0)
fact1(s1(x)) -> *2(s1(x), fact1(p1(s1(x))))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))

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:

*12(s1(x), y) -> +12(*2(x, y), y)
+12(x, s1(y)) -> +12(x, y)
*12(s1(x), y) -> *12(x, y)
FACT1(s1(x)) -> P1(s1(x))
FACT1(s1(x)) -> *12(s1(x), fact1(p1(s1(x))))
FACT1(s1(x)) -> FACT1(p1(s1(x)))

The TRS R consists of the following rules:

p1(s1(x)) -> x
fact1(0) -> s1(0)
fact1(s1(x)) -> *2(s1(x), fact1(p1(s1(x))))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

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

+12(x, s1(y)) -> +12(x, y)

The TRS R consists of the following rules:

p1(s1(x)) -> x
fact1(0) -> s1(0)
fact1(s1(x)) -> *2(s1(x), fact1(p1(s1(x))))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))

Q is empty.
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.


+12(x, s1(y)) -> +12(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( +12(x1, x2) ) = x1 + x2


POL( s1(x1) ) = x1 + 1



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

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

p1(s1(x)) -> x
fact1(0) -> s1(0)
fact1(s1(x)) -> *2(s1(x), fact1(p1(s1(x))))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))

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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

*12(s1(x), y) -> *12(x, y)

The TRS R consists of the following rules:

p1(s1(x)) -> x
fact1(0) -> s1(0)
fact1(s1(x)) -> *2(s1(x), fact1(p1(s1(x))))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))

Q is empty.
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.


*12(s1(x), y) -> *12(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( *12(x1, x2) ) = x1 + x2


POL( s1(x1) ) = x1 + 1



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

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

p1(s1(x)) -> x
fact1(0) -> s1(0)
fact1(s1(x)) -> *2(s1(x), fact1(p1(s1(x))))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))

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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof

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

FACT1(s1(x)) -> FACT1(p1(s1(x)))

The TRS R consists of the following rules:

p1(s1(x)) -> x
fact1(0) -> s1(0)
fact1(s1(x)) -> *2(s1(x), fact1(p1(s1(x))))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))

Q is empty.
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.


FACT1(s1(x)) -> FACT1(p1(s1(x)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s1(x1) ) = x1 + 1


POL( FACT1(x1) ) = x1


POL( p1(x1) ) = max{0, x1 - 1}



The following usable rules [14] were oriented:

p1(s1(x)) -> x



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof

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

p1(s1(x)) -> x
fact1(0) -> s1(0)
fact1(s1(x)) -> *2(s1(x), fact1(p1(s1(x))))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))

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.