Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

plus2(0, x) -> x
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(x, y))
p1(s1(x)) -> x
p1(0) -> 0
minus2(x, 0) -> x
minus2(0, x) -> 0
minus2(x, s1(y)) -> p1(minus2(x, y))
isZero1(0) -> true
isZero1(s1(x)) -> false
facIter2(x, y) -> if4(isZero1(x), minus2(x, s1(0)), y, times2(y, x))
if4(true, x, y, z) -> y
if4(false, x, y, z) -> facIter2(x, z)
factorial1(x) -> facIter2(x, s1(0))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

plus2(0, x) -> x
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(x, y))
p1(s1(x)) -> x
p1(0) -> 0
minus2(x, 0) -> x
minus2(0, x) -> 0
minus2(x, s1(y)) -> p1(minus2(x, y))
isZero1(0) -> true
isZero1(s1(x)) -> false
facIter2(x, y) -> if4(isZero1(x), minus2(x, s1(0)), y, times2(y, x))
if4(true, x, y, z) -> y
if4(false, x, y, z) -> facIter2(x, z)
factorial1(x) -> facIter2(x, s1(0))

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:

MINUS2(x, s1(y)) -> P1(minus2(x, y))
FACTORIAL1(x) -> FACITER2(x, s1(0))
PLUS2(s1(x), y) -> PLUS2(x, y)
TIMES2(s1(x), y) -> PLUS2(y, times2(x, y))
FACITER2(x, y) -> IF4(isZero1(x), minus2(x, s1(0)), y, times2(y, x))
IF4(false, x, y, z) -> FACITER2(x, z)
TIMES2(s1(x), y) -> TIMES2(x, y)
FACITER2(x, y) -> MINUS2(x, s1(0))
FACITER2(x, y) -> TIMES2(y, x)
MINUS2(x, s1(y)) -> MINUS2(x, y)
FACITER2(x, y) -> ISZERO1(x)

The TRS R consists of the following rules:

plus2(0, x) -> x
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(x, y))
p1(s1(x)) -> x
p1(0) -> 0
minus2(x, 0) -> x
minus2(0, x) -> 0
minus2(x, s1(y)) -> p1(minus2(x, y))
isZero1(0) -> true
isZero1(s1(x)) -> false
facIter2(x, y) -> if4(isZero1(x), minus2(x, s1(0)), y, times2(y, x))
if4(true, x, y, z) -> y
if4(false, x, y, z) -> facIter2(x, z)
factorial1(x) -> facIter2(x, s1(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:

MINUS2(x, s1(y)) -> P1(minus2(x, y))
FACTORIAL1(x) -> FACITER2(x, s1(0))
PLUS2(s1(x), y) -> PLUS2(x, y)
TIMES2(s1(x), y) -> PLUS2(y, times2(x, y))
FACITER2(x, y) -> IF4(isZero1(x), minus2(x, s1(0)), y, times2(y, x))
IF4(false, x, y, z) -> FACITER2(x, z)
TIMES2(s1(x), y) -> TIMES2(x, y)
FACITER2(x, y) -> MINUS2(x, s1(0))
FACITER2(x, y) -> TIMES2(y, x)
MINUS2(x, s1(y)) -> MINUS2(x, y)
FACITER2(x, y) -> ISZERO1(x)

The TRS R consists of the following rules:

plus2(0, x) -> x
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(x, y))
p1(s1(x)) -> x
p1(0) -> 0
minus2(x, 0) -> x
minus2(0, x) -> 0
minus2(x, s1(y)) -> p1(minus2(x, y))
isZero1(0) -> true
isZero1(s1(x)) -> false
facIter2(x, y) -> if4(isZero1(x), minus2(x, s1(0)), y, times2(y, x))
if4(true, x, y, z) -> y
if4(false, x, y, z) -> facIter2(x, z)
factorial1(x) -> facIter2(x, s1(0))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 6 less nodes.

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

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

MINUS2(x, s1(y)) -> MINUS2(x, y)

The TRS R consists of the following rules:

plus2(0, x) -> x
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(x, y))
p1(s1(x)) -> x
p1(0) -> 0
minus2(x, 0) -> x
minus2(0, x) -> 0
minus2(x, s1(y)) -> p1(minus2(x, y))
isZero1(0) -> true
isZero1(s1(x)) -> false
facIter2(x, y) -> if4(isZero1(x), minus2(x, s1(0)), y, times2(y, x))
if4(true, x, y, z) -> y
if4(false, x, y, z) -> facIter2(x, z)
factorial1(x) -> facIter2(x, s1(0))

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.


MINUS2(x, s1(y)) -> MINUS2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( MINUS2(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
          ↳ QDP

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

plus2(0, x) -> x
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(x, y))
p1(s1(x)) -> x
p1(0) -> 0
minus2(x, 0) -> x
minus2(0, x) -> 0
minus2(x, s1(y)) -> p1(minus2(x, y))
isZero1(0) -> true
isZero1(s1(x)) -> false
facIter2(x, y) -> if4(isZero1(x), minus2(x, s1(0)), y, times2(y, x))
if4(true, x, y, z) -> y
if4(false, x, y, z) -> facIter2(x, z)
factorial1(x) -> facIter2(x, s1(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.

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

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

PLUS2(s1(x), y) -> PLUS2(x, y)

The TRS R consists of the following rules:

plus2(0, x) -> x
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(x, y))
p1(s1(x)) -> x
p1(0) -> 0
minus2(x, 0) -> x
minus2(0, x) -> 0
minus2(x, s1(y)) -> p1(minus2(x, y))
isZero1(0) -> true
isZero1(s1(x)) -> false
facIter2(x, y) -> if4(isZero1(x), minus2(x, s1(0)), y, times2(y, x))
if4(true, x, y, z) -> y
if4(false, x, y, z) -> facIter2(x, z)
factorial1(x) -> facIter2(x, s1(0))

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.


PLUS2(s1(x), y) -> PLUS2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s1(x1) ) = x1 + 1


POL( PLUS2(x1, x2) ) = x1 + x2



The following usable rules [14] were oriented: none



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

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

plus2(0, x) -> x
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(x, y))
p1(s1(x)) -> x
p1(0) -> 0
minus2(x, 0) -> x
minus2(0, x) -> 0
minus2(x, s1(y)) -> p1(minus2(x, y))
isZero1(0) -> true
isZero1(s1(x)) -> false
facIter2(x, y) -> if4(isZero1(x), minus2(x, s1(0)), y, times2(y, x))
if4(true, x, y, z) -> y
if4(false, x, y, z) -> facIter2(x, z)
factorial1(x) -> facIter2(x, s1(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.

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

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

TIMES2(s1(x), y) -> TIMES2(x, y)

The TRS R consists of the following rules:

plus2(0, x) -> x
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(x, y))
p1(s1(x)) -> x
p1(0) -> 0
minus2(x, 0) -> x
minus2(0, x) -> 0
minus2(x, s1(y)) -> p1(minus2(x, y))
isZero1(0) -> true
isZero1(s1(x)) -> false
facIter2(x, y) -> if4(isZero1(x), minus2(x, s1(0)), y, times2(y, x))
if4(true, x, y, z) -> y
if4(false, x, y, z) -> facIter2(x, z)
factorial1(x) -> facIter2(x, s1(0))

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.


TIMES2(s1(x), y) -> TIMES2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s1(x1) ) = x1 + 1


POL( TIMES2(x1, x2) ) = x1 + x2



The following usable rules [14] were oriented: none



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

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

plus2(0, x) -> x
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(x, y))
p1(s1(x)) -> x
p1(0) -> 0
minus2(x, 0) -> x
minus2(0, x) -> 0
minus2(x, s1(y)) -> p1(minus2(x, y))
isZero1(0) -> true
isZero1(s1(x)) -> false
facIter2(x, y) -> if4(isZero1(x), minus2(x, s1(0)), y, times2(y, x))
if4(true, x, y, z) -> y
if4(false, x, y, z) -> facIter2(x, z)
factorial1(x) -> facIter2(x, s1(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.

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

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

FACITER2(x, y) -> IF4(isZero1(x), minus2(x, s1(0)), y, times2(y, x))
IF4(false, x, y, z) -> FACITER2(x, z)

The TRS R consists of the following rules:

plus2(0, x) -> x
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(x, y))
p1(s1(x)) -> x
p1(0) -> 0
minus2(x, 0) -> x
minus2(0, x) -> 0
minus2(x, s1(y)) -> p1(minus2(x, y))
isZero1(0) -> true
isZero1(s1(x)) -> false
facIter2(x, y) -> if4(isZero1(x), minus2(x, s1(0)), y, times2(y, x))
if4(true, x, y, z) -> y
if4(false, x, y, z) -> facIter2(x, z)
factorial1(x) -> facIter2(x, s1(0))

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