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:

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
<=2(0, y) -> true
<=2(s1(x), 0) -> false
<=2(s1(x), s1(y)) -> <=2(x, y)
if3(true, x, y) -> x
if3(false, x, y) -> y
perfectp1(0) -> false
perfectp1(s1(x)) -> f4(x, s1(0), s1(x), s1(x))
f4(0, y, 0, u) -> true
f4(0, y, s1(z), u) -> false
f4(s1(x), 0, z, u) -> f4(x, u, -2(z, s1(x)), u)
f4(s1(x), s1(y), z, u) -> if3(<=2(x, y), f4(s1(x), -2(y, x), z, u), f4(x, u, z, u))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
<=2(0, y) -> true
<=2(s1(x), 0) -> false
<=2(s1(x), s1(y)) -> <=2(x, y)
if3(true, x, y) -> x
if3(false, x, y) -> y
perfectp1(0) -> false
perfectp1(s1(x)) -> f4(x, s1(0), s1(x), s1(x))
f4(0, y, 0, u) -> true
f4(0, y, s1(z), u) -> false
f4(s1(x), 0, z, u) -> f4(x, u, -2(z, s1(x)), u)
f4(s1(x), s1(y), z, u) -> if3(<=2(x, y), f4(s1(x), -2(y, x), z, u), f4(x, u, z, u))

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:

F4(s1(x), s1(y), z, u) -> F4(s1(x), -2(y, x), z, u)
F4(s1(x), s1(y), z, u) -> F4(x, u, z, u)
F4(s1(x), 0, z, u) -> F4(x, u, -2(z, s1(x)), u)
-12(s1(x), s1(y)) -> -12(x, y)
PERFECTP1(s1(x)) -> F4(x, s1(0), s1(x), s1(x))
<=12(s1(x), s1(y)) -> <=12(x, y)
F4(s1(x), 0, z, u) -> -12(z, s1(x))
F4(s1(x), s1(y), z, u) -> <=12(x, y)
F4(s1(x), s1(y), z, u) -> -12(y, x)
F4(s1(x), s1(y), z, u) -> IF3(<=2(x, y), f4(s1(x), -2(y, x), z, u), f4(x, u, z, u))

The TRS R consists of the following rules:

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
<=2(0, y) -> true
<=2(s1(x), 0) -> false
<=2(s1(x), s1(y)) -> <=2(x, y)
if3(true, x, y) -> x
if3(false, x, y) -> y
perfectp1(0) -> false
perfectp1(s1(x)) -> f4(x, s1(0), s1(x), s1(x))
f4(0, y, 0, u) -> true
f4(0, y, s1(z), u) -> false
f4(s1(x), 0, z, u) -> f4(x, u, -2(z, s1(x)), u)
f4(s1(x), s1(y), z, u) -> if3(<=2(x, y), f4(s1(x), -2(y, x), z, u), f4(x, u, z, u))

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:

F4(s1(x), s1(y), z, u) -> F4(s1(x), -2(y, x), z, u)
F4(s1(x), s1(y), z, u) -> F4(x, u, z, u)
F4(s1(x), 0, z, u) -> F4(x, u, -2(z, s1(x)), u)
-12(s1(x), s1(y)) -> -12(x, y)
PERFECTP1(s1(x)) -> F4(x, s1(0), s1(x), s1(x))
<=12(s1(x), s1(y)) -> <=12(x, y)
F4(s1(x), 0, z, u) -> -12(z, s1(x))
F4(s1(x), s1(y), z, u) -> <=12(x, y)
F4(s1(x), s1(y), z, u) -> -12(y, x)
F4(s1(x), s1(y), z, u) -> IF3(<=2(x, y), f4(s1(x), -2(y, x), z, u), f4(x, u, z, u))

The TRS R consists of the following rules:

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
<=2(0, y) -> true
<=2(s1(x), 0) -> false
<=2(s1(x), s1(y)) -> <=2(x, y)
if3(true, x, y) -> x
if3(false, x, y) -> y
perfectp1(0) -> false
perfectp1(s1(x)) -> f4(x, s1(0), s1(x), s1(x))
f4(0, y, 0, u) -> true
f4(0, y, s1(z), u) -> false
f4(s1(x), 0, z, u) -> f4(x, u, -2(z, s1(x)), u)
f4(s1(x), s1(y), z, u) -> if3(<=2(x, y), f4(s1(x), -2(y, x), z, u), f4(x, u, z, u))

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 5 less nodes.

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

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

<=12(s1(x), s1(y)) -> <=12(x, y)

The TRS R consists of the following rules:

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
<=2(0, y) -> true
<=2(s1(x), 0) -> false
<=2(s1(x), s1(y)) -> <=2(x, y)
if3(true, x, y) -> x
if3(false, x, y) -> y
perfectp1(0) -> false
perfectp1(s1(x)) -> f4(x, s1(0), s1(x), s1(x))
f4(0, y, 0, u) -> true
f4(0, y, s1(z), u) -> false
f4(s1(x), 0, z, u) -> f4(x, u, -2(z, s1(x)), u)
f4(s1(x), s1(y), z, u) -> if3(<=2(x, y), f4(s1(x), -2(y, x), z, u), f4(x, u, z, u))

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), s1(y)) -> <=12(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(<=12(x1, x2)) = 2·x1 + 2·x2   
POL(s1(x1)) = 2 + x1   

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:

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
<=2(0, y) -> true
<=2(s1(x), 0) -> false
<=2(s1(x), s1(y)) -> <=2(x, y)
if3(true, x, y) -> x
if3(false, x, y) -> y
perfectp1(0) -> false
perfectp1(s1(x)) -> f4(x, s1(0), s1(x), s1(x))
f4(0, y, 0, u) -> true
f4(0, y, s1(z), u) -> false
f4(s1(x), 0, z, u) -> f4(x, u, -2(z, s1(x)), u)
f4(s1(x), s1(y), z, u) -> if3(<=2(x, y), f4(s1(x), -2(y, x), z, u), f4(x, u, z, u))

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

The TRS R consists of the following rules:

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
<=2(0, y) -> true
<=2(s1(x), 0) -> false
<=2(s1(x), s1(y)) -> <=2(x, y)
if3(true, x, y) -> x
if3(false, x, y) -> y
perfectp1(0) -> false
perfectp1(s1(x)) -> f4(x, s1(0), s1(x), s1(x))
f4(0, y, 0, u) -> true
f4(0, y, s1(z), u) -> false
f4(s1(x), 0, z, u) -> f4(x, u, -2(z, s1(x)), u)
f4(s1(x), s1(y), z, u) -> if3(<=2(x, y), f4(s1(x), -2(y, x), z, u), f4(x, u, z, u))

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

POL(-12(x1, x2)) = 2·x1 + 2·x2   
POL(s1(x1)) = 2 + x1   

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:

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
<=2(0, y) -> true
<=2(s1(x), 0) -> false
<=2(s1(x), s1(y)) -> <=2(x, y)
if3(true, x, y) -> x
if3(false, x, y) -> y
perfectp1(0) -> false
perfectp1(s1(x)) -> f4(x, s1(0), s1(x), s1(x))
f4(0, y, 0, u) -> true
f4(0, y, s1(z), u) -> false
f4(s1(x), 0, z, u) -> f4(x, u, -2(z, s1(x)), u)
f4(s1(x), s1(y), z, u) -> if3(<=2(x, y), f4(s1(x), -2(y, x), z, u), f4(x, u, z, u))

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:

F4(s1(x), s1(y), z, u) -> F4(s1(x), -2(y, x), z, u)
F4(s1(x), s1(y), z, u) -> F4(x, u, z, u)
F4(s1(x), 0, z, u) -> F4(x, u, -2(z, s1(x)), u)

The TRS R consists of the following rules:

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
<=2(0, y) -> true
<=2(s1(x), 0) -> false
<=2(s1(x), s1(y)) -> <=2(x, y)
if3(true, x, y) -> x
if3(false, x, y) -> y
perfectp1(0) -> false
perfectp1(s1(x)) -> f4(x, s1(0), s1(x), s1(x))
f4(0, y, 0, u) -> true
f4(0, y, s1(z), u) -> false
f4(s1(x), 0, z, u) -> f4(x, u, -2(z, s1(x)), u)
f4(s1(x), s1(y), z, u) -> if3(<=2(x, y), f4(s1(x), -2(y, x), z, u), f4(x, u, z, u))

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.


F4(s1(x), s1(y), z, u) -> F4(x, u, z, u)
F4(s1(x), 0, z, u) -> F4(x, u, -2(z, s1(x)), u)
The remaining pairs can at least be oriented weakly.

F4(s1(x), s1(y), z, u) -> F4(s1(x), -2(y, x), z, u)
Used ordering: Polynomial interpretation [21]:

POL(-2(x1, x2)) = 0   
POL(0) = 0   
POL(F4(x1, x2, x3, x4)) = 2·x1   
POL(s1(x1)) = 3 + 3·x1   

The following usable rules [14] were oriented: none



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

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

F4(s1(x), s1(y), z, u) -> F4(s1(x), -2(y, x), z, u)

The TRS R consists of the following rules:

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
<=2(0, y) -> true
<=2(s1(x), 0) -> false
<=2(s1(x), s1(y)) -> <=2(x, y)
if3(true, x, y) -> x
if3(false, x, y) -> y
perfectp1(0) -> false
perfectp1(s1(x)) -> f4(x, s1(0), s1(x), s1(x))
f4(0, y, 0, u) -> true
f4(0, y, s1(z), u) -> false
f4(s1(x), 0, z, u) -> f4(x, u, -2(z, s1(x)), u)
f4(s1(x), s1(y), z, u) -> if3(<=2(x, y), f4(s1(x), -2(y, x), z, u), f4(x, u, z, u))

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.


F4(s1(x), s1(y), z, u) -> F4(s1(x), -2(y, x), z, u)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(-2(x1, x2)) = x1   
POL(0) = 0   
POL(F4(x1, x2, x3, x4)) = 2·x2   
POL(s1(x1)) = 3 + 3·x1   

The following usable rules [14] were oriented:

-2(s1(x), s1(y)) -> -2(x, y)
-2(x, 0) -> x



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

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

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
<=2(0, y) -> true
<=2(s1(x), 0) -> false
<=2(s1(x), s1(y)) -> <=2(x, y)
if3(true, x, y) -> x
if3(false, x, y) -> y
perfectp1(0) -> false
perfectp1(s1(x)) -> f4(x, s1(0), s1(x), s1(x))
f4(0, y, 0, u) -> true
f4(0, y, s1(z), u) -> false
f4(s1(x), 0, z, u) -> f4(x, u, -2(z, s1(x)), u)
f4(s1(x), s1(y), z, u) -> if3(<=2(x, y), f4(s1(x), -2(y, x), z, u), f4(x, u, z, u))

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.