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:

minus2(0, y) -> 0
minus2(s1(x), 0) -> s1(x)
minus2(s1(x), s1(y)) -> minus2(x, y)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(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, minus2(z, s1(x)), u)
f4(s1(x), s1(y), z, u) -> if3(le2(x, y), f4(s1(x), minus2(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:

minus2(0, y) -> 0
minus2(s1(x), 0) -> s1(x)
minus2(s1(x), s1(y)) -> minus2(x, y)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(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, minus2(z, s1(x)), u)
f4(s1(x), s1(y), z, u) -> if3(le2(x, y), f4(s1(x), minus2(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), 0, z, u) -> MINUS2(z, s1(x))
F4(s1(x), s1(y), z, u) -> F4(s1(x), minus2(y, x), z, u)
F4(s1(x), s1(y), z, u) -> MINUS2(y, x)
F4(s1(x), s1(y), z, u) -> F4(x, u, z, u)
F4(s1(x), s1(y), z, u) -> LE2(x, y)
F4(s1(x), s1(y), z, u) -> IF3(le2(x, y), f4(s1(x), minus2(y, x), z, u), f4(x, u, z, u))
LE2(s1(x), s1(y)) -> LE2(x, y)
PERFECTP1(s1(x)) -> F4(x, s1(0), s1(x), s1(x))
MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
F4(s1(x), 0, z, u) -> F4(x, u, minus2(z, s1(x)), u)

The TRS R consists of the following rules:

minus2(0, y) -> 0
minus2(s1(x), 0) -> s1(x)
minus2(s1(x), s1(y)) -> minus2(x, y)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(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, minus2(z, s1(x)), u)
f4(s1(x), s1(y), z, u) -> if3(le2(x, y), f4(s1(x), minus2(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), 0, z, u) -> MINUS2(z, s1(x))
F4(s1(x), s1(y), z, u) -> F4(s1(x), minus2(y, x), z, u)
F4(s1(x), s1(y), z, u) -> MINUS2(y, x)
F4(s1(x), s1(y), z, u) -> F4(x, u, z, u)
F4(s1(x), s1(y), z, u) -> LE2(x, y)
F4(s1(x), s1(y), z, u) -> IF3(le2(x, y), f4(s1(x), minus2(y, x), z, u), f4(x, u, z, u))
LE2(s1(x), s1(y)) -> LE2(x, y)
PERFECTP1(s1(x)) -> F4(x, s1(0), s1(x), s1(x))
MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
F4(s1(x), 0, z, u) -> F4(x, u, minus2(z, s1(x)), u)

The TRS R consists of the following rules:

minus2(0, y) -> 0
minus2(s1(x), 0) -> s1(x)
minus2(s1(x), s1(y)) -> minus2(x, y)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(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, minus2(z, s1(x)), u)
f4(s1(x), s1(y), z, u) -> if3(le2(x, y), f4(s1(x), minus2(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:

LE2(s1(x), s1(y)) -> LE2(x, y)

The TRS R consists of the following rules:

minus2(0, y) -> 0
minus2(s1(x), 0) -> s1(x)
minus2(s1(x), s1(y)) -> minus2(x, y)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(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, minus2(z, s1(x)), u)
f4(s1(x), s1(y), z, u) -> if3(le2(x, y), f4(s1(x), minus2(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.


LE2(s1(x), s1(y)) -> LE2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(LE2(x1, x2)) = 2·x1 + x2   
POL(s1(x1)) = 2 + 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:

minus2(0, y) -> 0
minus2(s1(x), 0) -> s1(x)
minus2(s1(x), s1(y)) -> minus2(x, y)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(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, minus2(z, s1(x)), u)
f4(s1(x), s1(y), z, u) -> if3(le2(x, y), f4(s1(x), minus2(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:

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

The TRS R consists of the following rules:

minus2(0, y) -> 0
minus2(s1(x), 0) -> s1(x)
minus2(s1(x), s1(y)) -> minus2(x, y)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(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, minus2(z, s1(x)), u)
f4(s1(x), s1(y), z, u) -> if3(le2(x, y), f4(s1(x), minus2(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.


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

POL(MINUS2(x1, x2)) = 2·x1 + x2   
POL(s1(x1)) = 2 + 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:

minus2(0, y) -> 0
minus2(s1(x), 0) -> s1(x)
minus2(s1(x), s1(y)) -> minus2(x, y)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(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, minus2(z, s1(x)), u)
f4(s1(x), s1(y), z, u) -> if3(le2(x, y), f4(s1(x), minus2(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), minus2(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, minus2(z, s1(x)), u)

The TRS R consists of the following rules:

minus2(0, y) -> 0
minus2(s1(x), 0) -> s1(x)
minus2(s1(x), s1(y)) -> minus2(x, y)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(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, minus2(z, s1(x)), u)
f4(s1(x), s1(y), z, u) -> if3(le2(x, y), f4(s1(x), minus2(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, minus2(z, s1(x)), u)
The remaining pairs can at least be oriented weakly.

F4(s1(x), s1(y), z, u) -> F4(s1(x), minus2(y, x), z, u)
Used ordering: Polynomial interpretation [21]:

POL(0) = 1   
POL(F4(x1, x2, x3, x4)) = x1   
POL(minus2(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(s1(x1)) = 1 + 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), minus2(y, x), z, u)

The TRS R consists of the following rules:

minus2(0, y) -> 0
minus2(s1(x), 0) -> s1(x)
minus2(s1(x), s1(y)) -> minus2(x, y)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(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, minus2(z, s1(x)), u)
f4(s1(x), s1(y), z, u) -> if3(le2(x, y), f4(s1(x), minus2(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), minus2(y, x), z, u)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 1   
POL(F4(x1, x2, x3, x4)) = 2·x2   
POL(minus2(x1, x2)) = 2·x1   
POL(s1(x1)) = 2 + 2·x1   

The following usable rules [14] were oriented:

minus2(s1(x), s1(y)) -> minus2(x, y)
minus2(0, y) -> 0
minus2(s1(x), 0) -> s1(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:

minus2(0, y) -> 0
minus2(s1(x), 0) -> s1(x)
minus2(s1(x), s1(y)) -> minus2(x, y)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(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, minus2(z, s1(x)), u)
f4(s1(x), s1(y), z, u) -> if3(le2(x, y), f4(s1(x), minus2(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.