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:

min2(0, y) -> 0
min2(x, 0) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
max2(0, y) -> y
max2(x, 0) -> x
max2(s1(x), s1(y)) -> s1(max2(x, y))
p1(s1(x)) -> x
f3(s1(x), s1(y), s1(z)) -> f3(max2(s1(x), max2(s1(y), s1(z))), p1(min2(s1(x), max2(s1(y), s1(z)))), min2(s1(x), min2(s1(y), s1(z))))
f3(0, y, z) -> max2(y, z)
f3(x, 0, z) -> max2(x, z)
f3(x, y, 0) -> max2(x, y)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

min2(0, y) -> 0
min2(x, 0) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
max2(0, y) -> y
max2(x, 0) -> x
max2(s1(x), s1(y)) -> s1(max2(x, y))
p1(s1(x)) -> x
f3(s1(x), s1(y), s1(z)) -> f3(max2(s1(x), max2(s1(y), s1(z))), p1(min2(s1(x), max2(s1(y), s1(z)))), min2(s1(x), min2(s1(y), s1(z))))
f3(0, y, z) -> max2(y, z)
f3(x, 0, z) -> max2(x, z)
f3(x, y, 0) -> max2(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:

F3(s1(x), s1(y), s1(z)) -> MIN2(s1(x), max2(s1(y), s1(z)))
F3(x, y, 0) -> MAX2(x, y)
F3(x, 0, z) -> MAX2(x, z)
F3(0, y, z) -> MAX2(y, z)
MIN2(s1(x), s1(y)) -> MIN2(x, y)
F3(s1(x), s1(y), s1(z)) -> MAX2(s1(y), s1(z))
MAX2(s1(x), s1(y)) -> MAX2(x, y)
F3(s1(x), s1(y), s1(z)) -> F3(max2(s1(x), max2(s1(y), s1(z))), p1(min2(s1(x), max2(s1(y), s1(z)))), min2(s1(x), min2(s1(y), s1(z))))
F3(s1(x), s1(y), s1(z)) -> MIN2(s1(y), s1(z))
F3(s1(x), s1(y), s1(z)) -> P1(min2(s1(x), max2(s1(y), s1(z))))
F3(s1(x), s1(y), s1(z)) -> MAX2(s1(x), max2(s1(y), s1(z)))
F3(s1(x), s1(y), s1(z)) -> MIN2(s1(x), min2(s1(y), s1(z)))

The TRS R consists of the following rules:

min2(0, y) -> 0
min2(x, 0) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
max2(0, y) -> y
max2(x, 0) -> x
max2(s1(x), s1(y)) -> s1(max2(x, y))
p1(s1(x)) -> x
f3(s1(x), s1(y), s1(z)) -> f3(max2(s1(x), max2(s1(y), s1(z))), p1(min2(s1(x), max2(s1(y), s1(z)))), min2(s1(x), min2(s1(y), s1(z))))
f3(0, y, z) -> max2(y, z)
f3(x, 0, z) -> max2(x, z)
f3(x, y, 0) -> max2(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:

F3(s1(x), s1(y), s1(z)) -> MIN2(s1(x), max2(s1(y), s1(z)))
F3(x, y, 0) -> MAX2(x, y)
F3(x, 0, z) -> MAX2(x, z)
F3(0, y, z) -> MAX2(y, z)
MIN2(s1(x), s1(y)) -> MIN2(x, y)
F3(s1(x), s1(y), s1(z)) -> MAX2(s1(y), s1(z))
MAX2(s1(x), s1(y)) -> MAX2(x, y)
F3(s1(x), s1(y), s1(z)) -> F3(max2(s1(x), max2(s1(y), s1(z))), p1(min2(s1(x), max2(s1(y), s1(z)))), min2(s1(x), min2(s1(y), s1(z))))
F3(s1(x), s1(y), s1(z)) -> MIN2(s1(y), s1(z))
F3(s1(x), s1(y), s1(z)) -> P1(min2(s1(x), max2(s1(y), s1(z))))
F3(s1(x), s1(y), s1(z)) -> MAX2(s1(x), max2(s1(y), s1(z)))
F3(s1(x), s1(y), s1(z)) -> MIN2(s1(x), min2(s1(y), s1(z)))

The TRS R consists of the following rules:

min2(0, y) -> 0
min2(x, 0) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
max2(0, y) -> y
max2(x, 0) -> x
max2(s1(x), s1(y)) -> s1(max2(x, y))
p1(s1(x)) -> x
f3(s1(x), s1(y), s1(z)) -> f3(max2(s1(x), max2(s1(y), s1(z))), p1(min2(s1(x), max2(s1(y), s1(z)))), min2(s1(x), min2(s1(y), s1(z))))
f3(0, y, z) -> max2(y, z)
f3(x, 0, z) -> max2(x, z)
f3(x, y, 0) -> max2(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 9 less nodes.

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

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

MAX2(s1(x), s1(y)) -> MAX2(x, y)

The TRS R consists of the following rules:

min2(0, y) -> 0
min2(x, 0) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
max2(0, y) -> y
max2(x, 0) -> x
max2(s1(x), s1(y)) -> s1(max2(x, y))
p1(s1(x)) -> x
f3(s1(x), s1(y), s1(z)) -> f3(max2(s1(x), max2(s1(y), s1(z))), p1(min2(s1(x), max2(s1(y), s1(z)))), min2(s1(x), min2(s1(y), s1(z))))
f3(0, y, z) -> max2(y, z)
f3(x, 0, z) -> max2(x, z)
f3(x, y, 0) -> max2(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.


MAX2(s1(x), s1(y)) -> MAX2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( MAX2(x1, x2) ) = 3x2 + 3


POL( s1(x1) ) = x1 + 3



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:

min2(0, y) -> 0
min2(x, 0) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
max2(0, y) -> y
max2(x, 0) -> x
max2(s1(x), s1(y)) -> s1(max2(x, y))
p1(s1(x)) -> x
f3(s1(x), s1(y), s1(z)) -> f3(max2(s1(x), max2(s1(y), s1(z))), p1(min2(s1(x), max2(s1(y), s1(z)))), min2(s1(x), min2(s1(y), s1(z))))
f3(0, y, z) -> max2(y, z)
f3(x, 0, z) -> max2(x, z)
f3(x, y, 0) -> max2(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:

MIN2(s1(x), s1(y)) -> MIN2(x, y)

The TRS R consists of the following rules:

min2(0, y) -> 0
min2(x, 0) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
max2(0, y) -> y
max2(x, 0) -> x
max2(s1(x), s1(y)) -> s1(max2(x, y))
p1(s1(x)) -> x
f3(s1(x), s1(y), s1(z)) -> f3(max2(s1(x), max2(s1(y), s1(z))), p1(min2(s1(x), max2(s1(y), s1(z)))), min2(s1(x), min2(s1(y), s1(z))))
f3(0, y, z) -> max2(y, z)
f3(x, 0, z) -> max2(x, z)
f3(x, y, 0) -> max2(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.


MIN2(s1(x), s1(y)) -> MIN2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( MIN2(x1, x2) ) = 3x2 + 3


POL( s1(x1) ) = x1 + 3



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:

min2(0, y) -> 0
min2(x, 0) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
max2(0, y) -> y
max2(x, 0) -> x
max2(s1(x), s1(y)) -> s1(max2(x, y))
p1(s1(x)) -> x
f3(s1(x), s1(y), s1(z)) -> f3(max2(s1(x), max2(s1(y), s1(z))), p1(min2(s1(x), max2(s1(y), s1(z)))), min2(s1(x), min2(s1(y), s1(z))))
f3(0, y, z) -> max2(y, z)
f3(x, 0, z) -> max2(x, z)
f3(x, y, 0) -> max2(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

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

F3(s1(x), s1(y), s1(z)) -> F3(max2(s1(x), max2(s1(y), s1(z))), p1(min2(s1(x), max2(s1(y), s1(z)))), min2(s1(x), min2(s1(y), s1(z))))

The TRS R consists of the following rules:

min2(0, y) -> 0
min2(x, 0) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
max2(0, y) -> y
max2(x, 0) -> x
max2(s1(x), s1(y)) -> s1(max2(x, y))
p1(s1(x)) -> x
f3(s1(x), s1(y), s1(z)) -> f3(max2(s1(x), max2(s1(y), s1(z))), p1(min2(s1(x), max2(s1(y), s1(z)))), min2(s1(x), min2(s1(y), s1(z))))
f3(0, y, z) -> max2(y, z)
f3(x, 0, z) -> max2(x, z)
f3(x, y, 0) -> max2(x, y)

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