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
  ↳ Non-Overlap Check

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.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
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))

The set Q consists of the following terms:

minus2(0, x0)
minus2(s1(x0), 0)
minus2(s1(x0), s1(x1))
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
if3(true, x0, x1)
if3(false, x0, x1)
perfectp1(0)
perfectp1(s1(x0))
f4(0, x0, 0, x1)
f4(0, x0, s1(x1), x2)
f4(s1(x0), 0, x1, x2)
f4(s1(x0), s1(x1), x2, x3)


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))

The set Q consists of the following terms:

minus2(0, x0)
minus2(s1(x0), 0)
minus2(s1(x0), s1(x1))
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
if3(true, x0, x1)
if3(false, x0, x1)
perfectp1(0)
perfectp1(s1(x0))
f4(0, x0, 0, x1)
f4(0, x0, s1(x1), x2)
f4(s1(x0), 0, x1, x2)
f4(s1(x0), s1(x1), x2, x3)

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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))

The set Q consists of the following terms:

minus2(0, x0)
minus2(s1(x0), 0)
minus2(s1(x0), s1(x1))
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
if3(true, x0, x1)
if3(false, x0, x1)
perfectp1(0)
perfectp1(s1(x0))
f4(0, x0, 0, x1)
f4(0, x0, s1(x1), x2)
f4(s1(x0), 0, x1, x2)
f4(s1(x0), s1(x1), x2, x3)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs with 5 less nodes.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
QDP
                ↳ QDPAfsSolverProof
              ↳ 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))

The set Q consists of the following terms:

minus2(0, x0)
minus2(s1(x0), 0)
minus2(s1(x0), s1(x1))
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
if3(true, x0, x1)
if3(false, x0, x1)
perfectp1(0)
perfectp1(s1(x0))
f4(0, x0, 0, x1)
f4(0, x0, s1(x1), x2)
f4(s1(x0), 0, x1, x2)
f4(s1(x0), s1(x1), x2, x3)

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

LE2(s1(x), s1(y)) -> LE2(x, y)
Used argument filtering: LE2(x1, x2)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
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))

The set Q consists of the following terms:

minus2(0, x0)
minus2(s1(x0), 0)
minus2(s1(x0), s1(x1))
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
if3(true, x0, x1)
if3(false, x0, x1)
perfectp1(0)
perfectp1(s1(x0))
f4(0, x0, 0, x1)
f4(0, x0, s1(x1), x2)
f4(s1(x0), 0, x1, x2)
f4(s1(x0), s1(x1), x2, x3)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
QDP
                ↳ QDPAfsSolverProof
              ↳ 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))

The set Q consists of the following terms:

minus2(0, x0)
minus2(s1(x0), 0)
minus2(s1(x0), s1(x1))
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
if3(true, x0, x1)
if3(false, x0, x1)
perfectp1(0)
perfectp1(s1(x0))
f4(0, x0, 0, x1)
f4(0, x0, s1(x1), x2)
f4(s1(x0), 0, x1, x2)
f4(s1(x0), s1(x1), x2, x3)

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
Used argument filtering: MINUS2(x1, x2)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
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))

The set Q consists of the following terms:

minus2(0, x0)
minus2(s1(x0), 0)
minus2(s1(x0), s1(x1))
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
if3(true, x0, x1)
if3(false, x0, x1)
perfectp1(0)
perfectp1(s1(x0))
f4(0, x0, 0, x1)
f4(0, x0, s1(x1), x2)
f4(s1(x0), 0, x1, x2)
f4(s1(x0), s1(x1), x2, x3)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
QDP
                ↳ QDPAfsSolverProof

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))

The set Q consists of the following terms:

minus2(0, x0)
minus2(s1(x0), 0)
minus2(s1(x0), s1(x1))
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
if3(true, x0, x1)
if3(false, x0, x1)
perfectp1(0)
perfectp1(s1(x0))
f4(0, x0, 0, x1)
f4(0, x0, s1(x1), x2)
f4(s1(x0), 0, x1, x2)
f4(s1(x0), s1(x1), x2, x3)

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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)
Used argument filtering: F4(x1, x2, x3, x4)  =  x1
s1(x1)  =  s1(x1)
minus2(x1, x2)  =  x1
0  =  0
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
QDP
                    ↳ QDPAfsSolverProof

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))

The set Q consists of the following terms:

minus2(0, x0)
minus2(s1(x0), 0)
minus2(s1(x0), s1(x1))
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
if3(true, x0, x1)
if3(false, x0, x1)
perfectp1(0)
perfectp1(s1(x0))
f4(0, x0, 0, x1)
f4(0, x0, s1(x1), x2)
f4(s1(x0), 0, x1, x2)
f4(s1(x0), s1(x1), x2, x3)

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

F4(s1(x), s1(y), z, u) -> F4(s1(x), minus2(y, x), z, u)
Used argument filtering: F4(x1, x2, x3, x4)  =  x2
s1(x1)  =  s1(x1)
minus2(x1, x2)  =  x1
0  =  0
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ QDPAfsSolverProof
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))

The set Q consists of the following terms:

minus2(0, x0)
minus2(s1(x0), 0)
minus2(s1(x0), s1(x1))
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
if3(true, x0, x1)
if3(false, x0, x1)
perfectp1(0)
perfectp1(s1(x0))
f4(0, x0, 0, x1)
f4(0, x0, s1(x1), x2)
f4(s1(x0), 0, x1, x2)
f4(s1(x0), s1(x1), x2, x3)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.