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:

min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(X, Y), s1(Y)))
log1(s1(0)) -> 0
log1(s1(s1(X))) -> s1(log1(s1(quot2(X, s1(s1(0))))))

Q is empty.


QTRS
  ↳ Non-Overlap Check

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

min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(X, Y), s1(Y)))
log1(s1(0)) -> 0
log1(s1(s1(X))) -> s1(log1(s1(quot2(X, s1(s1(0))))))

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:

min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(X, Y), s1(Y)))
log1(s1(0)) -> 0
log1(s1(s1(X))) -> s1(log1(s1(quot2(X, s1(s1(0))))))

The set Q consists of the following terms:

min2(x0, 0)
min2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
log1(s1(0))
log1(s1(s1(x0)))


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

MIN2(s1(X), s1(Y)) -> MIN2(X, Y)
QUOT2(s1(X), s1(Y)) -> MIN2(X, Y)
LOG1(s1(s1(X))) -> QUOT2(X, s1(s1(0)))
QUOT2(s1(X), s1(Y)) -> QUOT2(min2(X, Y), s1(Y))
LOG1(s1(s1(X))) -> LOG1(s1(quot2(X, s1(s1(0)))))

The TRS R consists of the following rules:

min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(X, Y), s1(Y)))
log1(s1(0)) -> 0
log1(s1(s1(X))) -> s1(log1(s1(quot2(X, s1(s1(0))))))

The set Q consists of the following terms:

min2(x0, 0)
min2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
log1(s1(0))
log1(s1(s1(x0)))

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:

MIN2(s1(X), s1(Y)) -> MIN2(X, Y)
QUOT2(s1(X), s1(Y)) -> MIN2(X, Y)
LOG1(s1(s1(X))) -> QUOT2(X, s1(s1(0)))
QUOT2(s1(X), s1(Y)) -> QUOT2(min2(X, Y), s1(Y))
LOG1(s1(s1(X))) -> LOG1(s1(quot2(X, s1(s1(0)))))

The TRS R consists of the following rules:

min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(X, Y), s1(Y)))
log1(s1(0)) -> 0
log1(s1(s1(X))) -> s1(log1(s1(quot2(X, s1(s1(0))))))

The set Q consists of the following terms:

min2(x0, 0)
min2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
log1(s1(0))
log1(s1(s1(x0)))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs with 2 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:

MIN2(s1(X), s1(Y)) -> MIN2(X, Y)

The TRS R consists of the following rules:

min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(X, Y), s1(Y)))
log1(s1(0)) -> 0
log1(s1(s1(X))) -> s1(log1(s1(quot2(X, s1(s1(0))))))

The set Q consists of the following terms:

min2(x0, 0)
min2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
log1(s1(0))
log1(s1(s1(x0)))

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.

MIN2(s1(X), s1(Y)) -> MIN2(X, Y)
Used argument filtering: MIN2(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:

min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(X, Y), s1(Y)))
log1(s1(0)) -> 0
log1(s1(s1(X))) -> s1(log1(s1(quot2(X, s1(s1(0))))))

The set Q consists of the following terms:

min2(x0, 0)
min2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
log1(s1(0))
log1(s1(s1(x0)))

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:

QUOT2(s1(X), s1(Y)) -> QUOT2(min2(X, Y), s1(Y))

The TRS R consists of the following rules:

min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(X, Y), s1(Y)))
log1(s1(0)) -> 0
log1(s1(s1(X))) -> s1(log1(s1(quot2(X, s1(s1(0))))))

The set Q consists of the following terms:

min2(x0, 0)
min2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
log1(s1(0))
log1(s1(s1(x0)))

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.

QUOT2(s1(X), s1(Y)) -> QUOT2(min2(X, Y), s1(Y))
Used argument filtering: QUOT2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
min2(x1, x2)  =  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:

min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(X, Y), s1(Y)))
log1(s1(0)) -> 0
log1(s1(s1(X))) -> s1(log1(s1(quot2(X, s1(s1(0))))))

The set Q consists of the following terms:

min2(x0, 0)
min2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
log1(s1(0))
log1(s1(s1(x0)))

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:

LOG1(s1(s1(X))) -> LOG1(s1(quot2(X, s1(s1(0)))))

The TRS R consists of the following rules:

min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(X, Y), s1(Y)))
log1(s1(0)) -> 0
log1(s1(s1(X))) -> s1(log1(s1(quot2(X, s1(s1(0))))))

The set Q consists of the following terms:

min2(x0, 0)
min2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
log1(s1(0))
log1(s1(s1(x0)))

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.

LOG1(s1(s1(X))) -> LOG1(s1(quot2(X, s1(s1(0)))))
Used argument filtering: LOG1(x1)  =  x1
s1(x1)  =  s1(x1)
quot2(x1, x2)  =  x1
0  =  0
min2(x1, x2)  =  x1
Used ordering: Quasi Precedence: trivial


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

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

min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(X, Y), s1(Y)))
log1(s1(0)) -> 0
log1(s1(s1(X))) -> s1(log1(s1(quot2(X, s1(s1(0))))))

The set Q consists of the following terms:

min2(x0, 0)
min2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
log1(s1(0))
log1(s1(s1(x0)))

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