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:

pred1(s1(x)) -> x
minus2(x, 0) -> x
minus2(x, s1(y)) -> pred1(minus2(x, y))
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(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:

pred1(s1(x)) -> x
minus2(x, 0) -> x
minus2(x, s1(y)) -> pred1(minus2(x, y))
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(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:

pred1(s1(x)) -> x
minus2(x, 0) -> x
minus2(x, s1(y)) -> pred1(minus2(x, y))
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(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:

pred1(s1(x0))
minus2(x0, 0)
minus2(x0, s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
log1(s1(0))
log1(s1(s1(x0)))


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

QUOT2(s1(x), s1(y)) -> QUOT2(minus2(x, y), s1(y))
MINUS2(x, s1(y)) -> PRED1(minus2(x, y))
LOG1(s1(s1(x))) -> QUOT2(x, s1(s1(0)))
LOG1(s1(s1(x))) -> LOG1(s1(quot2(x, s1(s1(0)))))
QUOT2(s1(x), s1(y)) -> MINUS2(x, y)
MINUS2(x, s1(y)) -> MINUS2(x, y)

The TRS R consists of the following rules:

pred1(s1(x)) -> x
minus2(x, 0) -> x
minus2(x, s1(y)) -> pred1(minus2(x, y))
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(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:

pred1(s1(x0))
minus2(x0, 0)
minus2(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:

QUOT2(s1(x), s1(y)) -> QUOT2(minus2(x, y), s1(y))
MINUS2(x, s1(y)) -> PRED1(minus2(x, y))
LOG1(s1(s1(x))) -> QUOT2(x, s1(s1(0)))
LOG1(s1(s1(x))) -> LOG1(s1(quot2(x, s1(s1(0)))))
QUOT2(s1(x), s1(y)) -> MINUS2(x, y)
MINUS2(x, s1(y)) -> MINUS2(x, y)

The TRS R consists of the following rules:

pred1(s1(x)) -> x
minus2(x, 0) -> x
minus2(x, s1(y)) -> pred1(minus2(x, y))
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(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:

pred1(s1(x0))
minus2(x0, 0)
minus2(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 [13,14,18] contains 3 SCCs with 3 less nodes.

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

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

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

The TRS R consists of the following rules:

pred1(s1(x)) -> x
minus2(x, 0) -> x
minus2(x, s1(y)) -> pred1(minus2(x, y))
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(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:

pred1(s1(x0))
minus2(x0, 0)
minus2(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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


MINUS2(x, s1(y)) -> MINUS2(x, y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MINUS2(x1, x2)  =  MINUS1(x2)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
[MINUS1, s1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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:

pred1(s1(x)) -> x
minus2(x, 0) -> x
minus2(x, s1(y)) -> pred1(minus2(x, y))
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(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:

pred1(s1(x0))
minus2(x0, 0)
minus2(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
                ↳ QDPOrderProof
              ↳ QDP

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

QUOT2(s1(x), s1(y)) -> QUOT2(minus2(x, y), s1(y))

The TRS R consists of the following rules:

pred1(s1(x)) -> x
minus2(x, 0) -> x
minus2(x, s1(y)) -> pred1(minus2(x, y))
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(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:

pred1(s1(x0))
minus2(x0, 0)
minus2(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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


QUOT2(s1(x), s1(y)) -> QUOT2(minus2(x, y), s1(y))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
QUOT2(x1, x2)  =  QUOT1(x1)
s1(x1)  =  s1(x1)
minus2(x1, x2)  =  minus1(x1)
0  =  0
pred1(x1)  =  x1

Lexicographic Path Order [19].
Precedence:
s1 > [QUOT1, minus1]
0 > [QUOT1, minus1]


The following usable rules [14] were oriented:

minus2(x, 0) -> x
minus2(x, s1(y)) -> pred1(minus2(x, y))
pred1(s1(x)) -> x



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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:

pred1(s1(x)) -> x
minus2(x, 0) -> x
minus2(x, s1(y)) -> pred1(minus2(x, y))
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(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:

pred1(s1(x0))
minus2(x0, 0)
minus2(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
                ↳ QDPOrderProof

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:

pred1(s1(x)) -> x
minus2(x, 0) -> x
minus2(x, s1(y)) -> pred1(minus2(x, y))
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(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:

pred1(s1(x0))
minus2(x0, 0)
minus2(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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


LOG1(s1(s1(x))) -> LOG1(s1(quot2(x, s1(s1(0)))))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
LOG1(x1)  =  LOG1(x1)
s1(x1)  =  s1(x1)
quot2(x1, x2)  =  x1
0  =  0
minus2(x1, x2)  =  x1
pred1(x1)  =  x1

Lexicographic Path Order [19].
Precedence:
[LOG1, s1] > 0


The following usable rules [14] were oriented:

quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
minus2(x, 0) -> x
minus2(x, s1(y)) -> pred1(minus2(x, y))
pred1(s1(x)) -> x



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

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

pred1(s1(x)) -> x
minus2(x, 0) -> x
minus2(x, s1(y)) -> pred1(minus2(x, y))
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(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:

pred1(s1(x0))
minus2(x0, 0)
minus2(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.