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:

le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(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:

le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(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:

le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(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:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(0, x0)
minus2(s1(x0), x1)
if_minus3(true, s1(x0), x1)
if_minus3(false, s1(x0), 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(s1(x), y) -> IF_MINUS3(le2(s1(x), y), s1(x), y)
MINUS2(s1(x), y) -> LE2(s1(x), y)
LE2(s1(x), s1(y)) -> LE2(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)
IF_MINUS3(false, s1(x), y) -> MINUS2(x, y)

The TRS R consists of the following rules:

le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(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:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(0, x0)
minus2(s1(x0), x1)
if_minus3(true, s1(x0), x1)
if_minus3(false, s1(x0), 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(s1(x), y) -> IF_MINUS3(le2(s1(x), y), s1(x), y)
MINUS2(s1(x), y) -> LE2(s1(x), y)
LE2(s1(x), s1(y)) -> LE2(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)
IF_MINUS3(false, s1(x), y) -> MINUS2(x, y)

The TRS R consists of the following rules:

le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(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:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(0, x0)
minus2(s1(x0), x1)
if_minus3(true, s1(x0), x1)
if_minus3(false, s1(x0), 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 4 SCCs with 3 less nodes.

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

le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(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:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(0, x0)
minus2(s1(x0), x1)
if_minus3(true, s1(x0), x1)
if_minus3(false, s1(x0), 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.


LE2(s1(x), s1(y)) -> LE2(x, y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
LE2(x1, x2)  =  LE1(x1)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



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

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

le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(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:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(0, x0)
minus2(s1(x0), x1)
if_minus3(true, s1(x0), x1)
if_minus3(false, s1(x0), 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
              ↳ QDP

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

MINUS2(s1(x), y) -> IF_MINUS3(le2(s1(x), y), s1(x), y)
IF_MINUS3(false, s1(x), y) -> MINUS2(x, y)

The TRS R consists of the following rules:

le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(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:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(0, x0)
minus2(s1(x0), x1)
if_minus3(true, s1(x0), x1)
if_minus3(false, s1(x0), 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.


IF_MINUS3(false, s1(x), y) -> MINUS2(x, y)
The remaining pairs can at least by weakly be oriented.

MINUS2(s1(x), y) -> IF_MINUS3(le2(s1(x), y), s1(x), y)
Used ordering: Combined order from the following AFS and order.
MINUS2(x1, x2)  =  MINUS2(x1, x2)
s1(x1)  =  s1(x1)
IF_MINUS3(x1, x2, x3)  =  IF_MINUS2(x2, x3)
le2(x1, x2)  =  le2(x1, x2)
false  =  false
0  =  0
true  =  true

Lexicographic Path Order [19].
Precedence:
[s1, le2] > [MINUS2, IFMINUS2]
[s1, le2] > [0, true]
false > [MINUS2, IFMINUS2]


The following usable rules [14] were oriented: none



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

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

MINUS2(s1(x), y) -> IF_MINUS3(le2(s1(x), y), s1(x), y)

The TRS R consists of the following rules:

le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(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:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(0, x0)
minus2(s1(x0), x1)
if_minus3(true, s1(x0), x1)
if_minus3(false, s1(x0), 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 0 SCCs with 1 less node.

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

le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(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:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(0, x0)
minus2(s1(x0), x1)
if_minus3(true, s1(x0), x1)
if_minus3(false, s1(x0), 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)  =  x1
le2(x1, x2)  =  le
0  =  0
true  =  true
if_minus3(x1, x2, x3)  =  x2
false  =  false

Lexicographic Path Order [19].
Precedence:
le > true
false > [s1, 0] > QUOT1 > true


The following usable rules [14] were oriented:

minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(minus2(x, y))



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

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

le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(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:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(0, x0)
minus2(s1(x0), x1)
if_minus3(true, s1(x0), x1)
if_minus3(false, s1(x0), 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
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:

le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(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:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(0, x0)
minus2(s1(x0), x1)
if_minus3(true, s1(x0), x1)
if_minus3(false, s1(x0), 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
le2(x1, x2)  =  le2(x1, x2)
true  =  true
if_minus3(x1, x2, x3)  =  x2
minus2(x1, x2)  =  x1
false  =  false

Lexicographic Path Order [19].
Precedence:
LOG1 > [s1, 0, true] > le2
false > [s1, 0, true] > le2


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(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(minus2(x, y))



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

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

le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(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:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(0, x0)
minus2(s1(x0), x1)
if_minus3(true, s1(x0), x1)
if_minus3(false, s1(x0), 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.