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
  ↳ 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))))))

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:

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))))))

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:

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))))))

Q is empty.
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
  ↳ 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))))))

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.


LE2(s1(x), s1(y)) -> LE2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(LE2(x1, x2)) = 2·x1 + x2   
POL(s1(x1)) = 2 + 2·x1   

The following usable rules [14] were oriented: none



↳ 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))))))

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
          ↳ 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))))))

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.


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

MINUS2(s1(x), y) -> IF_MINUS3(le2(s1(x), y), s1(x), y)
Used ordering: Polynomial interpretation [21]:

POL(0) = 2   
POL(IF_MINUS3(x1, x2, x3)) = 2 + x2 + x3   
POL(MINUS2(x1, x2)) = 1 + 2·x1 + x2   
POL(false) = 2   
POL(le2(x1, x2)) = 2   
POL(s1(x1)) = 1 + 2·x1   
POL(true) = 2   

The following usable rules [14] were oriented: none



↳ 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))))))

Q is empty.
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
  ↳ 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))))))

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.


QUOT2(s1(x), s1(y)) -> QUOT2(minus2(x, y), s1(y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 2   
POL(QUOT2(x1, x2)) = x1   
POL(false) = 2   
POL(if_minus3(x1, x2, x3)) = 2·x2   
POL(le2(x1, x2)) = 2   
POL(minus2(x1, x2)) = 2·x1   
POL(s1(x1)) = 1 + 2·x1   
POL(true) = 1   

The following usable rules [14] were oriented:

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



↳ 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))))))

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
          ↳ 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))))))

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.


LOG1(s1(s1(x))) -> LOG1(s1(quot2(x, s1(s1(0)))))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(LOG1(x1)) = x1   
POL(false) = 0   
POL(if_minus3(x1, x2, x3)) = x2   
POL(le2(x1, x2)) = 2·x2   
POL(minus2(x1, x2)) = x1   
POL(quot2(x1, x2)) = x1   
POL(s1(x1)) = 1 + x1   
POL(true) = 1   

The following usable rules [14] were oriented:

quot2(0, s1(y)) -> 0
if_minus3(true, s1(x), y) -> 0
minus2(0, y) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
if_minus3(false, s1(x), y) -> s1(minus2(x, y))
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)



↳ 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))))))

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.