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
  ↳ 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))))))

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:

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))))))

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:

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))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ 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))))))

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.


MIN2(s1(X), s1(Y)) -> MIN2(X, Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(MIN2(x1, x2)) = 2·x1 + 2·x2   
POL(s1(x1)) = 2 + x1   

The following usable rules [14] were oriented: none



↳ 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:

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.
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

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))))))

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(min2(X, Y), s1(Y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(QUOT2(x1, x2)) = 2·x1   
POL(min2(x1, x2)) = x1   
POL(s1(x1)) = 3 + 3·x1   

The following usable rules [14] were oriented:

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



↳ 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:

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.
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
            ↳ 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:

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.
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) = 2   
POL(LOG1(x1)) = x1   
POL(min2(x1, x2)) = x1   
POL(quot2(x1, x2)) = 1 + x1   
POL(s1(x1)) = 2 + x1   

The following usable rules [14] were oriented:

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



↳ 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:

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