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:

quot3(0, s1(y), s1(z)) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
quot3(x, 0, s1(z)) -> s1(quot3(x, plus2(z, s1(0)), s1(z)))

Q is empty.


QTRS
  ↳ Non-Overlap Check

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

quot3(0, s1(y), s1(z)) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
quot3(x, 0, s1(z)) -> s1(quot3(x, plus2(z, s1(0)), s1(z)))

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:

quot3(0, s1(y), s1(z)) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
quot3(x, 0, s1(z)) -> s1(quot3(x, plus2(z, s1(0)), s1(z)))

The set Q consists of the following terms:

quot3(0, s1(x0), s1(x1))
quot3(s1(x0), s1(x1), x2)
plus2(0, x0)
plus2(s1(x0), x1)
quot3(x0, 0, s1(x1))


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

PLUS2(s1(x), y) -> PLUS2(x, y)
QUOT3(s1(x), s1(y), z) -> QUOT3(x, y, z)
QUOT3(x, 0, s1(z)) -> PLUS2(z, s1(0))
QUOT3(x, 0, s1(z)) -> QUOT3(x, plus2(z, s1(0)), s1(z))

The TRS R consists of the following rules:

quot3(0, s1(y), s1(z)) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
quot3(x, 0, s1(z)) -> s1(quot3(x, plus2(z, s1(0)), s1(z)))

The set Q consists of the following terms:

quot3(0, s1(x0), s1(x1))
quot3(s1(x0), s1(x1), x2)
plus2(0, x0)
plus2(s1(x0), x1)
quot3(x0, 0, s1(x1))

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:

PLUS2(s1(x), y) -> PLUS2(x, y)
QUOT3(s1(x), s1(y), z) -> QUOT3(x, y, z)
QUOT3(x, 0, s1(z)) -> PLUS2(z, s1(0))
QUOT3(x, 0, s1(z)) -> QUOT3(x, plus2(z, s1(0)), s1(z))

The TRS R consists of the following rules:

quot3(0, s1(y), s1(z)) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
quot3(x, 0, s1(z)) -> s1(quot3(x, plus2(z, s1(0)), s1(z)))

The set Q consists of the following terms:

quot3(0, s1(x0), s1(x1))
quot3(s1(x0), s1(x1), x2)
plus2(0, x0)
plus2(s1(x0), x1)
quot3(x0, 0, s1(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.

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

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

PLUS2(s1(x), y) -> PLUS2(x, y)

The TRS R consists of the following rules:

quot3(0, s1(y), s1(z)) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
quot3(x, 0, s1(z)) -> s1(quot3(x, plus2(z, s1(0)), s1(z)))

The set Q consists of the following terms:

quot3(0, s1(x0), s1(x1))
quot3(s1(x0), s1(x1), x2)
plus2(0, x0)
plus2(s1(x0), x1)
quot3(x0, 0, s1(x1))

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.


PLUS2(s1(x), y) -> PLUS2(x, y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
PLUS2(x1, x2)  =  PLUS1(x1)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
[PLUS1, s1]


The following usable rules [14] were oriented: none



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

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

quot3(0, s1(y), s1(z)) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
quot3(x, 0, s1(z)) -> s1(quot3(x, plus2(z, s1(0)), s1(z)))

The set Q consists of the following terms:

quot3(0, s1(x0), s1(x1))
quot3(s1(x0), s1(x1), x2)
plus2(0, x0)
plus2(s1(x0), x1)
quot3(x0, 0, s1(x1))

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

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

QUOT3(s1(x), s1(y), z) -> QUOT3(x, y, z)
QUOT3(x, 0, s1(z)) -> QUOT3(x, plus2(z, s1(0)), s1(z))

The TRS R consists of the following rules:

quot3(0, s1(y), s1(z)) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
quot3(x, 0, s1(z)) -> s1(quot3(x, plus2(z, s1(0)), s1(z)))

The set Q consists of the following terms:

quot3(0, s1(x0), s1(x1))
quot3(s1(x0), s1(x1), x2)
plus2(0, x0)
plus2(s1(x0), x1)
quot3(x0, 0, s1(x1))

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.


QUOT3(s1(x), s1(y), z) -> QUOT3(x, y, z)
The remaining pairs can at least by weakly be oriented.

QUOT3(x, 0, s1(z)) -> QUOT3(x, plus2(z, s1(0)), s1(z))
Used ordering: Combined order from the following AFS and order.
QUOT3(x1, x2, x3)  =  x1
s1(x1)  =  s1(x1)
0  =  0
plus2(x1, x2)  =  plus2(x1, x2)

Lexicographic Path Order [19].
Precedence:
[0, plus2] > s1


The following usable rules [14] were oriented:

plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))



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

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

QUOT3(x, 0, s1(z)) -> QUOT3(x, plus2(z, s1(0)), s1(z))

The TRS R consists of the following rules:

quot3(0, s1(y), s1(z)) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
quot3(x, 0, s1(z)) -> s1(quot3(x, plus2(z, s1(0)), s1(z)))

The set Q consists of the following terms:

quot3(0, s1(x0), s1(x1))
quot3(s1(x0), s1(x1), x2)
plus2(0, x0)
plus2(s1(x0), x1)
quot3(x0, 0, s1(x1))

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.


QUOT3(x, 0, s1(z)) -> QUOT3(x, plus2(z, s1(0)), s1(z))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
QUOT3(x1, x2, x3)  =  QUOT1(x2)
0  =  0
s1(x1)  =  s
plus2(x1, x2)  =  x2

Lexicographic Path Order [19].
Precedence:
QUOT1 > s
0 > s


The following usable rules [14] were oriented:

plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))



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

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

quot3(0, s1(y), s1(z)) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
quot3(x, 0, s1(z)) -> s1(quot3(x, plus2(z, s1(0)), s1(z)))

The set Q consists of the following terms:

quot3(0, s1(x0), s1(x1))
quot3(s1(x0), s1(x1), x2)
plus2(0, x0)
plus2(s1(x0), x1)
quot3(x0, 0, s1(x1))

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