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:

g(x, y) → x
g(x, y) → y
f(0, 1, x) → f(s(x), x, x)
f(x, y, s(z)) → s(f(0, 1, z))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

g(x, y) → x
g(x, y) → y
f(0, 1, x) → f(s(x), x, x)
f(x, y, s(z)) → s(f(0, 1, z))

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:

F(x, y, s(z)) → F(0, 1, z)
F(0, 1, x) → F(s(x), x, x)

The TRS R consists of the following rules:

g(x, y) → x
g(x, y) → y
f(0, 1, x) → f(s(x), x, x)
f(x, y, s(z)) → s(f(0, 1, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

F(x, y, s(z)) → F(0, 1, z)
F(0, 1, x) → F(s(x), x, x)

The TRS R consists of the following rules:

g(x, y) → x
g(x, y) → y
f(0, 1, x) → f(s(x), x, x)
f(x, y, s(z)) → s(f(0, 1, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ QDPOrderProof

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

F(x, y, s(z)) → F(0, 1, z)
F(0, 1, x) → F(s(x), x, x)

The TRS R consists of the following rules:

g(x, y) → x
g(x, y) → y
f(0, 1, x) → f(s(x), x, x)
f(x, y, s(z)) → s(f(0, 1, z))

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.


F(x, y, s(z)) → F(0, 1, z)
The remaining pairs can at least be oriented weakly.

F(0, 1, x) → F(s(x), x, x)
Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3)  =  F(x3)
s(x1)  =  s(x1)
0  =  0
1  =  1

Lexicographic path order with status [19].
Precedence:
s1 > F1 > 0
s1 > F1 > 1

Status:
1: multiset
0: multiset
s1: multiset
F1: [1]

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof

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

F(0, 1, x) → F(s(x), x, x)

The TRS R consists of the following rules:

g(x, y) → x
g(x, y) → y
f(0, 1, x) → f(s(x), x, x)
f(x, y, s(z)) → s(f(0, 1, z))

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.