Termination w.r.t. Q proof of TRS/nontermin/AG01#4.34.trs

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:

f(0) → true
f(1) → false
f(s(x)) → f(x)
if(true, x, y) → x
if(false, x, y) → y
g(s(x), s(y)) → if(f(x), s(x), s(y))
g(x, c(y)) → g(x, g(s(c(y)), y))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

f(0) → true
f(1) → false
f(s(x)) → f(x)
if(true, x, y) → x
if(false, x, y) → y
g(s(x), s(y)) → if(f(x), s(x), s(y))
g(x, c(y)) → g(x, g(s(c(y)), y))

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

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

f(0) → true
f(1) → false
f(s(x)) → f(x)
if(true, x, y) → x
if(false, x, y) → y
g(s(x), s(y)) → if(f(x), s(x), s(y))
g(x, c(y)) → g(x, g(s(c(y)), y))

The set Q consists of the following terms:

f(0)
f(1)
f(s(x0))
if(true, x0, x1)
if(false, x0, x1)
g(s(x0), s(x1))
g(x0, c(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:

G(x, c(y)) → G(x, g(s(c(y)), y))
G(s(x), s(y)) → IF(f(x), s(x), s(y))
G(s(x), s(y)) → F(x)
G(x, c(y)) → G(s(c(y)), y)
F(s(x)) → F(x)

The TRS R consists of the following rules:

f(0) → true
f(1) → false
f(s(x)) → f(x)
if(true, x, y) → x
if(false, x, y) → y
g(s(x), s(y)) → if(f(x), s(x), s(y))
g(x, c(y)) → g(x, g(s(c(y)), y))

The set Q consists of the following terms:

f(0)
f(1)
f(s(x0))
if(true, x0, x1)
if(false, x0, x1)
g(s(x0), s(x1))
g(x0, c(x1))

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

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

G(x, c(y)) → G(x, g(s(c(y)), y))
G(s(x), s(y)) → IF(f(x), s(x), s(y))
G(s(x), s(y)) → F(x)
G(x, c(y)) → G(s(c(y)), y)
F(s(x)) → F(x)

The TRS R consists of the following rules:

f(0) → true
f(1) → false
f(s(x)) → f(x)
if(true, x, y) → x
if(false, x, y) → y
g(s(x), s(y)) → if(f(x), s(x), s(y))
g(x, c(y)) → g(x, g(s(c(y)), y))

The set Q consists of the following terms:

f(0)
f(1)
f(s(x0))
if(true, x0, x1)
if(false, x0, x1)
g(s(x0), s(x1))
g(x0, c(x1))

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

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

G(s(x), s(y)) → IF(f(x), s(x), s(y))
G(x, c(y)) → G(x, g(s(c(y)), y))
G(s(x), s(y)) → F(x)
G(x, c(y)) → G(s(c(y)), y)
F(s(x)) → F(x)

The TRS R consists of the following rules:

f(0) → true
f(1) → false
f(s(x)) → f(x)
if(true, x, y) → x
if(false, x, y) → y
g(s(x), s(y)) → if(f(x), s(x), s(y))
g(x, c(y)) → g(x, g(s(c(y)), y))

The set Q consists of the following terms:

f(0)
f(1)
f(s(x0))
if(true, x0, x1)
if(false, x0, x1)
g(s(x0), s(x1))
g(x0, c(x1))

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

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

F(s(x)) → F(x)

The TRS R consists of the following rules:

f(0) → true
f(1) → false
f(s(x)) → f(x)
if(true, x, y) → x
if(false, x, y) → y
g(s(x), s(y)) → if(f(x), s(x), s(y))
g(x, c(y)) → g(x, g(s(c(y)), y))

The set Q consists of the following terms:

f(0)
f(1)
f(s(x0))
if(true, x0, x1)
if(false, x0, x1)
g(s(x0), s(x1))
g(x0, c(x1))

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(s(x)) → F(x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
F(x1) = x1
s(x1) = s(x1)

Recursive Path Order [2].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

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

f(0) → true
f(1) → false
f(s(x)) → f(x)
if(true, x, y) → x
if(false, x, y) → y
g(s(x), s(y)) → if(f(x), s(x), s(y))
g(x, c(y)) → g(x, g(s(c(y)), y))

The set Q consists of the following terms:

f(0)
f(1)
f(s(x0))
if(true, x0, x1)
if(false, x0, x1)
g(s(x0), s(x1))
g(x0, c(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

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

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

G(x, c(y)) → G(x, g(s(c(y)), y))
G(x, c(y)) → G(s(c(y)), y)

The TRS R consists of the following rules:

f(0) → true
f(1) → false
f(s(x)) → f(x)
if(true, x, y) → x
if(false, x, y) → y
g(s(x), s(y)) → if(f(x), s(x), s(y))
g(x, c(y)) → g(x, g(s(c(y)), y))

The set Q consists of the following terms:

f(0)
f(1)
f(s(x0))
if(true, x0, x1)
if(false, x0, x1)
g(s(x0), s(x1))
g(x0, c(x1))

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.

G(x, c(y)) → G(x, g(s(c(y)), y))
G(x, c(y)) → G(s(c(y)), y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
G(x1, x2) = G(x2)
c(x1) = c(x1)
g(x1, x2) = g
s(x1) = s
if(x1, x2, x3) = if(x2, x3)
false = false
f(x1) = f(x1)
0 = 0
true = true
1 = 1

Recursive Path Order [2].
Precedence:

c₁ > G₁ > g > [s, if_2]
false > [s, if_2]
f₁ > true > [s, if_2]
0 > [s, if_2]
1 > [s, if_2]

The following usable rules [14] were oriented:

g(x, c(y)) → g(x, g(s(c(y)), y))
if(false, x, y) → y
if(true, x, y) → x
g(s(x), s(y)) → if(f(x), s(x), s(y))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

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

f(0) → true
f(1) → false
f(s(x)) → f(x)
if(true, x, y) → x
if(false, x, y) → y
g(s(x), s(y)) → if(f(x), s(x), s(y))
g(x, c(y)) → g(x, g(s(c(y)), y))

The set Q consists of the following terms:

f(0)
f(1)
f(s(x0))
if(true, x0, x1)
if(false, x0, x1)
g(s(x0), s(x1))
g(x0, c(x1))

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