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:

app2(f, 0) -> true
app2(f, 1) -> false
app2(f, app2(s, x)) -> app2(f, x)
app2(app2(app2(if, true), app2(s, x)), app2(s, y)) -> app2(s, x)
app2(app2(app2(if, false), app2(s, x)), app2(s, y)) -> app2(s, y)
app2(app2(g, x), app2(c, y)) -> app2(c, app2(app2(g, x), y))
app2(app2(g, x), app2(c, y)) -> app2(app2(g, x), app2(app2(app2(if, app2(f, x)), app2(c, app2(app2(g, app2(s, x)), y))), app2(c, y)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

app2(f, 0) -> true
app2(f, 1) -> false
app2(f, app2(s, x)) -> app2(f, x)
app2(app2(app2(if, true), app2(s, x)), app2(s, y)) -> app2(s, x)
app2(app2(app2(if, false), app2(s, x)), app2(s, y)) -> app2(s, y)
app2(app2(g, x), app2(c, y)) -> app2(c, app2(app2(g, x), y))
app2(app2(g, x), app2(c, y)) -> app2(app2(g, x), app2(app2(app2(if, app2(f, x)), app2(c, app2(app2(g, app2(s, x)), y))), app2(c, y)))

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:

APP2(app2(g, x), app2(c, y)) -> APP2(g, app2(s, x))
APP2(f, app2(s, x)) -> APP2(f, x)
APP2(app2(g, x), app2(c, y)) -> APP2(f, x)
APP2(app2(g, x), app2(c, y)) -> APP2(s, x)
APP2(app2(g, x), app2(c, y)) -> APP2(app2(g, x), y)
APP2(app2(g, x), app2(c, y)) -> APP2(c, app2(app2(g, x), y))
APP2(app2(g, x), app2(c, y)) -> APP2(app2(if, app2(f, x)), app2(c, app2(app2(g, app2(s, x)), y)))
APP2(app2(g, x), app2(c, y)) -> APP2(app2(g, app2(s, x)), y)
APP2(app2(g, x), app2(c, y)) -> APP2(app2(app2(if, app2(f, x)), app2(c, app2(app2(g, app2(s, x)), y))), app2(c, y))
APP2(app2(g, x), app2(c, y)) -> APP2(app2(g, x), app2(app2(app2(if, app2(f, x)), app2(c, app2(app2(g, app2(s, x)), y))), app2(c, y)))
APP2(app2(g, x), app2(c, y)) -> APP2(c, app2(app2(g, app2(s, x)), y))
APP2(app2(g, x), app2(c, y)) -> APP2(if, app2(f, x))

The TRS R consists of the following rules:

app2(f, 0) -> true
app2(f, 1) -> false
app2(f, app2(s, x)) -> app2(f, x)
app2(app2(app2(if, true), app2(s, x)), app2(s, y)) -> app2(s, x)
app2(app2(app2(if, false), app2(s, x)), app2(s, y)) -> app2(s, y)
app2(app2(g, x), app2(c, y)) -> app2(c, app2(app2(g, x), y))
app2(app2(g, x), app2(c, y)) -> app2(app2(g, x), app2(app2(app2(if, app2(f, x)), app2(c, app2(app2(g, app2(s, x)), y))), app2(c, y)))

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:

APP2(app2(g, x), app2(c, y)) -> APP2(g, app2(s, x))
APP2(f, app2(s, x)) -> APP2(f, x)
APP2(app2(g, x), app2(c, y)) -> APP2(f, x)
APP2(app2(g, x), app2(c, y)) -> APP2(s, x)
APP2(app2(g, x), app2(c, y)) -> APP2(app2(g, x), y)
APP2(app2(g, x), app2(c, y)) -> APP2(c, app2(app2(g, x), y))
APP2(app2(g, x), app2(c, y)) -> APP2(app2(if, app2(f, x)), app2(c, app2(app2(g, app2(s, x)), y)))
APP2(app2(g, x), app2(c, y)) -> APP2(app2(g, app2(s, x)), y)
APP2(app2(g, x), app2(c, y)) -> APP2(app2(app2(if, app2(f, x)), app2(c, app2(app2(g, app2(s, x)), y))), app2(c, y))
APP2(app2(g, x), app2(c, y)) -> APP2(app2(g, x), app2(app2(app2(if, app2(f, x)), app2(c, app2(app2(g, app2(s, x)), y))), app2(c, y)))
APP2(app2(g, x), app2(c, y)) -> APP2(c, app2(app2(g, app2(s, x)), y))
APP2(app2(g, x), app2(c, y)) -> APP2(if, app2(f, x))

The TRS R consists of the following rules:

app2(f, 0) -> true
app2(f, 1) -> false
app2(f, app2(s, x)) -> app2(f, x)
app2(app2(app2(if, true), app2(s, x)), app2(s, y)) -> app2(s, x)
app2(app2(app2(if, false), app2(s, x)), app2(s, y)) -> app2(s, y)
app2(app2(g, x), app2(c, y)) -> app2(c, app2(app2(g, x), y))
app2(app2(g, x), app2(c, y)) -> app2(app2(g, x), app2(app2(app2(if, app2(f, x)), app2(c, app2(app2(g, app2(s, x)), y))), app2(c, y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 9 less nodes.

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

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

APP2(f, app2(s, x)) -> APP2(f, x)

The TRS R consists of the following rules:

app2(f, 0) -> true
app2(f, 1) -> false
app2(f, app2(s, x)) -> app2(f, x)
app2(app2(app2(if, true), app2(s, x)), app2(s, y)) -> app2(s, x)
app2(app2(app2(if, false), app2(s, x)), app2(s, y)) -> app2(s, y)
app2(app2(g, x), app2(c, y)) -> app2(c, app2(app2(g, x), y))
app2(app2(g, x), app2(c, y)) -> app2(app2(g, x), app2(app2(app2(if, app2(f, x)), app2(c, app2(app2(g, app2(s, x)), y))), app2(c, y)))

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 strictly oriented and are deleted.


APP2(f, app2(s, x)) -> APP2(f, x)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP2(x1, x2)  =  APP1(x2)
f  =  f
app2(x1, x2)  =  app1(x2)
s  =  s

Lexicographic Path Order [19].
Precedence:
[APP1, app1] > f
s > f


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

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

app2(f, 0) -> true
app2(f, 1) -> false
app2(f, app2(s, x)) -> app2(f, x)
app2(app2(app2(if, true), app2(s, x)), app2(s, y)) -> app2(s, x)
app2(app2(app2(if, false), app2(s, x)), app2(s, y)) -> app2(s, y)
app2(app2(g, x), app2(c, y)) -> app2(c, app2(app2(g, x), y))
app2(app2(g, x), app2(c, y)) -> app2(app2(g, x), app2(app2(app2(if, app2(f, x)), app2(c, app2(app2(g, app2(s, x)), y))), app2(c, y)))

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

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

APP2(app2(g, x), app2(c, y)) -> APP2(app2(g, x), y)
APP2(app2(g, x), app2(c, y)) -> APP2(app2(g, app2(s, x)), y)

The TRS R consists of the following rules:

app2(f, 0) -> true
app2(f, 1) -> false
app2(f, app2(s, x)) -> app2(f, x)
app2(app2(app2(if, true), app2(s, x)), app2(s, y)) -> app2(s, x)
app2(app2(app2(if, false), app2(s, x)), app2(s, y)) -> app2(s, y)
app2(app2(g, x), app2(c, y)) -> app2(c, app2(app2(g, x), y))
app2(app2(g, x), app2(c, y)) -> app2(app2(g, x), app2(app2(app2(if, app2(f, x)), app2(c, app2(app2(g, app2(s, x)), y))), app2(c, y)))

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 strictly oriented and are deleted.


APP2(app2(g, x), app2(c, y)) -> APP2(app2(g, x), y)
APP2(app2(g, x), app2(c, y)) -> APP2(app2(g, app2(s, x)), y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP2(x1, x2)  =  APP1(x2)
app2(x1, x2)  =  app1(x2)
g  =  g
c  =  c
s  =  s

Lexicographic Path Order [19].
Precedence:
APP1 > app1 > g
c > app1 > g
s > g


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof

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

app2(f, 0) -> true
app2(f, 1) -> false
app2(f, app2(s, x)) -> app2(f, x)
app2(app2(app2(if, true), app2(s, x)), app2(s, y)) -> app2(s, x)
app2(app2(app2(if, false), app2(s, x)), app2(s, y)) -> app2(s, y)
app2(app2(g, x), app2(c, y)) -> app2(c, app2(app2(g, x), y))
app2(app2(g, x), app2(c, y)) -> app2(app2(g, x), app2(app2(app2(if, app2(f, x)), app2(c, app2(app2(g, app2(s, x)), y))), app2(c, y)))

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.