Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 PisEmptyProof (⇔)
    ↳11 TRUE
  ↳12 QDP
    ↳13 QDPOrderProof (⇔)
    ↳14 QDP
    ↳15 PisEmptyProof (⇔)
    ↳16 TRUE

(0) Obligation:

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

app(app(append, nil), ys) → ys
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))
app(app(flatwith, f), app(leaf, x)) → app(app(cons, app(f, x)), nil)
app(app(flatwith, f), app(node, xs)) → app(app(flatwithsub, f), xs)
app(app(flatwithsub, f), nil) → nil
app(app(flatwithsub, f), app(app(cons, x), xs)) → app(app(append, app(app(flatwith, f), x)), app(app(flatwithsub, f), xs))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

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

(2) Obligation:

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

app(app(append, nil), ys) → ys
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))
app(app(flatwith, f), app(leaf, x)) → app(app(cons, app(f, x)), nil)
app(app(flatwith, f), app(node, xs)) → app(app(flatwithsub, f), xs)
app(app(flatwithsub, f), nil) → nil
app(app(flatwithsub, f), app(app(cons, x), xs)) → app(app(append, app(app(flatwith, f), x)), app(app(flatwithsub, f), xs))

The set Q consists of the following terms:

app(app(append, nil), x0)
app(app(append, app(app(cons, x0), x1)), x2)
app(app(flatwith, x0), app(leaf, x1))
app(app(flatwith, x0), app(node, x1))
app(app(flatwithsub, x0), nil)
app(app(flatwithsub, x0), app(app(cons, x1), x2))

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

APP(app(append, app(app(cons, x), xs)), ys) → APP(app(cons, x), app(app(append, xs), ys))
APP(app(append, app(app(cons, x), xs)), ys) → APP(app(append, xs), ys)
APP(app(append, app(app(cons, x), xs)), ys) → APP(append, xs)
APP(app(flatwith, f), app(leaf, x)) → APP(app(cons, app(f, x)), nil)
APP(app(flatwith, f), app(leaf, x)) → APP(cons, app(f, x))
APP(app(flatwith, f), app(leaf, x)) → APP(f, x)
APP(app(flatwith, f), app(node, xs)) → APP(app(flatwithsub, f), xs)
APP(app(flatwith, f), app(node, xs)) → APP(flatwithsub, f)
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(app(append, app(app(flatwith, f), x)), app(app(flatwithsub, f), xs))
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(append, app(app(flatwith, f), x))
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(app(flatwith, f), x)
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(flatwith, f)
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(app(flatwithsub, f), xs)

The TRS R consists of the following rules:

app(app(append, nil), ys) → ys
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))
app(app(flatwith, f), app(leaf, x)) → app(app(cons, app(f, x)), nil)
app(app(flatwith, f), app(node, xs)) → app(app(flatwithsub, f), xs)
app(app(flatwithsub, f), nil) → nil
app(app(flatwithsub, f), app(app(cons, x), xs)) → app(app(append, app(app(flatwith, f), x)), app(app(flatwithsub, f), xs))

The set Q consists of the following terms:

app(app(append, nil), x0)
app(app(append, app(app(cons, x0), x1)), x2)
app(app(flatwith, x0), app(leaf, x1))
app(app(flatwith, x0), app(node, x1))
app(app(flatwithsub, x0), nil)
app(app(flatwithsub, x0), app(app(cons, x1), x2))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 8 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

APP(app(append, app(app(cons, x), xs)), ys) → APP(app(append, xs), ys)

The TRS R consists of the following rules:

app(app(append, nil), ys) → ys
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))
app(app(flatwith, f), app(leaf, x)) → app(app(cons, app(f, x)), nil)
app(app(flatwith, f), app(node, xs)) → app(app(flatwithsub, f), xs)
app(app(flatwithsub, f), nil) → nil
app(app(flatwithsub, f), app(app(cons, x), xs)) → app(app(append, app(app(flatwith, f), x)), app(app(flatwithsub, f), xs))

The set Q consists of the following terms:

app(app(append, nil), x0)
app(app(append, app(app(cons, x0), x1)), x2)
app(app(flatwith, x0), app(leaf, x1))
app(app(flatwith, x0), app(node, x1))
app(app(flatwithsub, x0), nil)
app(app(flatwithsub, x0), app(app(cons, x1), x2))

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04]. Here, we combined the reduction pair processor with the A-transformation [FROCOS05] which results in the following intermediate Q-DP Problem.
The a-transformed P is

append1(cons(x, xs), ys) → append1(xs, ys)

The a-transformed usable rules are
none


The following pairs can be oriented strictly and are deleted.


APP(app(append, app(app(cons, x), xs)), ys) → APP(app(append, xs), ys)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
append1(x1, x2)  =  x1
cons(x1, x2)  =  cons(x2)

Recursive Path Order [RPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

app(app(append, nil), ys) → ys
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))
app(app(flatwith, f), app(leaf, x)) → app(app(cons, app(f, x)), nil)
app(app(flatwith, f), app(node, xs)) → app(app(flatwithsub, f), xs)
app(app(flatwithsub, f), nil) → nil
app(app(flatwithsub, f), app(app(cons, x), xs)) → app(app(append, app(app(flatwith, f), x)), app(app(flatwithsub, f), xs))

The set Q consists of the following terms:

app(app(append, nil), x0)
app(app(append, app(app(cons, x0), x1)), x2)
app(app(flatwith, x0), app(leaf, x1))
app(app(flatwith, x0), app(node, x1))
app(app(flatwithsub, x0), nil)
app(app(flatwithsub, x0), app(app(cons, x1), x2))

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

(10) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(11) TRUE

(12) Obligation:

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

APP(app(flatwith, f), app(node, xs)) → APP(app(flatwithsub, f), xs)
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(app(flatwith, f), x)
APP(app(flatwith, f), app(leaf, x)) → APP(f, x)
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(app(flatwithsub, f), xs)

The TRS R consists of the following rules:

app(app(append, nil), ys) → ys
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))
app(app(flatwith, f), app(leaf, x)) → app(app(cons, app(f, x)), nil)
app(app(flatwith, f), app(node, xs)) → app(app(flatwithsub, f), xs)
app(app(flatwithsub, f), nil) → nil
app(app(flatwithsub, f), app(app(cons, x), xs)) → app(app(append, app(app(flatwith, f), x)), app(app(flatwithsub, f), xs))

The set Q consists of the following terms:

app(app(append, nil), x0)
app(app(append, app(app(cons, x0), x1)), x2)
app(app(flatwith, x0), app(leaf, x1))
app(app(flatwith, x0), app(node, x1))
app(app(flatwithsub, x0), nil)
app(app(flatwithsub, x0), app(app(cons, x1), x2))

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


APP(app(flatwith, f), app(node, xs)) → APP(app(flatwithsub, f), xs)
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(app(flatwith, f), x)
APP(app(flatwith, f), app(leaf, x)) → APP(f, x)
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(app(flatwithsub, f), xs)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  x2
app(x1, x2)  =  app(x1, x2)
node  =  node
cons  =  cons
leaf  =  leaf

Recursive Path Order [RPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

app(app(append, nil), ys) → ys
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))
app(app(flatwith, f), app(leaf, x)) → app(app(cons, app(f, x)), nil)
app(app(flatwith, f), app(node, xs)) → app(app(flatwithsub, f), xs)
app(app(flatwithsub, f), nil) → nil
app(app(flatwithsub, f), app(app(cons, x), xs)) → app(app(append, app(app(flatwith, f), x)), app(app(flatwithsub, f), xs))

The set Q consists of the following terms:

app(app(append, nil), x0)
app(app(append, app(app(cons, x0), x1)), x2)
app(app(flatwith, x0), app(leaf, x1))
app(app(flatwith, x0), app(node, x1))
app(app(flatwithsub, x0), nil)
app(app(flatwithsub, x0), app(app(cons, x1), x2))

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

(15) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(16) TRUE