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 QDP
↳7 QDPOrderProof (⇔)
↳8 QDP
↳9 QDPOrderProof (⇔)
↳10 QDP
↳11 PisEmptyProof (⇔)
↳12 TRUE

(0) Obligation:

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

app(app(mapt, f), app(leaf, x)) → app(leaf, app(f, x))
app(app(mapt, f), app(node, xs)) → app(node, app(app(maptlist, f), xs))
app(app(maptlist, f), nil) → nil
app(app(maptlist, f), app(app(cons, x), xs)) → app(app(cons, app(app(mapt, f), x)), app(app(maptlist, 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(mapt, f), app(leaf, x)) → app(leaf, app(f, x))
app(app(mapt, f), app(node, xs)) → app(node, app(app(maptlist, f), xs))
app(app(maptlist, f), nil) → nil
app(app(maptlist, f), app(app(cons, x), xs)) → app(app(cons, app(app(mapt, f), x)), app(app(maptlist, f), xs))

The set Q consists of the following terms:

app(app(mapt, x0), app(leaf, x1))
app(app(mapt, x0), app(node, x1))
app(app(maptlist, x0), nil)
app(app(maptlist, 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(mapt, f), app(leaf, x)) → APP(leaf, app(f, x))
APP(app(mapt, f), app(leaf, x)) → APP(f, x)
APP(app(mapt, f), app(node, xs)) → APP(node, app(app(maptlist, f), xs))
APP(app(mapt, f), app(node, xs)) → APP(app(maptlist, f), xs)
APP(app(mapt, f), app(node, xs)) → APP(maptlist, f)
APP(app(maptlist, f), app(app(cons, x), xs)) → APP(app(cons, app(app(mapt, f), x)), app(app(maptlist, f), xs))
APP(app(maptlist, f), app(app(cons, x), xs)) → APP(cons, app(app(mapt, f), x))
APP(app(maptlist, f), app(app(cons, x), xs)) → APP(app(mapt, f), x)
APP(app(maptlist, f), app(app(cons, x), xs)) → APP(mapt, f)
APP(app(maptlist, f), app(app(cons, x), xs)) → APP(app(maptlist, f), xs)

The TRS R consists of the following rules:

app(app(mapt, f), app(leaf, x)) → app(leaf, app(f, x))
app(app(mapt, f), app(node, xs)) → app(node, app(app(maptlist, f), xs))
app(app(maptlist, f), nil) → nil
app(app(maptlist, f), app(app(cons, x), xs)) → app(app(cons, app(app(mapt, f), x)), app(app(maptlist, f), xs))

The set Q consists of the following terms:

app(app(mapt, x0), app(leaf, x1))
app(app(mapt, x0), app(node, x1))
app(app(maptlist, x0), nil)
app(app(maptlist, 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 1 SCC with 6 less nodes.

(6) Obligation:

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

APP(app(mapt, f), app(node, xs)) → APP(app(maptlist, f), xs)
APP(app(maptlist, f), app(app(cons, x), xs)) → APP(app(mapt, f), x)
APP(app(mapt, f), app(leaf, x)) → APP(f, x)
APP(app(maptlist, f), app(app(cons, x), xs)) → APP(app(maptlist, f), xs)

The TRS R consists of the following rules:

app(app(mapt, f), app(leaf, x)) → app(leaf, app(f, x))
app(app(mapt, f), app(node, xs)) → app(node, app(app(maptlist, f), xs))
app(app(maptlist, f), nil) → nil
app(app(maptlist, f), app(app(cons, x), xs)) → app(app(cons, app(app(mapt, f), x)), app(app(maptlist, f), xs))

The set Q consists of the following terms:

app(app(mapt, x0), app(leaf, x1))
app(app(mapt, x0), app(node, x1))
app(app(maptlist, x0), nil)
app(app(maptlist, x0), app(app(cons, x1), x2))

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

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic path order with status [LPO].
Precedence:
mapt > app1
node > app1
maptlist > app1
leaf > APP1 > app1

Status:
APP1: [1]
app1: [1]
mapt: []
node: []
maptlist: []
cons: []
leaf: []

The following usable rules [FROCOS05] were oriented: none

(8) Obligation:

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

APP(app(mapt, f), app(node, xs)) → APP(app(maptlist, f), xs)
APP(app(maptlist, f), app(app(cons, x), xs)) → APP(app(mapt, f), x)
APP(app(maptlist, f), app(app(cons, x), xs)) → APP(app(maptlist, f), xs)

The TRS R consists of the following rules:

app(app(mapt, f), app(leaf, x)) → app(leaf, app(f, x))
app(app(mapt, f), app(node, xs)) → app(node, app(app(maptlist, f), xs))
app(app(maptlist, f), nil) → nil
app(app(maptlist, f), app(app(cons, x), xs)) → app(app(cons, app(app(mapt, f), x)), app(app(maptlist, f), xs))

The set Q consists of the following terms:

app(app(mapt, x0), app(leaf, x1))
app(app(mapt, x0), app(node, x1))
app(app(maptlist, x0), nil)
app(app(maptlist, x0), app(app(cons, x1), x2))

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

(9) 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

mapt1(f, node(xs)) → maptlist1(f, xs)
maptlist1(f, cons(x, xs)) → mapt1(f, x)
maptlist1(f, cons(x, xs)) → maptlist1(f, xs)

The a-transformed usable rules are
none


The following pairs can be oriented strictly and are deleted.


APP(app(mapt, f), app(node, xs)) → APP(app(maptlist, f), xs)
APP(app(maptlist, f), app(app(cons, x), xs)) → APP(app(mapt, f), x)
APP(app(maptlist, f), app(app(cons, x), xs)) → APP(app(maptlist, f), xs)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
mapt1(x1, x2)  =  mapt1(x2)
node(x1)  =  node(x1)
maptlist1(x1, x2)  =  maptlist1(x2)
cons(x1, x2)  =  cons(x1, x2)

Lexicographic path order with status [LPO].
Precedence:
node1 > maptlist11
cons2 > mapt11 > maptlist11

Status:
mapt11: [1]
node1: [1]
maptlist11: [1]
cons2: [1,2]

The following usable rules [FROCOS05] were oriented: none

(10) Obligation:

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

app(app(mapt, f), app(leaf, x)) → app(leaf, app(f, x))
app(app(mapt, f), app(node, xs)) → app(node, app(app(maptlist, f), xs))
app(app(maptlist, f), nil) → nil
app(app(maptlist, f), app(app(cons, x), xs)) → app(app(cons, app(app(mapt, f), x)), app(app(maptlist, f), xs))

The set Q consists of the following terms:

app(app(mapt, x0), app(leaf, x1))
app(app(mapt, x0), app(node, x1))
app(app(maptlist, x0), nil)
app(app(maptlist, x0), app(app(cons, x1), x2))

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

(11) PisEmptyProof (EQUIVALENT transformation)

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

(12) TRUE