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(app(consif, true), x), ys) → app(app(cons, x), ys)
app(app(app(consif, false), x), ys) → ys
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(consif, app(f, x)), x), app(app(filter, 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(app(consif, true), x), ys) → app(app(cons, x), ys)
app(app(app(consif, false), x), ys) → ys
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(consif, app(f, x)), x), app(app(filter, f), xs))

The set Q consists of the following terms:

app(app(app(consif, true), x0), x1)
app(app(app(consif, false), x0), x1)
app(app(filter, x0), nil)
app(app(filter, 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(app(consif, true), x), ys) → APP(app(cons, x), ys)
APP(app(app(consif, true), x), ys) → APP(cons, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(consif, app(f, x)), x), app(app(filter, f), xs))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(consif, app(f, x)), x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(consif, app(f, x))
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter, f), xs)

The TRS R consists of the following rules:

app(app(app(consif, true), x), ys) → app(app(cons, x), ys)
app(app(app(consif, false), x), ys) → ys
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(consif, app(f, x)), x), app(app(filter, f), xs))

The set Q consists of the following terms:

app(app(app(consif, true), x0), x1)
app(app(app(consif, false), x0), x1)
app(app(filter, x0), nil)
app(app(filter, 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 5 less nodes.

(6) Obligation:

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

APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)

The TRS R consists of the following rules:

app(app(app(consif, true), x), ys) → app(app(cons, x), ys)
app(app(app(consif, false), x), ys) → ys
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(consif, app(f, x)), x), app(app(filter, f), xs))

The set Q consists of the following terms:

app(app(app(consif, true), x0), x1)
app(app(app(consif, false), x0), x1)
app(app(filter, x0), nil)
app(app(filter, 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(filter, f), app(app(cons, x), xs)) → 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)
filter  =  filter
cons  =  cons
consif  =  consif
true  =  true
false  =  false
nil  =  nil

Lexicographic path order with status [LPO].
Precedence:
APP1 > app1
APP1 > filter
cons > app1
cons > filter
consif > app1
true > app1

Status:
APP1: [1]
consif: []
cons: []
true: []
false: []
app1: [1]
filter: []
nil: []

The following usable rules [FROCOS05] were oriented:

app(app(app(consif, true), x), ys) → app(app(cons, x), ys)
app(app(app(consif, false), x), ys) → ys
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(consif, app(f, x)), x), app(app(filter, f), xs))

(8) Obligation:

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

APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter, f), xs)

The TRS R consists of the following rules:

app(app(app(consif, true), x), ys) → app(app(cons, x), ys)
app(app(app(consif, false), x), ys) → ys
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(consif, app(f, x)), x), app(app(filter, f), xs))

The set Q consists of the following terms:

app(app(app(consif, true), x0), x1)
app(app(app(consif, false), x0), x1)
app(app(filter, x0), nil)
app(app(filter, 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].


The following pairs can be oriented strictly and are deleted.


APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter, 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(x2)
filter  =  filter
cons  =  cons
consif  =  consif
true  =  true
false  =  false
nil  =  nil

Lexicographic path order with status [LPO].
Precedence:
cons > app1
cons > filter
consif > app1

Status:
consif: []
cons: []
true: []
false: []
app1: [1]
nil: []
filter: []

The following usable rules [FROCOS05] were oriented:

app(app(app(consif, true), x), ys) → app(app(cons, x), ys)
app(app(app(consif, false), x), ys) → ys
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(consif, app(f, x)), x), app(app(filter, f), xs))

(10) Obligation:

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

app(app(app(consif, true), x), ys) → app(app(cons, x), ys)
app(app(app(consif, false), x), ys) → ys
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(consif, app(f, x)), x), app(app(filter, f), xs))

The set Q consists of the following terms:

app(app(app(consif, true), x0), x1)
app(app(app(consif, false), x0), x1)
app(app(filter, x0), nil)
app(app(filter, 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