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 QDPOrderProof (⇔)
    ↳16 QDP
    ↳17 DependencyGraphProof (⇔)
    ↳18 TRUE

(0) Obligation:

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

app(app(neq, 0), 0) → false
app(app(neq, 0), app(s, y)) → true
app(app(neq, app(s, x)), 0) → true
app(app(neq, app(s, x)), app(s, y)) → app(app(neq, x), y)
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, y), ys)) → app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) → app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) → app(app(filter, f), ys)
nonzero → app(filter, app(neq, 0))

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(neq, 0), 0) → false
app(app(neq, 0), app(s, y)) → true
app(app(neq, app(s, x)), 0) → true
app(app(neq, app(s, x)), app(s, y)) → app(app(neq, x), y)
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, y), ys)) → app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) → app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) → app(app(filter, f), ys)
nonzero → app(filter, app(neq, 0))

The set Q consists of the following terms:

app(app(neq, 0), 0)
app(app(neq, 0), app(s, x0))
app(app(neq, app(s, x0)), 0)
app(app(neq, app(s, x0)), app(s, x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(filtersub, true), x0), app(app(cons, x1), x2))
app(app(app(filtersub, false), x0), app(app(cons, x1), x2))
nonzero

(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(neq, app(s, x)), app(s, y)) → APP(app(neq, x), y)
APP(app(neq, app(s, x)), app(s, y)) → APP(neq, x)
APP(app(filter, f), app(app(cons, y), ys)) → APP(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
APP(app(filter, f), app(app(cons, y), ys)) → APP(app(filtersub, app(f, y)), f)
APP(app(filter, f), app(app(cons, y), ys)) → APP(filtersub, app(f, y))
APP(app(filter, f), app(app(cons, y), ys)) → APP(f, y)
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) → APP(app(cons, y), app(app(filter, f), ys))
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) → APP(app(filter, f), ys)
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) → APP(filter, f)
APP(app(app(filtersub, false), f), app(app(cons, y), ys)) → APP(app(filter, f), ys)
APP(app(app(filtersub, false), f), app(app(cons, y), ys)) → APP(filter, f)
NONZERO → APP(filter, app(neq, 0))
NONZERO → APP(neq, 0)

The TRS R consists of the following rules:

app(app(neq, 0), 0) → false
app(app(neq, 0), app(s, y)) → true
app(app(neq, app(s, x)), 0) → true
app(app(neq, app(s, x)), app(s, y)) → app(app(neq, x), y)
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, y), ys)) → app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) → app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) → app(app(filter, f), ys)
nonzero → app(filter, app(neq, 0))

The set Q consists of the following terms:

app(app(neq, 0), 0)
app(app(neq, 0), app(s, x0))
app(app(neq, app(s, x0)), 0)
app(app(neq, app(s, x0)), app(s, x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(filtersub, true), x0), app(app(cons, x1), x2))
app(app(app(filtersub, false), x0), app(app(cons, x1), x2))
nonzero

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(neq, app(s, x)), app(s, y)) → APP(app(neq, x), y)

The TRS R consists of the following rules:

app(app(neq, 0), 0) → false
app(app(neq, 0), app(s, y)) → true
app(app(neq, app(s, x)), 0) → true
app(app(neq, app(s, x)), app(s, y)) → app(app(neq, x), y)
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, y), ys)) → app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) → app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) → app(app(filter, f), ys)
nonzero → app(filter, app(neq, 0))

The set Q consists of the following terms:

app(app(neq, 0), 0)
app(app(neq, 0), app(s, x0))
app(app(neq, app(s, x0)), 0)
app(app(neq, app(s, x0)), app(s, x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(filtersub, true), x0), app(app(cons, x1), x2))
app(app(app(filtersub, false), x0), app(app(cons, x1), x2))
nonzero

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


APP(app(neq, app(s, x)), app(s, y)) → APP(app(neq, x), y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  APP(x1, x2)
app(x1, x2)  =  app(x2)
neq  =  neq
s  =  s
0  =  0
false  =  false
true  =  true
filter  =  filter
nil  =  nil
cons  =  cons
filtersub  =  filtersub
nonzero  =  nonzero

Lexicographic path order with status [LPO].
Precedence:
s > app1 > true > APP2
s > neq > true > APP2
nil > APP2
cons > app1 > true > APP2
filtersub > app1 > true > APP2
nonzero > neq > true > APP2
nonzero > 0 > false > app1 > true > APP2
nonzero > filter > app1 > true > APP2

Status:
true: []
app1: [1]
s: []
0: []
filter: []
APP2: [1,2]
cons: []
false: []
neq: []
filtersub: []
nonzero: []
nil: []

The following usable rules [FROCOS05] were oriented:

app(app(neq, 0), 0) → false
app(app(neq, 0), app(s, y)) → true
app(app(neq, app(s, x)), 0) → true
app(app(neq, app(s, x)), app(s, y)) → app(app(neq, x), y)
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, y), ys)) → app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) → app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) → app(app(filter, f), ys)
nonzero → app(filter, app(neq, 0))

(9) Obligation:

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

app(app(neq, 0), 0) → false
app(app(neq, 0), app(s, y)) → true
app(app(neq, app(s, x)), 0) → true
app(app(neq, app(s, x)), app(s, y)) → app(app(neq, x), y)
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, y), ys)) → app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) → app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) → app(app(filter, f), ys)
nonzero → app(filter, app(neq, 0))

The set Q consists of the following terms:

app(app(neq, 0), 0)
app(app(neq, 0), app(s, x0))
app(app(neq, app(s, x0)), 0)
app(app(neq, app(s, x0)), app(s, x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(filtersub, true), x0), app(app(cons, x1), x2))
app(app(app(filtersub, false), x0), app(app(cons, x1), x2))
nonzero

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(filter, f), app(app(cons, y), ys)) → APP(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) → APP(app(filter, f), ys)
APP(app(filter, f), app(app(cons, y), ys)) → APP(f, y)
APP(app(app(filtersub, false), f), app(app(cons, y), ys)) → APP(app(filter, f), ys)

The TRS R consists of the following rules:

app(app(neq, 0), 0) → false
app(app(neq, 0), app(s, y)) → true
app(app(neq, app(s, x)), 0) → true
app(app(neq, app(s, x)), app(s, y)) → app(app(neq, x), y)
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, y), ys)) → app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) → app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) → app(app(filter, f), ys)
nonzero → app(filter, app(neq, 0))

The set Q consists of the following terms:

app(app(neq, 0), 0)
app(app(neq, 0), app(s, x0))
app(app(neq, app(s, x0)), 0)
app(app(neq, app(s, x0)), app(s, x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(filtersub, true), x0), app(app(cons, x1), x2))
app(app(app(filtersub, false), x0), app(app(cons, x1), x2))
nonzero

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(filter, f), app(app(cons, y), ys)) → APP(f, y)
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
filtersub  =  filtersub
true  =  true
false  =  false
neq  =  neq
0  =  0
s  =  s
nil  =  nil
nonzero  =  nonzero

Lexicographic path order with status [LPO].
Precedence:
APP1 > cons > app1 > false > nil
APP1 > cons > app1 > neq > true > nil
APP1 > cons > filter > nil
APP1 > cons > filtersub > nil
s > app1 > false > nil
s > app1 > neq > true > nil
nonzero > app1 > false > nil
nonzero > app1 > neq > true > nil
nonzero > filter > nil
nonzero > 0 > true > nil

Status:
APP1: [1]
true: []
app1: [1]
s: []
0: []
filter: []
cons: []
false: []
filtersub: []
neq: []
nonzero: []
nil: []

The following usable rules [FROCOS05] were oriented:

app(app(neq, 0), 0) → false
app(app(neq, 0), app(s, y)) → true
app(app(neq, app(s, x)), 0) → true
app(app(neq, app(s, x)), app(s, y)) → app(app(neq, x), y)
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, y), ys)) → app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) → app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) → app(app(filter, f), ys)
nonzero → app(filter, app(neq, 0))

(14) Obligation:

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

APP(app(filter, f), app(app(cons, y), ys)) → APP(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) → APP(app(filter, f), ys)
APP(app(app(filtersub, false), f), app(app(cons, y), ys)) → APP(app(filter, f), ys)

The TRS R consists of the following rules:

app(app(neq, 0), 0) → false
app(app(neq, 0), app(s, y)) → true
app(app(neq, app(s, x)), 0) → true
app(app(neq, app(s, x)), app(s, y)) → app(app(neq, x), y)
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, y), ys)) → app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) → app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) → app(app(filter, f), ys)
nonzero → app(filter, app(neq, 0))

The set Q consists of the following terms:

app(app(neq, 0), 0)
app(app(neq, 0), app(s, x0))
app(app(neq, app(s, x0)), 0)
app(app(neq, app(s, x0)), app(s, x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(filtersub, true), x0), app(app(cons, x1), x2))
app(app(app(filtersub, false), x0), app(app(cons, x1), x2))
nonzero

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

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


APP(app(app(filtersub, true), f), app(app(cons, y), ys)) → APP(app(filter, f), ys)
APP(app(app(filtersub, false), f), app(app(cons, y), ys)) → APP(app(filter, f), ys)
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
filtersub  =  filtersub
true  =  true
false  =  false
neq  =  neq
0  =  0
s  =  s
nil  =  nil
nonzero  =  nonzero

Lexicographic path order with status [LPO].
Precedence:
filtersub > filter > app1 > true > neq
filtersub > cons > app1 > true > neq
s > neq
nil > neq
nonzero > filter > app1 > true > neq
nonzero > 0 > true > neq
nonzero > 0 > false > neq

Status:
cons: []
true: []
false: []
app1: [1]
s: []
filtersub: []
neq: []
0: []
nonzero: []
filter: []
nil: []

The following usable rules [FROCOS05] were oriented:

app(app(neq, 0), 0) → false
app(app(neq, 0), app(s, y)) → true
app(app(neq, app(s, x)), 0) → true
app(app(neq, app(s, x)), app(s, y)) → app(app(neq, x), y)
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, y), ys)) → app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) → app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) → app(app(filter, f), ys)
nonzero → app(filter, app(neq, 0))

(16) Obligation:

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

APP(app(filter, f), app(app(cons, y), ys)) → APP(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))

The TRS R consists of the following rules:

app(app(neq, 0), 0) → false
app(app(neq, 0), app(s, y)) → true
app(app(neq, app(s, x)), 0) → true
app(app(neq, app(s, x)), app(s, y)) → app(app(neq, x), y)
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, y), ys)) → app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) → app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) → app(app(filter, f), ys)
nonzero → app(filter, app(neq, 0))

The set Q consists of the following terms:

app(app(neq, 0), 0)
app(app(neq, 0), app(s, x0))
app(app(neq, app(s, x0)), 0)
app(app(neq, app(s, x0)), app(s, x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(filtersub, true), x0), app(app(cons, x1), x2))
app(app(app(filtersub, false), x0), app(app(cons, x1), x2))
nonzero

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

(17) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(18) TRUE