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 DependencyGraphProof (⇔)
    ↳16 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]. 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

neq1(s(x), s(y)) → neq1(x, y)

The a-transformed usable rules are
none


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.
neq1(x1, x2)  =  x2
s(x1)  =  s(x1)

Lexicographic Path Order [LPO].
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(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(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 remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  APP(x2)
app(x1, x2)  =  app(x1, x2)
filter  =  filter
cons  =  cons
filtersub  =  filtersub
true  =  true
false  =  false
neq  =  neq
0  =  0
s  =  s
nil  =  nil

Lexicographic Path Order [LPO].
Precedence:
APP1 > filter > app2
APP1 > cons > app2
APP1 > filtersub > app2
true > app2
false > app2
neq > app2

The following usable rules [FROCOS05] were oriented: none

(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))

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) DependencyGraphProof (EQUIVALENT transformation)

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

(16) TRUE