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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 UsableRulesProof (⇔)
    ↳7 QDP
    ↳8 UsableRulesReductionPairsProof (⇔)
    ↳9 QDP
    ↳10 MRRProof (⇔)
    ↳11 QDP
    ↳12 MRRProof (⇔)
    ↳13 QDP
    ↳14 DependencyGraphProof (⇔)
    ↳15 TRUE
  ↳16 QDP
    ↳17 UsableRulesProof (⇔)
    ↳18 QDP
    ↳19 ATransformationProof (⇔)
    ↳20 QDP
    ↳21 QDPSizeChangeProof (⇔)
    ↳22 TRUE
  ↳23 QDP
    ↳24 QDPApplicativeOrderProof (⇔)
    ↳25 QDP
    ↳26 PisEmptyProof (⇔)
    ↳27 TRUE
  ↳28 QDP
    ↳29 QDPSizeChangeProof (⇔)
    ↳30 TRUE

(0) Obligation:

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

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

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

(2) Obligation:

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

APP(app(minus, app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(app(minus, app(s, x)), app(s, y)) → APP(minus, x)
APP(app(quot, app(s, x)), app(s, y)) → APP(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
APP(app(quot, app(s, x)), app(s, y)) → APP(app(quot, app(app(minus, x), y)), app(s, y))
APP(app(quot, app(s, x)), app(s, y)) → APP(quot, app(app(minus, x), y))
APP(app(quot, app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(app(quot, app(s, x)), app(s, y)) → APP(minus, x)
APP(app(plus, app(s, x)), y) → APP(s, app(app(plus, x), y))
APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)
APP(app(plus, app(s, x)), y) → APP(plus, x)
APP(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → APP(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
APP(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → APP(plus, app(app(minus, y), app(s, app(s, z))))
APP(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → APP(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
APP(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → APP(plus, app(app(plus, y), app(s, app(s, z))))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), f), x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter2, app(f, x)), f)
APP(app(filter, f), app(app(cons, x), xs)) → APP(filter2, app(f, x))
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(app(app(filter2, false), f), x), xs) → APP(filter, f)

The TRS R consists of the following rules:

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 18 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

APP(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → APP(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)
APP(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → APP(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))

The TRS R consists of the following rules:

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

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

(6) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(7) Obligation:

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

APP(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → APP(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)
APP(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → APP(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))

The TRS R consists of the following rules:

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(minus, x), 0) → x

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

(8) UsableRulesReductionPairsProof (EQUIVALENT transformation)

First, we A-transformed [FROCOS05] the QDP-Problem. Then we obtain the following A-transformed DP problem.
The pairs P are:

plus1(minus(x, s(0)), minus(y, s(s(z)))) → plus1(minus(y, s(s(z))), minus(x, s(0)))
plus1(s(x), y) → plus1(x, y)
plus1(plus(x, s(0)), plus(y, s(s(z)))) → plus1(plus(y, s(s(z))), plus(x, s(0)))

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

plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(plus(x, s(0)), plus(y, s(s(z)))) → plus(plus(y, s(s(z))), plus(x, s(0)))
plus(minus(x, s(0)), minus(y, s(s(z)))) → plus(minus(y, s(s(z))), minus(x, s(0)))
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x

Q is empty.

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(minus(x1, x2)) = x1 + 2·x2   
POL(plus(x1, x2)) = 2·x1 + 2·x2   
POL(plus1(x1, x2)) = x1 + x2   
POL(s(x1)) = x1   

(9) Obligation:

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

plus1(minus(x, s(0)), minus(y, s(s(z)))) → plus1(minus(y, s(s(z))), minus(x, s(0)))
plus1(s(x), y) → plus1(x, y)
plus1(plus(x, s(0)), plus(y, s(s(z)))) → plus1(plus(y, s(s(z))), plus(x, s(0)))

The TRS R consists of the following rules:

plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(plus(x, s(0)), plus(y, s(s(z)))) → plus(plus(y, s(s(z))), plus(x, s(0)))
plus(minus(x, s(0)), minus(y, s(s(z)))) → plus(minus(y, s(s(z))), minus(x, s(0)))
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x

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

(10) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

plus(0, y) → y
minus(x, 0) → x

Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(minus(x1, x2)) = 1 + x1 + x2   
POL(plus(x1, x2)) = 1 + x1 + x2   
POL(plus1(x1, x2)) = x1 + x2   
POL(s(x1)) = x1   

(11) Obligation:

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

plus1(minus(x, s(0)), minus(y, s(s(z)))) → plus1(minus(y, s(s(z))), minus(x, s(0)))
plus1(s(x), y) → plus1(x, y)
plus1(plus(x, s(0)), plus(y, s(s(z)))) → plus1(plus(y, s(s(z))), plus(x, s(0)))

The TRS R consists of the following rules:

plus(s(x), y) → s(plus(x, y))
plus(plus(x, s(0)), plus(y, s(s(z)))) → plus(plus(y, s(s(z))), plus(x, s(0)))
plus(minus(x, s(0)), minus(y, s(s(z)))) → plus(minus(y, s(s(z))), minus(x, s(0)))
minus(s(x), s(y)) → minus(x, y)

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

(12) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

plus1(s(x), y) → plus1(x, y)

Strictly oriented rules of the TRS R:

minus(s(x), s(y)) → minus(x, y)

Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(minus(x1, x2)) = x1 + x2   
POL(plus(x1, x2)) = x1 + x2   
POL(plus1(x1, x2)) = x1 + x2   
POL(s(x1)) = 1 + x1   

(13) Obligation:

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

plus1(minus(x, s(0)), minus(y, s(s(z)))) → plus1(minus(y, s(s(z))), minus(x, s(0)))
plus1(plus(x, s(0)), plus(y, s(s(z)))) → plus1(plus(y, s(s(z))), plus(x, s(0)))

The TRS R consists of the following rules:

plus(s(x), y) → s(plus(x, y))
plus(plus(x, s(0)), plus(y, s(s(z)))) → plus(plus(y, s(s(z))), plus(x, s(0)))
plus(minus(x, s(0)), minus(y, s(s(z)))) → plus(minus(y, s(s(z))), minus(x, s(0)))

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

(14) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(15) TRUE

(16) Obligation:

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

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

The TRS R consists of the following rules:

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

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

(17) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(18) Obligation:

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

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

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

(19) ATransformationProof (EQUIVALENT transformation)

We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem.

(20) Obligation:

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

minus(s(x), s(y)) → minus(x, y)

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

(21) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • minus(s(x), s(y)) → minus(x, y)
    The graph contains the following edges 1 > 1, 2 > 2

(22) TRUE

(23) Obligation:

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

APP(app(quot, app(s, x)), app(s, y)) → APP(app(quot, app(app(minus, x), y)), app(s, y))

The TRS R consists of the following rules:

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

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

(24) QDPApplicativeOrderProof (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

quot1(s(x), s(y)) → quot1(minus(x, y), s(y))

The a-transformed usable rules are

minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x


The following pairs can be oriented strictly and are deleted.


APP(app(quot, app(s, x)), app(s, y)) → APP(app(quot, app(app(minus, x), y)), app(s, y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(minus(x1, x2)) = x1   
POL(quot1(x1, x2)) = x1   
POL(s(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented:

app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(minus, x), 0) → x

(25) Obligation:

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

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

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

(26) PisEmptyProof (EQUIVALENT transformation)

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

(27) TRUE

(28) Obligation:

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

APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)

The TRS R consists of the following rules:

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

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

(29) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
    The graph contains the following edges 1 > 1, 2 > 2

  • APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
    The graph contains the following edges 1 > 1, 2 > 2

  • APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
    The graph contains the following edges 1 >= 1, 2 > 2

  • APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
    The graph contains the following edges 2 > 2

  • APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
    The graph contains the following edges 2 >= 2

  • APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
    The graph contains the following edges 2 >= 2

(30) TRUE