(0) Obligation:

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

app(p, app(s, x)) → x
app(fact, 0) → app(s, 0)
app(fact, app(s, x)) → app(app(*, app(s, x)), app(fact, app(p, app(s, x))))
app(app(*, 0), y) → 0
app(app(*, app(s, x)), y) → app(app(+, app(app(*, x), y)), y)
app(app(+, x), 0) → x
app(app(+, x), app(s, y)) → app(s, app(app(+, x), y))
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) 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(p, app(s, x)) → x
app(fact, 0) → app(s, 0)
app(fact, app(s, x)) → app(app(*, app(s, x)), app(fact, app(p, app(s, x))))
app(app(*, 0), y) → 0
app(app(*, app(s, x)), y) → app(app(+, app(app(*, x), y)), y)
app(app(+, x), 0) → x
app(app(+, x), app(s, y)) → app(s, app(app(+, x), y))
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)

The set Q consists of the following terms:

app(p, app(s, x0))
app(fact, 0)
app(fact, app(s, x0))
app(app(*, 0), x0)
app(app(*, app(s, x0)), x1)
app(app(+, x0), 0)
app(app(+, x0), app(s, x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), 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(fact, 0) → APP(s, 0)
APP(fact, app(s, x)) → APP(app(*, app(s, x)), app(fact, app(p, app(s, x))))
APP(fact, app(s, x)) → APP(*, app(s, x))
APP(fact, app(s, x)) → APP(fact, app(p, app(s, x)))
APP(fact, app(s, x)) → APP(p, app(s, x))
APP(app(*, app(s, x)), y) → APP(app(+, app(app(*, x), y)), y)
APP(app(*, app(s, x)), y) → APP(+, app(app(*, x), y))
APP(app(*, app(s, x)), y) → APP(app(*, x), y)
APP(app(*, app(s, x)), y) → APP(*, x)
APP(app(+, x), app(s, y)) → APP(s, app(app(+, x), y))
APP(app(+, x), app(s, y)) → APP(app(+, x), y)
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(p, app(s, x)) → x
app(fact, 0) → app(s, 0)
app(fact, app(s, x)) → app(app(*, app(s, x)), app(fact, app(p, app(s, x))))
app(app(*, 0), y) → 0
app(app(*, app(s, x)), y) → app(app(+, app(app(*, x), y)), y)
app(app(+, x), 0) → x
app(app(+, x), app(s, y)) → app(s, app(app(+, x), y))
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)

The set Q consists of the following terms:

app(p, app(s, x0))
app(fact, 0)
app(fact, app(s, x0))
app(app(*, 0), x0)
app(app(*, app(s, x0)), x1)
app(app(+, x0), 0)
app(app(+, x0), app(s, x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), 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 4 SCCs with 17 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

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

The TRS R consists of the following rules:

app(p, app(s, x)) → x
app(fact, 0) → app(s, 0)
app(fact, app(s, x)) → app(app(*, app(s, x)), app(fact, app(p, app(s, x))))
app(app(*, 0), y) → 0
app(app(*, app(s, x)), y) → app(app(+, app(app(*, x), y)), y)
app(app(+, x), 0) → x
app(app(+, x), app(s, y)) → app(s, app(app(+, x), y))
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)

The set Q consists of the following terms:

app(p, app(s, x0))
app(fact, 0)
app(fact, app(s, x0))
app(app(*, 0), x0)
app(app(*, app(s, x0)), x1)
app(app(+, x0), 0)
app(app(+, x0), app(s, x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

(8) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(9) Obligation:

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

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

R is empty.
The set Q consists of the following terms:

app(p, app(s, x0))
app(fact, 0)
app(fact, app(s, x0))
app(app(*, 0), x0)
app(app(*, app(s, x0)), x1)
app(app(+, x0), 0)
app(app(+, x0), app(s, x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

(10) ATransformationProof (EQUIVALENT transformation)

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

(11) Obligation:

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

+1(x, s(y)) → +1(x, y)

R is empty.
The set Q consists of the following terms:

p(s(x0))
fact(0)
fact(s(x0))
*(0, x0)
*(s(x0), x1)
+(x0, 0)
+(x0, s(x1))
map(x0, nil)
map(x0, cons(x1, x2))
filter(x0, nil)
filter(x0, cons(x1, x2))
filter2(true, x0, x1, x2)
filter2(false, x0, x1, x2)

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

(12) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

p(s(x0))
fact(0)
fact(s(x0))
*(0, x0)
*(s(x0), x1)
+(x0, 0)
+(x0, s(x1))
map(x0, nil)
map(x0, cons(x1, x2))
filter(x0, nil)
filter(x0, cons(x1, x2))
filter2(true, x0, x1, x2)
filter2(false, x0, x1, x2)

(13) Obligation:

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

+1(x, s(y)) → +1(x, y)

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

(14) 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:

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

(15) TRUE

(16) Obligation:

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

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

The TRS R consists of the following rules:

app(p, app(s, x)) → x
app(fact, 0) → app(s, 0)
app(fact, app(s, x)) → app(app(*, app(s, x)), app(fact, app(p, app(s, x))))
app(app(*, 0), y) → 0
app(app(*, app(s, x)), y) → app(app(+, app(app(*, x), y)), y)
app(app(+, x), 0) → x
app(app(+, x), app(s, y)) → app(s, app(app(+, x), y))
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)

The set Q consists of the following terms:

app(p, app(s, x0))
app(fact, 0)
app(fact, app(s, x0))
app(app(*, 0), x0)
app(app(*, app(s, x0)), x1)
app(app(+, x0), 0)
app(app(+, x0), app(s, x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

(17) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] 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(*, app(s, x)), y) → APP(app(*, x), y)

R is empty.
The set Q consists of the following terms:

app(p, app(s, x0))
app(fact, 0)
app(fact, app(s, x0))
app(app(*, 0), x0)
app(app(*, app(s, x0)), x1)
app(app(+, x0), 0)
app(app(+, x0), app(s, x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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:

*1(s(x), y) → *1(x, y)

R is empty.
The set Q consists of the following terms:

p(s(x0))
fact(0)
fact(s(x0))
*(0, x0)
*(s(x0), x1)
+(x0, 0)
+(x0, s(x1))
map(x0, nil)
map(x0, cons(x1, x2))
filter(x0, nil)
filter(x0, cons(x1, x2))
filter2(true, x0, x1, x2)
filter2(false, x0, x1, x2)

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

(21) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

p(s(x0))
fact(0)
fact(s(x0))
*(0, x0)
*(s(x0), x1)
+(x0, 0)
+(x0, s(x1))
map(x0, nil)
map(x0, cons(x1, x2))
filter(x0, nil)
filter(x0, cons(x1, x2))
filter2(true, x0, x1, x2)
filter2(false, x0, x1, x2)

(22) Obligation:

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

*1(s(x), y) → *1(x, y)

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

(23) 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:

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

(24) TRUE

(25) Obligation:

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

APP(fact, app(s, x)) → APP(fact, app(p, app(s, x)))

The TRS R consists of the following rules:

app(p, app(s, x)) → x
app(fact, 0) → app(s, 0)
app(fact, app(s, x)) → app(app(*, app(s, x)), app(fact, app(p, app(s, x))))
app(app(*, 0), y) → 0
app(app(*, app(s, x)), y) → app(app(+, app(app(*, x), y)), y)
app(app(+, x), 0) → x
app(app(+, x), app(s, y)) → app(s, app(app(+, x), y))
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)

The set Q consists of the following terms:

app(p, app(s, x0))
app(fact, 0)
app(fact, app(s, x0))
app(app(*, 0), x0)
app(app(*, app(s, x0)), x1)
app(app(+, x0), 0)
app(app(+, x0), app(s, x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

(26) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(27) Obligation:

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

APP(fact, app(s, x)) → APP(fact, app(p, app(s, x)))

The TRS R consists of the following rules:

app(p, app(s, x)) → x

The set Q consists of the following terms:

app(p, app(s, x0))
app(fact, 0)
app(fact, app(s, x0))
app(app(*, 0), x0)
app(app(*, app(s, x0)), x1)
app(app(+, x0), 0)
app(app(+, x0), app(s, x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

(28) ATransformationProof (EQUIVALENT transformation)

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

(29) Obligation:

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

fact1(s(x)) → fact1(p(s(x)))

The TRS R consists of the following rules:

p(s(x)) → x

The set Q consists of the following terms:

p(s(x0))
fact(0)
fact(s(x0))
*(0, x0)
*(s(x0), x1)
+(x0, 0)
+(x0, s(x1))
map(x0, nil)
map(x0, cons(x1, x2))
filter(x0, nil)
filter(x0, cons(x1, x2))
filter2(true, x0, x1, x2)
filter2(false, x0, x1, x2)

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

(30) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

fact(0)
fact(s(x0))
*(0, x0)
*(s(x0), x1)
+(x0, 0)
+(x0, s(x1))
map(x0, nil)
map(x0, cons(x1, x2))
filter(x0, nil)
filter(x0, cons(x1, x2))
filter2(true, x0, x1, x2)
filter2(false, x0, x1, x2)

(31) Obligation:

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

fact1(s(x)) → fact1(p(s(x)))

The TRS R consists of the following rules:

p(s(x)) → x

The set Q consists of the following terms:

p(s(x0))

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

(32) 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:

p(s(x)) → x

Used ordering: Polynomial interpretation [POLO]:

POL(fact1(x1)) = x1   
POL(p(x1)) = x1   
POL(s(x1)) = 1 + x1   

(33) Obligation:

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

fact1(s(x)) → fact1(p(s(x)))

R is empty.
The set Q consists of the following terms:

p(s(x0))

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

(34) DependencyGraphProof (EQUIVALENT transformation)

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

(35) TRUE

(36) 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(p, app(s, x)) → x
app(fact, 0) → app(s, 0)
app(fact, app(s, x)) → app(app(*, app(s, x)), app(fact, app(p, app(s, x))))
app(app(*, 0), y) → 0
app(app(*, app(s, x)), y) → app(app(+, app(app(*, x), y)), y)
app(app(+, x), 0) → x
app(app(+, x), app(s, y)) → app(s, app(app(+, x), y))
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)

The set Q consists of the following terms:

app(p, app(s, x0))
app(fact, 0)
app(fact, app(s, x0))
app(app(*, 0), x0)
app(app(*, app(s, x0)), x1)
app(app(+, x0), 0)
app(app(+, x0), app(s, x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

(37) 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

(38) TRUE