YES Termination w.r.t. Q proof of Applicative_first_order_05_3.6.ari

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

0 QTRS
↳1 Overlay + Local Confluence (⇔, 13 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 0 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 AND
    ↳7 QDP
        ↳8 UsableRulesProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 ATransformationProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 QReductionProof (⇔, 0 ms)
        ↳13 QDP
        ↳14 QDPSizeChangeProof (⇔, 0 ms)
        ↳15 YES
    ↳16 QDP
        ↳17 UsableRulesProof (⇔, 0 ms)
        ↳18 QDP
        ↳19 ATransformationProof (⇔, 0 ms)
        ↳20 QDP
        ↳21 QReductionProof (⇔, 0 ms)
        ↳22 QDP
        ↳23 QDPSizeChangeProof (⇔, 0 ms)
        ↳24 YES
    ↳25 QDP
        ↳26 UsableRulesProof (⇔, 0 ms)
        ↳27 QDP
        ↳28 ATransformationProof (⇔, 0 ms)
        ↳29 QDP
        ↳30 QReductionProof (⇔, 0 ms)
        ↳31 QDP
        ↳32 QDPOrderProof (⇔, 0 ms)
        ↳33 QDP
        ↳34 PisEmptyProof (⇔, 0 ms)
        ↳35 YES
    ↳36 QDP
        ↳37 QDPSizeChangeProof (⇔, 0 ms)
        ↳38 YES

(0) Obligation:

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

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(gcd, 0), y) → y
app(app(gcd, app(s, x)), 0) → app(s, x)
app(app(gcd, app(s, x)), app(s, y)) → app(app(app(if_gcd, app(app(le, y), x)), app(s, x)), app(s, y))
app(app(app(if_gcd, true), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, x), y)), app(s, y))
app(app(app(if_gcd, false), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, y), x)), app(s, x))
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(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(gcd, 0), y) → y
app(app(gcd, app(s, x)), 0) → app(s, x)
app(app(gcd, app(s, x)), app(s, y)) → app(app(app(if_gcd, app(app(le, y), x)), app(s, x)), app(s, y))
app(app(app(if_gcd, true), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, x), y)), app(s, y))
app(app(app(if_gcd, false), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, y), x)), app(s, x))
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(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(gcd, 0), x0)
app(app(gcd, app(s, x0)), 0)
app(app(gcd, app(s, x0)), app(s, x1))
app(app(app(if_gcd, true), app(s, x0)), app(s, x1))
app(app(app(if_gcd, false), app(s, 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(app(le, app(s, x)), app(s, y)) → APP(app(le, x), y)
APP(app(le, app(s, x)), app(s, y)) → APP(le, x)
APP(app(minus, x), app(s, y)) → APP(pred, app(app(minus, x), y))
APP(app(minus, x), app(s, y)) → APP(app(minus, x), y)
APP(app(gcd, app(s, x)), app(s, y)) → APP(app(app(if_gcd, app(app(le, y), x)), app(s, x)), app(s, y))
APP(app(gcd, app(s, x)), app(s, y)) → APP(app(if_gcd, app(app(le, y), x)), app(s, x))
APP(app(gcd, app(s, x)), app(s, y)) → APP(if_gcd, app(app(le, y), x))
APP(app(gcd, app(s, x)), app(s, y)) → APP(app(le, y), x)
APP(app(gcd, app(s, x)), app(s, y)) → APP(le, y)
APP(app(app(if_gcd, true), app(s, x)), app(s, y)) → APP(app(gcd, app(app(minus, x), y)), app(s, y))
APP(app(app(if_gcd, true), app(s, x)), app(s, y)) → APP(gcd, app(app(minus, x), y))
APP(app(app(if_gcd, true), app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(app(app(if_gcd, true), app(s, x)), app(s, y)) → APP(minus, x)
APP(app(app(if_gcd, false), app(s, x)), app(s, y)) → APP(app(gcd, app(app(minus, y), x)), app(s, x))
APP(app(app(if_gcd, false), app(s, x)), app(s, y)) → APP(gcd, app(app(minus, y), x))
APP(app(app(if_gcd, false), app(s, x)), app(s, y)) → APP(app(minus, y), x)
APP(app(app(if_gcd, false), app(s, x)), app(s, y)) → APP(minus, 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(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(gcd, 0), y) → y
app(app(gcd, app(s, x)), 0) → app(s, x)
app(app(gcd, app(s, x)), app(s, y)) → app(app(app(if_gcd, app(app(le, y), x)), app(s, x)), app(s, y))
app(app(app(if_gcd, true), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, x), y)), app(s, y))
app(app(app(if_gcd, false), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, y), x)), app(s, x))
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(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(gcd, 0), x0)
app(app(gcd, app(s, x0)), 0)
app(app(gcd, app(s, x0)), app(s, x1))
app(app(app(if_gcd, true), app(s, x0)), app(s, x1))
app(app(app(if_gcd, false), app(s, 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 21 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

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

The TRS R consists of the following rules:

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(gcd, 0), y) → y
app(app(gcd, app(s, x)), 0) → app(s, x)
app(app(gcd, app(s, x)), app(s, y)) → app(app(app(if_gcd, app(app(le, y), x)), app(s, x)), app(s, y))
app(app(app(if_gcd, true), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, x), y)), app(s, y))
app(app(app(if_gcd, false), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, y), x)), app(s, x))
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(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(gcd, 0), x0)
app(app(gcd, app(s, x0)), 0)
app(app(gcd, app(s, x0)), app(s, x1))
app(app(app(if_gcd, true), app(s, x0)), app(s, x1))
app(app(app(if_gcd, false), app(s, 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(minus, x), app(s, y)) → APP(app(minus, x), y)

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

app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(gcd, 0), x0)
app(app(gcd, app(s, x0)), 0)
app(app(gcd, app(s, x0)), app(s, x1))
app(app(app(if_gcd, true), app(s, x0)), app(s, x1))
app(app(app(if_gcd, false), app(s, 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:

minus1(x, s(y)) → minus1(x, y)

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

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
pred(s(x0))
minus(x0, 0)
minus(x0, s(x1))
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if_gcd(true, s(x0), s(x1))
if_gcd(false, s(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].

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
pred(s(x0))
minus(x0, 0)
minus(x0, s(x1))
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if_gcd(true, s(x0), s(x1))
if_gcd(false, s(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:

minus1(x, s(y)) → minus1(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:

  • minus1(x, s(y)) → minus1(x, y)
    The graph contains the following edges 1 >= 1, 2 > 2

(15) YES

(16) Obligation:

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

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

The TRS R consists of the following rules:

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(gcd, 0), y) → y
app(app(gcd, app(s, x)), 0) → app(s, x)
app(app(gcd, app(s, x)), app(s, y)) → app(app(app(if_gcd, app(app(le, y), x)), app(s, x)), app(s, y))
app(app(app(if_gcd, true), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, x), y)), app(s, y))
app(app(app(if_gcd, false), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, y), x)), app(s, x))
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(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(gcd, 0), x0)
app(app(gcd, app(s, x0)), 0)
app(app(gcd, app(s, x0)), app(s, x1))
app(app(app(if_gcd, true), app(s, x0)), app(s, x1))
app(app(app(if_gcd, false), app(s, 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(le, app(s, x)), app(s, y)) → APP(app(le, x), y)

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

app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(gcd, 0), x0)
app(app(gcd, app(s, x0)), 0)
app(app(gcd, app(s, x0)), app(s, x1))
app(app(app(if_gcd, true), app(s, x0)), app(s, x1))
app(app(app(if_gcd, false), app(s, 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:

le1(s(x), s(y)) → le1(x, y)

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

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
pred(s(x0))
minus(x0, 0)
minus(x0, s(x1))
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if_gcd(true, s(x0), s(x1))
if_gcd(false, s(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].

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
pred(s(x0))
minus(x0, 0)
minus(x0, s(x1))
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if_gcd(true, s(x0), s(x1))
if_gcd(false, s(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:

le1(s(x), s(y)) → le1(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:

  • le1(s(x), s(y)) → le1(x, y)
    The graph contains the following edges 1 > 1, 2 > 2

(24) YES

(25) Obligation:

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

APP(app(gcd, app(s, x)), app(s, y)) → APP(app(app(if_gcd, app(app(le, y), x)), app(s, x)), app(s, y))
APP(app(app(if_gcd, true), app(s, x)), app(s, y)) → APP(app(gcd, app(app(minus, x), y)), app(s, y))
APP(app(app(if_gcd, false), app(s, x)), app(s, y)) → APP(app(gcd, app(app(minus, y), x)), app(s, x))

The TRS R consists of the following rules:

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(gcd, 0), y) → y
app(app(gcd, app(s, x)), 0) → app(s, x)
app(app(gcd, app(s, x)), app(s, y)) → app(app(app(if_gcd, app(app(le, y), x)), app(s, x)), app(s, y))
app(app(app(if_gcd, true), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, x), y)), app(s, y))
app(app(app(if_gcd, false), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, y), x)), app(s, x))
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(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(gcd, 0), x0)
app(app(gcd, app(s, x0)), 0)
app(app(gcd, app(s, x0)), app(s, x1))
app(app(app(if_gcd, true), app(s, x0)), app(s, x1))
app(app(app(if_gcd, false), app(s, 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(app(gcd, app(s, x)), app(s, y)) → APP(app(app(if_gcd, app(app(le, y), x)), app(s, x)), app(s, y))
APP(app(app(if_gcd, true), app(s, x)), app(s, y)) → APP(app(gcd, app(app(minus, x), y)), app(s, y))
APP(app(app(if_gcd, false), app(s, x)), app(s, y)) → APP(app(gcd, app(app(minus, y), x)), app(s, x))

The TRS R consists of the following rules:

app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(pred, app(s, x)) → x
app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)

The set Q consists of the following terms:

app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(gcd, 0), x0)
app(app(gcd, app(s, x0)), 0)
app(app(gcd, app(s, x0)), app(s, x1))
app(app(app(if_gcd, true), app(s, x0)), app(s, x1))
app(app(app(if_gcd, false), app(s, 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:

gcd1(s(x), s(y)) → if_gcd1(le(y, x), s(x), s(y))
if_gcd1(true, s(x), s(y)) → gcd1(minus(x, y), s(y))
if_gcd1(false, s(x), s(y)) → gcd1(minus(y, x), s(x))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → x
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
pred(s(x0))
minus(x0, 0)
minus(x0, s(x1))
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if_gcd(true, s(x0), s(x1))
if_gcd(false, s(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].

gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if_gcd(true, s(x0), s(x1))
if_gcd(false, s(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:

gcd1(s(x), s(y)) → if_gcd1(le(y, x), s(x), s(y))
if_gcd1(true, s(x), s(y)) → gcd1(minus(x, y), s(y))
if_gcd1(false, s(x), s(y)) → gcd1(minus(y, x), s(x))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → x
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
pred(s(x0))
minus(x0, 0)
minus(x0, s(x1))

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

(32) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


gcd1(s(x), s(y)) → if_gcd1(le(y, x), s(x), s(y))
if_gcd1(true, s(x), s(y)) → gcd1(minus(x, y), s(y))
if_gcd1(false, s(x), s(y)) → gcd1(minus(y, x), s(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
gcd1(x1, x2)  =  gcd1(x1, x2)
s(x1)  =  s(x1)
if_gcd1(x1, x2, x3)  =  if_gcd1(x2, x3)
minus(x1, x2)  =  x1
pred(x1)  =  x1

Knuth-Bendix order [KBO] with precedence:
trivial

and weight map:

s_1=2
dummyConstant=1
gcd1_2=2
if_gcd1_2=1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

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

(33) Obligation:

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

minus(x, 0) → x
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → x
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
pred(s(x0))
minus(x0, 0)
minus(x0, s(x1))

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

(34) PisEmptyProof (EQUIVALENT transformation)

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

(35) YES

(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(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(gcd, 0), y) → y
app(app(gcd, app(s, x)), 0) → app(s, x)
app(app(gcd, app(s, x)), app(s, y)) → app(app(app(if_gcd, app(app(le, y), x)), app(s, x)), app(s, y))
app(app(app(if_gcd, true), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, x), y)), app(s, y))
app(app(app(if_gcd, false), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, y), x)), app(s, x))
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(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(gcd, 0), x0)
app(app(gcd, app(s, x0)), 0)
app(app(gcd, app(s, x0)), app(s, x1))
app(app(app(if_gcd, true), app(s, x0)), app(s, x1))
app(app(app(if_gcd, false), app(s, 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) YES