YES Termination w.r.t. Q proof of Applicative_first_order_05_3.36.ari

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

0 QTRS
↳1 Overlay + Local Confluence (⇔, 0 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 24 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 QDPOrderProof (⇔, 14 ms)
        ↳24 QDP
        ↳25 PisEmptyProof (⇔, 0 ms)
        ↳26 YES
    ↳27 QDP
        ↳28 QDPSizeChangeProof (⇔, 0 ms)
        ↳29 YES

(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(f, 0) → app(s, 0)
app(f, app(s, x)) → app(app(minus, app(s, x)), app(g, app(f, x)))
app(g, 0) → 0
app(g, app(s, x)) → app(app(minus, app(s, x)), app(f, app(g, x)))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), 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(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(f, 0) → app(s, 0)
app(f, app(s, x)) → app(app(minus, app(s, x)), app(g, app(f, x)))
app(g, 0) → 0
app(g, app(s, x)) → app(app(minus, app(s, x)), app(f, app(g, x)))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

The set Q consists of the following terms:

app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(f, 0)
app(f, app(s, x0))
app(g, 0)
app(g, app(s, x0))
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(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(f, 0) → APP(s, 0)
APP(f, app(s, x)) → APP(app(minus, app(s, x)), app(g, app(f, x)))
APP(f, app(s, x)) → APP(minus, app(s, x))
APP(f, app(s, x)) → APP(g, app(f, x))
APP(f, app(s, x)) → APP(f, x)
APP(g, app(s, x)) → APP(app(minus, app(s, x)), app(f, app(g, x)))
APP(g, app(s, x)) → APP(minus, app(s, x))
APP(g, app(s, x)) → APP(f, app(g, x))
APP(g, app(s, x)) → APP(g, x)
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(cons, app(fun, x)), app(app(map, fun), xs))
APP(app(map, fun), app(app(cons, x), xs)) → APP(cons, app(fun, x))
APP(app(map, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(map, fun), xs)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(fun, x)), fun), x), xs)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(app(filter2, app(fun, x)), fun), x)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(filter2, app(fun, x)), fun)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(filter2, app(fun, x))
APP(app(filter, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(app(app(filter2, true), fun), x), xs) → APP(app(cons, x), app(app(filter, fun), xs))
APP(app(app(app(filter2, true), fun), x), xs) → APP(cons, x)
APP(app(app(app(filter2, true), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(app(app(filter2, true), fun), x), xs) → APP(filter, fun)
APP(app(app(app(filter2, false), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(app(app(filter2, false), fun), x), xs) → APP(filter, fun)

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(f, 0) → app(s, 0)
app(f, app(s, x)) → app(app(minus, app(s, x)), app(g, app(f, x)))
app(g, 0) → 0
app(g, app(s, x)) → app(app(minus, app(s, x)), app(f, app(g, x)))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

The set Q consists of the following terms:

app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(f, 0)
app(f, app(s, x0))
app(g, 0)
app(g, app(s, x0))
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 3 SCCs with 15 less nodes.

(6) Complex Obligation (AND)

(7) 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(f, 0) → app(s, 0)
app(f, app(s, x)) → app(app(minus, app(s, x)), app(g, app(f, x)))
app(g, 0) → 0
app(g, app(s, x)) → app(app(minus, app(s, x)), app(f, app(g, x)))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

The set Q consists of the following terms:

app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(f, 0)
app(f, app(s, x0))
app(g, 0)
app(g, app(s, x0))
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, app(s, x)), app(s, y)) → APP(app(minus, x), y)

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

app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(f, 0)
app(f, app(s, x0))
app(g, 0)
app(g, app(s, x0))
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(s(x), s(y)) → minus1(x, y)

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

minus(x0, 0)
minus(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))
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].

minus(x0, 0)
minus(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))
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(s(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(s(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(f, app(s, x)) → APP(g, app(f, x))
APP(g, app(s, x)) → APP(f, app(g, x))
APP(f, app(s, x)) → APP(f, x)
APP(g, app(s, x)) → APP(g, x)

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(f, 0) → app(s, 0)
app(f, app(s, x)) → app(app(minus, app(s, x)), app(g, app(f, x)))
app(g, 0) → 0
app(g, app(s, x)) → app(app(minus, app(s, x)), app(f, app(g, x)))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

The set Q consists of the following terms:

app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(f, 0)
app(f, app(s, x0))
app(g, 0)
app(g, app(s, x0))
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(f, app(s, x)) → APP(g, app(f, x))
APP(g, app(s, x)) → APP(f, app(g, x))
APP(f, app(s, x)) → APP(f, x)
APP(g, app(s, x)) → APP(g, x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(f, 0)
app(f, app(s, x0))
app(g, 0)
app(g, app(s, x0))
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:

f1(s(x)) → g1(f(x))
g1(s(x)) → f1(g(x))
f1(s(x)) → f1(x)
g1(s(x)) → g1(x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))
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].

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:

f1(s(x)) → g1(f(x))
g1(s(x)) → f1(g(x))
f1(s(x)) → f1(x)
g1(s(x)) → g1(x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))

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

(23) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


f1(s(x)) → g1(f(x))
g1(s(x)) → f1(g(x))
f1(s(x)) → f1(x)
g1(s(x)) → g1(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
f1(x1)  =  f1(x1)
s(x1)  =  s(x1)
g1(x1)  =  g1(x1)
f(x1)  =  f(x1)
g(x1)  =  x1
0  =  0
minus(x1, x2)  =  x1

Recursive path order with status [RPO].
Quasi-Precedence:
0 > [f11, s1, f1] > g11

Status:
f11: multiset
s1: multiset
g11: multiset
f1: multiset
0: multiset


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

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

(24) Obligation:

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

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

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))

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

(25) PisEmptyProof (EQUIVALENT transformation)

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

(26) YES

(27) Obligation:

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

APP(app(map, fun), app(app(cons, x), xs)) → APP(app(map, fun), xs)
APP(app(map, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(fun, x)), fun), x), xs)
APP(app(app(app(filter2, true), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(app(app(filter2, false), fun), x), xs) → APP(app(filter, fun), 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(f, 0) → app(s, 0)
app(f, app(s, x)) → app(app(minus, app(s, x)), app(g, app(f, x)))
app(g, 0) → 0
app(g, app(s, x)) → app(app(minus, app(s, x)), app(f, app(g, x)))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

The set Q consists of the following terms:

app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(f, 0)
app(f, app(s, x0))
app(g, 0)
app(g, app(s, x0))
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) 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, fun), app(app(cons, x), xs)) → APP(fun, x)
    The graph contains the following edges 1 > 1, 2 > 2

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

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

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

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

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

(29) YES