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 UsableRulesProof (⇔)
    ↳9 QDP
    ↳10 ATransformationProof (⇔)
    ↳11 QDP
    ↳12 QReductionProof (⇔)
    ↳13 QDP
    ↳14 QDPSizeChangeProof (⇔)
    ↳15 TRUE
  ↳16 QDP
    ↳17 UsableRulesProof (⇔)
    ↳18 QDP
    ↳19 ATransformationProof (⇔)
    ↳20 QDP
    ↳21 QReductionProof (⇔)
    ↳22 QDP
    ↳23 QDPOrderProof (⇔)
    ↳24 QDP
    ↳25 DependencyGraphProof (⇔)
    ↳26 TRUE
  ↳27 QDP
    ↳28 QDPSizeChangeProof (⇔)
    ↳29 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(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) TRUE

(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].


The following pairs can be oriented strictly and are deleted.


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: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(f(x1)) = 1 + x1   
POL(f1(x1)) = 1 + x1   
POL(g(x1)) = x1   
POL(g1(x1)) = 1 + x1   
POL(minus(x1, x2)) = x1   
POL(s(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented:

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)

(24) Obligation:

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

f1(s(x)) → g1(f(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.

(25) DependencyGraphProof (EQUIVALENT transformation)

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

(26) TRUE

(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) TRUE