YES Termination w.r.t. Q proof of Applicative_first_order_05_12.ari

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

0 QTRS
↳1 DependencyPairsProof (⇔, 0 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 AND
    ↳5 QDP
        ↳6 UsableRulesProof (⇔, 0 ms)
        ↳7 QDP
        ↳8 ATransformationProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 QDPSizeChangeProof (⇔, 0 ms)
        ↳11 YES
    ↳12 QDP
        ↳13 UsableRulesProof (⇔, 0 ms)
        ↳14 QDP
        ↳15 ATransformationProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 QDPSizeChangeProof (⇔, 0 ms)
        ↳18 YES
    ↳19 QDP
        ↳20 QDPSizeChangeProof (⇔, 0 ms)
        ↳21 YES

(0) Obligation:

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

app(not, app(not, x)) → x
app(not, app(app(or, x), y)) → app(app(and, app(not, x)), app(not, y))
app(not, app(app(and, x), y)) → app(app(or, app(not, x)), app(not, y))
app(app(and, x), app(app(or, y), z)) → app(app(or, app(app(and, x), y)), app(app(and, x), z))
app(app(and, app(app(or, y), z)), x) → app(app(or, app(app(and, x), y)), app(app(and, x), z))
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(not, app(app(or, x), y)) → APP(app(and, app(not, x)), app(not, y))
APP(not, app(app(or, x), y)) → APP(and, app(not, x))
APP(not, app(app(or, x), y)) → APP(not, x)
APP(not, app(app(or, x), y)) → APP(not, y)
APP(not, app(app(and, x), y)) → APP(app(or, app(not, x)), app(not, y))
APP(not, app(app(and, x), y)) → APP(or, app(not, x))
APP(not, app(app(and, x), y)) → APP(not, x)
APP(not, app(app(and, x), y)) → APP(not, y)
APP(app(and, x), app(app(or, y), z)) → APP(app(or, app(app(and, x), y)), app(app(and, x), z))
APP(app(and, x), app(app(or, y), z)) → APP(or, app(app(and, x), y))
APP(app(and, x), app(app(or, y), z)) → APP(app(and, x), y)
APP(app(and, x), app(app(or, y), z)) → APP(app(and, x), z)
APP(app(and, app(app(or, y), z)), x) → APP(app(or, app(app(and, x), y)), app(app(and, x), z))
APP(app(and, app(app(or, y), z)), x) → APP(or, app(app(and, x), y))
APP(app(and, app(app(or, y), z)), x) → APP(app(and, x), y)
APP(app(and, app(app(or, y), z)), x) → APP(and, x)
APP(app(and, app(app(or, y), z)), x) → APP(app(and, x), 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(not, app(not, x)) → x
app(not, app(app(or, x), y)) → app(app(and, app(not, x)), app(not, y))
app(not, app(app(and, x), y)) → app(app(or, app(not, x)), app(not, y))
app(app(and, x), app(app(or, y), z)) → app(app(or, app(app(and, x), y)), app(app(and, x), z))
app(app(and, app(app(or, y), z)), x) → app(app(or, app(app(and, x), y)), app(app(and, x), z))
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 3 SCCs with 18 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

APP(app(and, x), app(app(or, y), z)) → APP(app(and, x), z)
APP(app(and, x), app(app(or, y), z)) → APP(app(and, x), y)
APP(app(and, app(app(or, y), z)), x) → APP(app(and, x), y)
APP(app(and, app(app(or, y), z)), x) → APP(app(and, x), z)

The TRS R consists of the following rules:

app(not, app(not, x)) → x
app(not, app(app(or, x), y)) → app(app(and, app(not, x)), app(not, y))
app(not, app(app(and, x), y)) → app(app(or, app(not, x)), app(not, y))
app(app(and, x), app(app(or, y), z)) → app(app(or, app(app(and, x), y)), app(app(and, x), z))
app(app(and, app(app(or, y), z)), x) → app(app(or, app(app(and, x), y)), app(app(and, x), z))
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(and, x), app(app(or, y), z)) → APP(app(and, x), z)
APP(app(and, x), app(app(or, y), z)) → APP(app(and, x), y)
APP(app(and, app(app(or, y), z)), x) → APP(app(and, x), y)
APP(app(and, app(app(or, y), z)), x) → APP(app(and, x), z)

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

(8) ATransformationProof (EQUIVALENT transformation)

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

(9) Obligation:

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

and(x, or(y, z)) → and(x, z)
and(x, or(y, z)) → and(x, y)
and(or(y, z), x) → and(x, y)
and(or(y, z), x) → and(x, z)

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

(10) 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:

  • and(x, or(y, z)) → and(x, z)
    The graph contains the following edges 1 >= 1, 2 > 2

  • and(x, or(y, z)) → and(x, y)
    The graph contains the following edges 1 >= 1, 2 > 2

  • and(or(y, z), x) → and(x, y)
    The graph contains the following edges 2 >= 1, 1 > 2

  • and(or(y, z), x) → and(x, z)
    The graph contains the following edges 2 >= 1, 1 > 2

(11) YES

(12) Obligation:

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

APP(not, app(app(or, x), y)) → APP(not, y)
APP(not, app(app(or, x), y)) → APP(not, x)
APP(not, app(app(and, x), y)) → APP(not, x)
APP(not, app(app(and, x), y)) → APP(not, y)

The TRS R consists of the following rules:

app(not, app(not, x)) → x
app(not, app(app(or, x), y)) → app(app(and, app(not, x)), app(not, y))
app(not, app(app(and, x), y)) → app(app(or, app(not, x)), app(not, y))
app(app(and, x), app(app(or, y), z)) → app(app(or, app(app(and, x), y)), app(app(and, x), z))
app(app(and, app(app(or, y), z)), x) → app(app(or, app(app(and, x), y)), app(app(and, x), z))
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.

(13) 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.

(14) Obligation:

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

APP(not, app(app(or, x), y)) → APP(not, y)
APP(not, app(app(or, x), y)) → APP(not, x)
APP(not, app(app(and, x), y)) → APP(not, x)
APP(not, app(app(and, x), y)) → APP(not, y)

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

(15) ATransformationProof (EQUIVALENT transformation)

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

(16) Obligation:

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

not(or(x, y)) → not(y)
not(or(x, y)) → not(x)
not(and(x, y)) → not(x)
not(and(x, y)) → not(y)

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

(17) 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:

  • not(or(x, y)) → not(y)
    The graph contains the following edges 1 > 1

  • not(or(x, y)) → not(x)
    The graph contains the following edges 1 > 1

  • not(and(x, y)) → not(x)
    The graph contains the following edges 1 > 1

  • not(and(x, y)) → not(y)
    The graph contains the following edges 1 > 1

(18) YES

(19) 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(not, app(not, x)) → x
app(not, app(app(or, x), y)) → app(app(and, app(not, x)), app(not, y))
app(not, app(app(and, x), y)) → app(app(or, app(not, x)), app(not, y))
app(app(and, x), app(app(or, y), z)) → app(app(or, app(app(and, x), y)), app(app(and, x), z))
app(app(and, app(app(or, y), z)), x) → app(app(or, app(app(and, x), y)), app(app(and, x), z))
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.

(20) 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

(21) YES