YES Termination w.r.t. Q proof of Applicative_first_order_05_3.57.ari

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

0 QTRS
↳1 DependencyPairsProof (⇔, 18 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 UsableRulesProof (⇔, 0 ms)
        ↳21 QDP
        ↳22 MNOCProof (⇔, 0 ms)
        ↳23 QDP
        ↳24 ATransformationProof (⇔, 0 ms)
        ↳25 QDP
        ↳26 UsableRulesReductionPairsProof (⇔, 0 ms)
        ↳27 QDP
        ↳28 PisEmptyProof (⇔, 0 ms)
        ↳29 YES
    ↳30 QDP
        ↳31 UsableRulesProof (⇔, 0 ms)
        ↳32 QDP
        ↳33 UsableRulesReductionPairsProof (⇔, 0 ms)
        ↳34 QDP
        ↳35 UsableRulesProof (⇔, 0 ms)
        ↳36 QDP
        ↳37 UsableRulesReductionPairsProof (⇔, 4 ms)
        ↳38 QDP
        ↳39 DependencyGraphProof (⇔, 0 ms)
        ↳40 TRUE
    ↳41 QDP
        ↳42 UsableRulesProof (⇔, 0 ms)
        ↳43 QDP
        ↳44 MNOCProof (⇔, 0 ms)
        ↳45 QDP
        ↳46 ATransformationProof (⇔, 0 ms)
        ↳47 QDP
        ↳48 QDPSizeChangeProof (⇔, 0 ms)
        ↳49 YES
    ↳50 QDP
        ↳51 QDPOrderProof (⇔, 22 ms)
        ↳52 QDP
        ↳53 PisEmptyProof (⇔, 0 ms)
        ↳54 YES
    ↳55 QDP
        ↳56 QDPSizeChangeProof (⇔, 0 ms)
        ↳57 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'(app'(minus, app'(app'(minus, x), y)), z) → app'(app'(minus, x), app'(app'(plus, y), z))
app'(app'(quot, 0), app'(s, y)) → 0
app'(app'(quot, app'(s, x)), app'(s, y)) → app'(s, app'(app'(quot, app'(app'(minus, x), y)), app'(s, y)))
app'(app'(plus, 0), y) → y
app'(app'(plus, app'(s, x)), y) → app'(s, app'(app'(plus, x), y))
app'(app'(app, nil), k) → k
app'(app'(app, l), nil) → l
app'(app'(app, app'(app'(cons, x), l)), k) → app'(app'(cons, x), app'(app'(app, l), k))
app'(sum, app'(app'(cons, x), nil)) → app'(app'(cons, x), nil)
app'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → app'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))
app'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) → app'(sum, app'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))))
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'(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'(app'(minus, app'(app'(minus, x), y)), z) → APP'(app'(minus, x), app'(app'(plus, y), z))
APP'(app'(minus, app'(app'(minus, x), y)), z) → APP'(app'(plus, y), z)
APP'(app'(minus, app'(app'(minus, x), y)), z) → APP'(plus, y)
APP'(app'(quot, app'(s, x)), app'(s, y)) → APP'(s, app'(app'(quot, app'(app'(minus, x), y)), app'(s, y)))
APP'(app'(quot, app'(s, x)), app'(s, y)) → APP'(app'(quot, app'(app'(minus, x), y)), app'(s, y))
APP'(app'(quot, app'(s, x)), app'(s, y)) → APP'(quot, app'(app'(minus, x), y))
APP'(app'(quot, app'(s, x)), app'(s, y)) → APP'(app'(minus, x), y)
APP'(app'(quot, app'(s, x)), app'(s, y)) → APP'(minus, x)
APP'(app'(plus, app'(s, x)), y) → APP'(s, app'(app'(plus, x), y))
APP'(app'(plus, app'(s, x)), y) → APP'(app'(plus, x), y)
APP'(app'(plus, app'(s, x)), y) → APP'(plus, x)
APP'(app'(app, app'(app'(cons, x), l)), k) → APP'(app'(cons, x), app'(app'(app, l), k))
APP'(app'(app, app'(app'(cons, x), l)), k) → APP'(app'(app, l), k)
APP'(app'(app, app'(app'(cons, x), l)), k) → APP'(app, l)
APP'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → APP'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))
APP'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → APP'(app'(cons, app'(app'(plus, x), y)), l)
APP'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → APP'(cons, app'(app'(plus, x), y))
APP'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → APP'(app'(plus, x), y)
APP'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → APP'(plus, x)
APP'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) → APP'(sum, app'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))))
APP'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) → APP'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k))))
APP'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) → APP'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))
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'(minus, x), 0) → x
app'(app'(minus, app'(s, x)), app'(s, y)) → app'(app'(minus, x), y)
app'(app'(minus, app'(app'(minus, x), y)), z) → app'(app'(minus, x), app'(app'(plus, y), z))
app'(app'(quot, 0), app'(s, y)) → 0
app'(app'(quot, app'(s, x)), app'(s, y)) → app'(s, app'(app'(quot, app'(app'(minus, x), y)), app'(s, y)))
app'(app'(plus, 0), y) → y
app'(app'(plus, app'(s, x)), y) → app'(s, app'(app'(plus, x), y))
app'(app'(app, nil), k) → k
app'(app'(app, l), nil) → l
app'(app'(app, app'(app'(cons, x), l)), k) → app'(app'(cons, x), app'(app'(app, l), k))
app'(sum, app'(app'(cons, x), nil)) → app'(app'(cons, x), nil)
app'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → app'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))
app'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) → app'(sum, app'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))))
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 7 SCCs with 26 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

APP'(app'(app, app'(app'(cons, x), l)), k) → APP'(app'(app, l), k)

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'(app'(minus, app'(app'(minus, x), y)), z) → app'(app'(minus, x), app'(app'(plus, y), z))
app'(app'(quot, 0), app'(s, y)) → 0
app'(app'(quot, app'(s, x)), app'(s, y)) → app'(s, app'(app'(quot, app'(app'(minus, x), y)), app'(s, y)))
app'(app'(plus, 0), y) → y
app'(app'(plus, app'(s, x)), y) → app'(s, app'(app'(plus, x), y))
app'(app'(app, nil), k) → k
app'(app'(app, l), nil) → l
app'(app'(app, app'(app'(cons, x), l)), k) → app'(app'(cons, x), app'(app'(app, l), k))
app'(sum, app'(app'(cons, x), nil)) → app'(app'(cons, x), nil)
app'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → app'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))
app'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) → app'(sum, app'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))))
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'(app, app'(app'(cons, x), l)), k) → APP'(app'(app, l), k)

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:

app(cons(x, l), k) → app(l, k)

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:

  • app(cons(x, l), k) → app(l, k)
    The graph contains the following edges 1 > 1, 2 >= 2

(11) YES

(12) Obligation:

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

APP'(app'(plus, app'(s, x)), y) → APP'(app'(plus, 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'(app'(minus, app'(app'(minus, x), y)), z) → app'(app'(minus, x), app'(app'(plus, y), z))
app'(app'(quot, 0), app'(s, y)) → 0
app'(app'(quot, app'(s, x)), app'(s, y)) → app'(s, app'(app'(quot, app'(app'(minus, x), y)), app'(s, y)))
app'(app'(plus, 0), y) → y
app'(app'(plus, app'(s, x)), y) → app'(s, app'(app'(plus, x), y))
app'(app'(app, nil), k) → k
app'(app'(app, l), nil) → l
app'(app'(app, app'(app'(cons, x), l)), k) → app'(app'(cons, x), app'(app'(app, l), k))
app'(sum, app'(app'(cons, x), nil)) → app'(app'(cons, x), nil)
app'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → app'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))
app'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) → app'(sum, app'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))))
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'(app'(plus, app'(s, x)), y) → APP'(app'(plus, x), 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:

plus(s(x), y) → plus(x, 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:

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

(18) YES

(19) Obligation:

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

APP'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → APP'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))

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'(app'(minus, app'(app'(minus, x), y)), z) → app'(app'(minus, x), app'(app'(plus, y), z))
app'(app'(quot, 0), app'(s, y)) → 0
app'(app'(quot, app'(s, x)), app'(s, y)) → app'(s, app'(app'(quot, app'(app'(minus, x), y)), app'(s, y)))
app'(app'(plus, 0), y) → y
app'(app'(plus, app'(s, x)), y) → app'(s, app'(app'(plus, x), y))
app'(app'(app, nil), k) → k
app'(app'(app, l), nil) → l
app'(app'(app, app'(app'(cons, x), l)), k) → app'(app'(cons, x), app'(app'(app, l), k))
app'(sum, app'(app'(cons, x), nil)) → app'(app'(cons, x), nil)
app'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → app'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))
app'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) → app'(sum, app'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))))
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) 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.

(21) Obligation:

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

APP'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → APP'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))

The TRS R consists of the following rules:

app'(app'(plus, 0), y) → y
app'(app'(plus, app'(s, x)), y) → app'(s, app'(app'(plus, x), y))

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

(22) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.

(23) Obligation:

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

APP'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → APP'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))

The TRS R consists of the following rules:

app'(app'(plus, 0), y) → y
app'(app'(plus, app'(s, x)), y) → app'(s, app'(app'(plus, x), y))

The set Q consists of the following terms:

app'(app'(plus, 0), x0)
app'(app'(plus, app'(s, x0)), x1)

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

(24) ATransformationProof (EQUIVALENT transformation)

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

(25) Obligation:

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

sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))

The TRS R consists of the following rules:

plus(0, y) → y
plus(s(x), y) → s(plus(x, y))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)

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

(26) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
The following rules are removed from R:

plus(0, y) → y
Used ordering: POLO with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(cons(x1, x2)) = 1 + x1 + 2·x2   
POL(plus(x1, x2)) = x1 + x2   
POL(s(x1)) = x1   
POL(sum(x1)) = 2·x1   

(27) Obligation:

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

plus(s(x), y) → s(plus(x, y))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)

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

(28) PisEmptyProof (EQUIVALENT transformation)

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

(29) YES

(30) Obligation:

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

APP'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) → APP'(sum, app'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))))

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'(app'(minus, app'(app'(minus, x), y)), z) → app'(app'(minus, x), app'(app'(plus, y), z))
app'(app'(quot, 0), app'(s, y)) → 0
app'(app'(quot, app'(s, x)), app'(s, y)) → app'(s, app'(app'(quot, app'(app'(minus, x), y)), app'(s, y)))
app'(app'(plus, 0), y) → y
app'(app'(plus, app'(s, x)), y) → app'(s, app'(app'(plus, x), y))
app'(app'(app, nil), k) → k
app'(app'(app, l), nil) → l
app'(app'(app, app'(app'(cons, x), l)), k) → app'(app'(cons, x), app'(app'(app, l), k))
app'(sum, app'(app'(cons, x), nil)) → app'(app'(cons, x), nil)
app'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → app'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))
app'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) → app'(sum, app'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))))
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.

(31) 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.

(32) Obligation:

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

APP'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) → APP'(sum, app'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))))

The TRS R consists of the following rules:

app'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → app'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))
app'(app'(app, nil), k) → k
app'(app'(app, l), nil) → l
app'(app'(app, app'(app'(cons, x), l)), k) → app'(app'(cons, x), app'(app'(app, l), k))
app'(app'(plus, 0), y) → y
app'(app'(plus, app'(s, x)), y) → app'(s, app'(app'(plus, x), y))
app'(sum, app'(app'(cons, x), nil)) → app'(app'(cons, x), nil)

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

(33) UsableRulesReductionPairsProof (EQUIVALENT transformation)

First, we A-transformed [FROCOS05] the QDP-Problem. Then we obtain the following A-transformed DP problem.
The pairs P are:

sum1(app(l, cons(x, cons(y, k)))) → sum1(app(l, sum(cons(x, cons(y, k)))))

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

sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
sum(cons(x, nil)) → cons(x, nil)

Q is empty.

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

The following rules are removed from R:

app'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → app'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))
app'(app'(app, nil), k) → k
app'(app'(app, l), nil) → l
app'(app'(app, app'(app'(cons, x), l)), k) → app'(app'(cons, x), app'(app'(app, l), k))
app'(app'(plus, 0), y) → y
app'(app'(plus, app'(s, x)), y) → app'(s, app'(app'(plus, x), y))
app'(sum, app'(app'(cons, x), nil)) → app'(app'(cons, x), nil)
Used ordering: POLO with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(app(x1, x2)) = 2·x1 + x2   
POL(cons(x1, x2)) = 2 + x1 + x2   
POL(nil) = 0   
POL(plus(x1, x2)) = x1 + x2   
POL(s(x1)) = x1   
POL(sum(x1)) = x1   
POL(sum1(x1)) = 2·x1   

(34) Obligation:

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

sum1(app(l, cons(x, cons(y, k)))) → sum1(app(l, sum(cons(x, cons(y, k)))))

The TRS R consists of the following rules:

app(nil, k) → k
app(l, nil) → l
plus(s(x), y) → s(plus(x, y))
sum(cons(x, nil)) → cons(x, nil)

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

(35) 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.

(36) Obligation:

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

sum1(app(l, cons(x, cons(y, k)))) → sum1(app(l, sum(cons(x, cons(y, k)))))

The TRS R consists of the following rules:

app(nil, k) → k

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

(37) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

The following rules are removed from R:

app(nil, k) → k
Used ordering: POLO with Polynomial interpretation [POLO]:

POL(app(x1, x2)) = x1 + 2·x2   
POL(cons(x1, x2)) = 1 + x1 + 2·x2   
POL(nil) = 0   
POL(sum(x1)) = x1   
POL(sum1(x1)) = x1   

(38) Obligation:

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

sum1(app(l, cons(x, cons(y, k)))) → sum1(app(l, sum(cons(x, cons(y, k)))))

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

(39) DependencyGraphProof (EQUIVALENT transformation)

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

(40) TRUE

(41) Obligation:

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

APP'(app'(minus, app'(app'(minus, x), y)), z) → APP'(app'(minus, x), app'(app'(plus, y), z))
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'(app'(minus, app'(app'(minus, x), y)), z) → app'(app'(minus, x), app'(app'(plus, y), z))
app'(app'(quot, 0), app'(s, y)) → 0
app'(app'(quot, app'(s, x)), app'(s, y)) → app'(s, app'(app'(quot, app'(app'(minus, x), y)), app'(s, y)))
app'(app'(plus, 0), y) → y
app'(app'(plus, app'(s, x)), y) → app'(s, app'(app'(plus, x), y))
app'(app'(app, nil), k) → k
app'(app'(app, l), nil) → l
app'(app'(app, app'(app'(cons, x), l)), k) → app'(app'(cons, x), app'(app'(app, l), k))
app'(sum, app'(app'(cons, x), nil)) → app'(app'(cons, x), nil)
app'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → app'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))
app'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) → app'(sum, app'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))))
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.

(42) 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.

(43) Obligation:

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

APP'(app'(minus, app'(app'(minus, x), y)), z) → APP'(app'(minus, x), app'(app'(plus, y), z))
APP'(app'(minus, app'(s, x)), app'(s, y)) → APP'(app'(minus, x), y)

The TRS R consists of the following rules:

app'(app'(plus, 0), y) → y
app'(app'(plus, app'(s, x)), y) → app'(s, app'(app'(plus, x), y))

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

(44) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.

(45) Obligation:

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

APP'(app'(minus, app'(app'(minus, x), y)), z) → APP'(app'(minus, x), app'(app'(plus, y), z))
APP'(app'(minus, app'(s, x)), app'(s, y)) → APP'(app'(minus, x), y)

The TRS R consists of the following rules:

app'(app'(plus, 0), y) → y
app'(app'(plus, app'(s, x)), y) → app'(s, app'(app'(plus, x), y))

The set Q consists of the following terms:

app'(app'(plus, 0), x0)
app'(app'(plus, app'(s, x0)), x1)

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

(46) ATransformationProof (EQUIVALENT transformation)

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

(47) Obligation:

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

minus(minus1(x, y), z) → minus(x, plus(y, z))
minus(s(x), s(y)) → minus(x, y)

The TRS R consists of the following rules:

plus(0, y) → y
plus(s(x), y) → s(plus(x, y))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)

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

(48) 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:

  • minus(minus1(x, y), z) → minus(x, plus(y, z))
    The graph contains the following edges 1 > 1

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

(49) YES

(50) Obligation:

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

APP'(app'(quot, app'(s, x)), app'(s, y)) → APP'(app'(quot, app'(app'(minus, x), y)), app'(s, 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'(app'(minus, app'(app'(minus, x), y)), z) → app'(app'(minus, x), app'(app'(plus, y), z))
app'(app'(quot, 0), app'(s, y)) → 0
app'(app'(quot, app'(s, x)), app'(s, y)) → app'(s, app'(app'(quot, app'(app'(minus, x), y)), app'(s, y)))
app'(app'(plus, 0), y) → y
app'(app'(plus, app'(s, x)), y) → app'(s, app'(app'(plus, x), y))
app'(app'(app, nil), k) → k
app'(app'(app, l), nil) → l
app'(app'(app, app'(app'(cons, x), l)), k) → app'(app'(cons, x), app'(app'(app, l), k))
app'(sum, app'(app'(cons, x), nil)) → app'(app'(cons, x), nil)
app'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → app'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))
app'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) → app'(sum, app'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))))
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.

(51) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06]. Here, we combined the reduction pair processor with the A-transformation [FROCOS05] which results in the following intermediate Q-DP Problem.
The a-transformed P is

quot1(s(x), s(y)) → quot1(minus(x, y), s(y))

The a-transformed usable rules are

minus(x, 0) → x
minus(minus(x, y), z) → minus(x, plus(y, z))
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))


The following pairs can be oriented strictly and are deleted.


APP'(app'(quot, app'(s, x)), app'(s, y)) → APP'(app'(quot, app'(app'(minus, x), y)), app'(s, y))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
quot1(x1, x2)  =  quot1(x1)
s(x1)  =  s(x1)
minus(x1, x2)  =  minus(x1)
0  =  0
plus(x1, x2)  =  plus(x1, x2)

Recursive path order with status [RPO].
Quasi-Precedence:
0 > minus1
plus2 > [quot11, s1] > minus1

Status:
quot11: [1]
s1: multiset
minus1: multiset
0: multiset
plus2: [2,1]


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

app'(app'(minus, x), 0) → x
app'(app'(minus, app'(app'(minus, x), y)), z) → app'(app'(minus, x), app'(app'(plus, y), z))
app'(app'(minus, app'(s, x)), app'(s, y)) → app'(app'(minus, x), y)
app'(app'(plus, 0), y) → y
app'(app'(plus, app'(s, x)), y) → app'(s, app'(app'(plus, x), y))

(52) Obligation:

Q DP problem:
P is empty.
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'(app'(minus, app'(app'(minus, x), y)), z) → app'(app'(minus, x), app'(app'(plus, y), z))
app'(app'(quot, 0), app'(s, y)) → 0
app'(app'(quot, app'(s, x)), app'(s, y)) → app'(s, app'(app'(quot, app'(app'(minus, x), y)), app'(s, y)))
app'(app'(plus, 0), y) → y
app'(app'(plus, app'(s, x)), y) → app'(s, app'(app'(plus, x), y))
app'(app'(app, nil), k) → k
app'(app'(app, l), nil) → l
app'(app'(app, app'(app'(cons, x), l)), k) → app'(app'(cons, x), app'(app'(app, l), k))
app'(sum, app'(app'(cons, x), nil)) → app'(app'(cons, x), nil)
app'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → app'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))
app'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) → app'(sum, app'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))))
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.

(53) PisEmptyProof (EQUIVALENT transformation)

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

(54) YES

(55) 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'(minus, x), 0) → x
app'(app'(minus, app'(s, x)), app'(s, y)) → app'(app'(minus, x), y)
app'(app'(minus, app'(app'(minus, x), y)), z) → app'(app'(minus, x), app'(app'(plus, y), z))
app'(app'(quot, 0), app'(s, y)) → 0
app'(app'(quot, app'(s, x)), app'(s, y)) → app'(s, app'(app'(quot, app'(app'(minus, x), y)), app'(s, y)))
app'(app'(plus, 0), y) → y
app'(app'(plus, app'(s, x)), y) → app'(s, app'(app'(plus, x), y))
app'(app'(app, nil), k) → k
app'(app'(app, l), nil) → l
app'(app'(app, app'(app'(cons, x), l)), k) → app'(app'(cons, x), app'(app'(app, l), k))
app'(sum, app'(app'(cons, x), nil)) → app'(app'(cons, x), nil)
app'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → app'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))
app'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) → app'(sum, app'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))))
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.

(56) 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

(57) YES