YES
proof of Applicative_first_order_05_29.trs
# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 jera 20211004 unpublished


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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) QDPOrderProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) PisEmptyProof [EQUIVALENT, 0 ms]
        (9) YES
    (10) QDP
        (11) UsableRulesProof [EQUIVALENT, 0 ms]
        (12) QDP
        (13) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (14) YES


----------------------------------------

(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   app(app(ack, 0), y) -> app(succ, y)
   app(app(ack, app(succ, x)), y) -> app(app(ack, x), app(succ, 0))
   app(app(ack, app(succ, x)), app(succ, y)) -> app(app(ack, x), app(app(ack, app(succ, x)), y))
   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)
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(ack, 0), y) -> APP(succ, y)
   APP(app(ack, app(succ, x)), y) -> APP(app(ack, x), app(succ, 0))
   APP(app(ack, app(succ, x)), y) -> APP(ack, x)
   APP(app(ack, app(succ, x)), y) -> APP(succ, 0)
   APP(app(ack, app(succ, x)), app(succ, y)) -> APP(app(ack, x), app(app(ack, app(succ, x)), y))
   APP(app(ack, app(succ, x)), app(succ, y)) -> APP(ack, x)
   APP(app(ack, app(succ, x)), app(succ, y)) -> APP(app(ack, app(succ, x)), 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(ack, 0), y) -> app(succ, y)
   app(app(ack, app(succ, x)), y) -> app(app(ack, x), app(succ, 0))
   app(app(ack, app(succ, x)), app(succ, y)) -> app(app(ack, x), app(app(ack, app(succ, x)), y))
   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)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 14 less nodes.
----------------------------------------

(4)
Complex Obligation (AND)

----------------------------------------

(5)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   APP(app(ack, app(succ, x)), app(succ, y)) -> APP(app(ack, x), app(app(ack, app(succ, x)), y))
   APP(app(ack, app(succ, x)), y) -> APP(app(ack, x), app(succ, 0))
   APP(app(ack, app(succ, x)), app(succ, y)) -> APP(app(ack, app(succ, x)), y)

The TRS R consists of the following rules:

   app(app(ack, 0), y) -> app(succ, y)
   app(app(ack, app(succ, x)), y) -> app(app(ack, x), app(succ, 0))
   app(app(ack, app(succ, x)), app(succ, y)) -> app(app(ack, x), app(app(ack, app(succ, x)), y))
   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) QDPOrderProof (EQUIVALENT)
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 

   ack1(succ(x), succ(y)) -> ack1(x, ack(succ(x), y))
   ack1(succ(x), y) -> ack1(x, succ(0))
   ack1(succ(x), succ(y)) -> ack1(succ(x), y)

 The a-transformed usable rules are 

   ack(succ(x), succ(y)) -> ack(x, ack(succ(x), y))
   ack(succ(x), y) -> ack(x, succ(0))
   ack(0, y) -> succ(y)




The following pairs can be oriented strictly and are deleted.

   APP(app(ack, app(succ, x)), app(succ, y)) -> APP(app(ack, x), app(app(ack, app(succ, x)), y))
   APP(app(ack, app(succ, x)), y) -> APP(app(ack, x), app(succ, 0))
   APP(app(ack, app(succ, x)), app(succ, y)) -> APP(app(ack, app(succ, x)), y)
The remaining pairs can at least be oriented weakly.
Used ordering:  Recursive path order with status [RPO].
Quasi-Precedence: ack1_2 > ack_2 > succ_1 > 0

Status: ack1_2: [1,2]
succ_1: multiset status
ack_2: [1,2]
0: multiset status


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

   app(app(ack, app(succ, x)), app(succ, y)) -> app(app(ack, x), app(app(ack, app(succ, x)), y))
   app(app(ack, app(succ, x)), y) -> app(app(ack, x), app(succ, 0))
   app(app(ack, 0), y) -> app(succ, y)


----------------------------------------

(7)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   app(app(ack, 0), y) -> app(succ, y)
   app(app(ack, app(succ, x)), y) -> app(app(ack, x), app(succ, 0))
   app(app(ack, app(succ, x)), app(succ, y)) -> app(app(ack, x), app(app(ack, app(succ, x)), y))
   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.
----------------------------------------

(8) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
----------------------------------------

(9)
YES

----------------------------------------

(10)
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(f, x)
   APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs)
   APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs)

The TRS R consists of the following rules:

   app(app(ack, 0), y) -> app(succ, y)
   app(app(ack, app(succ, x)), y) -> app(app(ack, x), app(succ, 0))
   app(app(ack, app(succ, x)), app(succ, y)) -> app(app(ack, x), app(app(ack, app(succ, x)), y))
   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.
----------------------------------------

(11) UsableRulesProof (EQUIVALENT)
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.
----------------------------------------

(12)
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(f, x)
   APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs)
   APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs)

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

(13) QDPSizeChangeProof (EQUIVALENT)
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(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


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(14)
YES