YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 4 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) ATransformationProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QReductionProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) ATransformationProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) QReductionProof [EQUIVALENT, 0 ms]
        (20) QDP
        (21) QDPOrderProof [EQUIVALENT, 15 ms]
        (22) QDP
        (23) PisEmptyProof [EQUIVALENT, 0 ms]
        (24) YES
    (25) QDP
        (26) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (27) YES


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

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

   app(f, 0) -> true
   app(f, 1) -> false
   app(f, app(s, x)) -> app(f, x)
   app(app(app(if, true), x), y) -> x
   app(app(app(if, false), x), y) -> y
   app(app(g, app(s, x)), app(s, y)) -> app(app(app(if, app(f, x)), app(s, x)), app(s, y))
   app(app(g, x), app(c, y)) -> app(app(g, x), app(app(g, app(s, app(c, y))), y))
   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(f, 0)
   app(f, 1)
   app(f, app(s, x0))
   app(app(app(if, true), x0), x1)
   app(app(app(if, false), x0), x1)
   app(app(g, app(s, x0)), app(s, x1))
   app(app(g, x0), app(c, x1))
   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)


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

(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(f, app(s, x)) -> APP(f, x)
   APP(app(g, app(s, x)), app(s, y)) -> APP(app(app(if, app(f, x)), app(s, x)), app(s, y))
   APP(app(g, app(s, x)), app(s, y)) -> APP(app(if, app(f, x)), app(s, x))
   APP(app(g, app(s, x)), app(s, y)) -> APP(if, app(f, x))
   APP(app(g, app(s, x)), app(s, y)) -> APP(f, x)
   APP(app(g, x), app(c, y)) -> APP(app(g, x), app(app(g, app(s, app(c, y))), y))
   APP(app(g, x), app(c, y)) -> APP(app(g, app(s, app(c, y))), y)
   APP(app(g, x), app(c, y)) -> APP(g, app(s, app(c, y)))
   APP(app(g, x), app(c, y)) -> APP(s, app(c, y))
   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(f, 0) -> true
   app(f, 1) -> false
   app(f, app(s, x)) -> app(f, x)
   app(app(app(if, true), x), y) -> x
   app(app(app(if, false), x), y) -> y
   app(app(g, app(s, x)), app(s, y)) -> app(app(app(if, app(f, x)), app(s, x)), app(s, y))
   app(app(g, x), app(c, y)) -> app(app(g, x), app(app(g, app(s, app(c, y))), y))
   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(f, 0)
   app(f, 1)
   app(f, app(s, x0))
   app(app(app(if, true), x0), x1)
   app(app(app(if, false), x0), x1)
   app(app(g, app(s, x0)), app(s, x1))
   app(app(g, x0), app(c, x1))
   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.
----------------------------------------

(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 15 less nodes.
----------------------------------------

(4)
Complex Obligation (AND)

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

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

   APP(f, app(s, x)) -> APP(f, x)

The TRS R consists of the following rules:

   app(f, 0) -> true
   app(f, 1) -> false
   app(f, app(s, x)) -> app(f, x)
   app(app(app(if, true), x), y) -> x
   app(app(app(if, false), x), y) -> y
   app(app(g, app(s, x)), app(s, y)) -> app(app(app(if, app(f, x)), app(s, x)), app(s, y))
   app(app(g, x), app(c, y)) -> app(app(g, x), app(app(g, app(s, app(c, y))), y))
   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(f, 0)
   app(f, 1)
   app(f, app(s, x0))
   app(app(app(if, true), x0), x1)
   app(app(app(if, false), x0), x1)
   app(app(g, app(s, x0)), app(s, x1))
   app(app(g, x0), app(c, x1))
   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.
----------------------------------------

(6) UsableRulesProof (EQUIVALENT)
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.
----------------------------------------

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

   APP(f, app(s, x)) -> APP(f, x)

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

   app(f, 0)
   app(f, 1)
   app(f, app(s, x0))
   app(app(app(if, true), x0), x1)
   app(app(app(if, false), x0), x1)
   app(app(g, app(s, x0)), app(s, x1))
   app(app(g, x0), app(c, x1))
   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) ATransformationProof (EQUIVALENT)
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:

   f1(s(x)) -> f1(x)

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

   f(0)
   f(1)
   f(s(x0))
   if(true, x0, x1)
   if(false, x0, x1)
   g(s(x0), s(x1))
   g(x0, c(x1))
   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.
----------------------------------------

(10) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   f(0)
   f(1)
   f(s(x0))
   if(true, x0, x1)
   if(false, x0, x1)
   g(s(x0), s(x1))
   g(x0, c(x1))
   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)


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

(11)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   f1(s(x)) -> f1(x)

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

(12) 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:
*f1(s(x)) -> f1(x)
The graph contains the following edges 1 > 1


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

(13)
YES

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

(14)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   APP(app(g, x), app(c, y)) -> APP(app(g, app(s, app(c, y))), y)
   APP(app(g, x), app(c, y)) -> APP(app(g, x), app(app(g, app(s, app(c, y))), y))

The TRS R consists of the following rules:

   app(f, 0) -> true
   app(f, 1) -> false
   app(f, app(s, x)) -> app(f, x)
   app(app(app(if, true), x), y) -> x
   app(app(app(if, false), x), y) -> y
   app(app(g, app(s, x)), app(s, y)) -> app(app(app(if, app(f, x)), app(s, x)), app(s, y))
   app(app(g, x), app(c, y)) -> app(app(g, x), app(app(g, app(s, app(c, y))), y))
   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(f, 0)
   app(f, 1)
   app(f, app(s, x0))
   app(app(app(if, true), x0), x1)
   app(app(app(if, false), x0), x1)
   app(app(g, app(s, x0)), app(s, x1))
   app(app(g, x0), app(c, x1))
   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.
----------------------------------------

(15) UsableRulesProof (EQUIVALENT)
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.
----------------------------------------

(16)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   APP(app(g, x), app(c, y)) -> APP(app(g, app(s, app(c, y))), y)
   APP(app(g, x), app(c, y)) -> APP(app(g, x), app(app(g, app(s, app(c, y))), y))

The TRS R consists of the following rules:

   app(app(g, app(s, x)), app(s, y)) -> app(app(app(if, app(f, x)), app(s, x)), app(s, y))
   app(app(g, x), app(c, y)) -> app(app(g, x), app(app(g, app(s, app(c, y))), y))
   app(f, 0) -> true
   app(f, 1) -> false
   app(f, app(s, x)) -> app(f, x)
   app(app(app(if, true), x), y) -> x
   app(app(app(if, false), x), y) -> y

The set Q consists of the following terms:

   app(f, 0)
   app(f, 1)
   app(f, app(s, x0))
   app(app(app(if, true), x0), x1)
   app(app(app(if, false), x0), x1)
   app(app(g, app(s, x0)), app(s, x1))
   app(app(g, x0), app(c, x1))
   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) ATransformationProof (EQUIVALENT)
We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. 
----------------------------------------

(18)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   g1(x, c(y)) -> g1(s(c(y)), y)
   g1(x, c(y)) -> g1(x, g(s(c(y)), y))

The TRS R consists of the following rules:

   g(s(x), s(y)) -> if(f(x), s(x), s(y))
   g(x, c(y)) -> g(x, g(s(c(y)), y))
   f(0) -> true
   f(1) -> false
   f(s(x)) -> f(x)
   if(true, x, y) -> x
   if(false, x, y) -> y

The set Q consists of the following terms:

   f(0)
   f(1)
   f(s(x0))
   if(true, x0, x1)
   if(false, x0, x1)
   g(s(x0), s(x1))
   g(x0, c(x1))
   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.
----------------------------------------

(19) QReductionProof (EQUIVALENT)
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)


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

(20)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   g1(x, c(y)) -> g1(s(c(y)), y)
   g1(x, c(y)) -> g1(x, g(s(c(y)), y))

The TRS R consists of the following rules:

   g(s(x), s(y)) -> if(f(x), s(x), s(y))
   g(x, c(y)) -> g(x, g(s(c(y)), y))
   f(0) -> true
   f(1) -> false
   f(s(x)) -> f(x)
   if(true, x, y) -> x
   if(false, x, y) -> y

The set Q consists of the following terms:

   f(0)
   f(1)
   f(s(x0))
   if(true, x0, x1)
   if(false, x0, x1)
   g(s(x0), s(x1))
   g(x0, c(x1))

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

(21) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   g1(x, c(y)) -> g1(s(c(y)), y)
   g1(x, c(y)) -> g1(x, g(s(c(y)), y))
The remaining pairs can at least be oriented weakly.
Used ordering:  Combined order from the following AFS and order.
g1(x1, x2)  =  x2

c(x1)  =  c(x1)

g(x1, x2)  =  g

if(x1, x2, x3)  =  if(x2, x3)

s(x1)  =  s


Knuth-Bendix order [KBO] with precedence:trivial

and weight map:

   s=1
   c_1=4
   g=4
   if_2=1

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

   g(s(x), s(y)) -> if(f(x), s(x), s(y))
   g(x, c(y)) -> g(x, g(s(c(y)), y))
   if(true, x, y) -> x
   if(false, x, y) -> y


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

(22)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   g(s(x), s(y)) -> if(f(x), s(x), s(y))
   g(x, c(y)) -> g(x, g(s(c(y)), y))
   f(0) -> true
   f(1) -> false
   f(s(x)) -> f(x)
   if(true, x, y) -> x
   if(false, x, y) -> y

The set Q consists of the following terms:

   f(0)
   f(1)
   f(s(x0))
   if(true, x0, x1)
   if(false, x0, x1)
   g(s(x0), s(x1))
   g(x0, c(x1))

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

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

(24)
YES

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

(25)
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(f, 0) -> true
   app(f, 1) -> false
   app(f, app(s, x)) -> app(f, x)
   app(app(app(if, true), x), y) -> x
   app(app(app(if, false), x), y) -> y
   app(app(g, app(s, x)), app(s, y)) -> app(app(app(if, app(f, x)), app(s, x)), app(s, y))
   app(app(g, x), app(c, y)) -> app(app(g, x), app(app(g, app(s, app(c, y))), y))
   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(f, 0)
   app(f, 1)
   app(f, app(s, x0))
   app(app(app(if, true), x0), x1)
   app(app(app(if, false), x0), x1)
   app(app(g, app(s, x0)), app(s, x1))
   app(app(g, x0), app(c, x1))
   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.
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(26) 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, 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


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(27)
YES