YES
proof of /export/starexec/sandbox/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 18 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 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, 51 ms]
        (22) QDP
        (23) DependencyGraphProof [EQUIVALENT, 0 ms]
        (24) TRUE
    (25) QDP
        (26) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (27) YES


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

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

   app(p, 0) -> 0
   app(p, app(s, x)) -> x
   app(app(le, 0), y) -> true
   app(app(le, app(s, x)), 0) -> false
   app(app(le, app(s, x)), app(s, y)) -> app(app(le, x), y)
   app(app(minus, x), y) -> app(app(app(if, app(app(le, x), y)), x), y)
   app(app(app(if, true), x), y) -> 0
   app(app(app(if, false), x), y) -> app(s, app(app(minus, app(p, 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)

The set Q consists of the following terms:

   app(p, 0)
   app(p, app(s, x0))
   app(app(le, 0), x0)
   app(app(le, app(s, x0)), 0)
   app(app(le, app(s, x0)), app(s, x1))
   app(app(minus, x0), x1)
   app(app(app(if, true), x0), x1)
   app(app(app(if, false), x0), 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(app(le, app(s, x)), app(s, y)) -> APP(app(le, x), y)
   APP(app(le, app(s, x)), app(s, y)) -> APP(le, x)
   APP(app(minus, x), y) -> APP(app(app(if, app(app(le, x), y)), x), y)
   APP(app(minus, x), y) -> APP(app(if, app(app(le, x), y)), x)
   APP(app(minus, x), y) -> APP(if, app(app(le, x), y))
   APP(app(minus, x), y) -> APP(app(le, x), y)
   APP(app(minus, x), y) -> APP(le, x)
   APP(app(app(if, false), x), y) -> APP(s, app(app(minus, app(p, x)), y))
   APP(app(app(if, false), x), y) -> APP(app(minus, app(p, x)), y)
   APP(app(app(if, false), x), y) -> APP(minus, app(p, x))
   APP(app(app(if, false), x), y) -> APP(p, x)
   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(p, 0) -> 0
   app(p, app(s, x)) -> x
   app(app(le, 0), y) -> true
   app(app(le, app(s, x)), 0) -> false
   app(app(le, app(s, x)), app(s, y)) -> app(app(le, x), y)
   app(app(minus, x), y) -> app(app(app(if, app(app(le, x), y)), x), y)
   app(app(app(if, true), x), y) -> 0
   app(app(app(if, false), x), y) -> app(s, app(app(minus, app(p, 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)

The set Q consists of the following terms:

   app(p, 0)
   app(p, app(s, x0))
   app(app(le, 0), x0)
   app(app(le, app(s, x0)), 0)
   app(app(le, app(s, x0)), app(s, x1))
   app(app(minus, x0), x1)
   app(app(app(if, true), x0), x1)
   app(app(app(if, false), x0), 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 17 less nodes.
----------------------------------------

(4)
Complex Obligation (AND)

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

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

   APP(app(le, app(s, x)), app(s, y)) -> APP(app(le, x), y)

The TRS R consists of the following rules:

   app(p, 0) -> 0
   app(p, app(s, x)) -> x
   app(app(le, 0), y) -> true
   app(app(le, app(s, x)), 0) -> false
   app(app(le, app(s, x)), app(s, y)) -> app(app(le, x), y)
   app(app(minus, x), y) -> app(app(app(if, app(app(le, x), y)), x), y)
   app(app(app(if, true), x), y) -> 0
   app(app(app(if, false), x), y) -> app(s, app(app(minus, app(p, 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)

The set Q consists of the following terms:

   app(p, 0)
   app(p, app(s, x0))
   app(app(le, 0), x0)
   app(app(le, app(s, x0)), 0)
   app(app(le, app(s, x0)), app(s, x1))
   app(app(minus, x0), x1)
   app(app(app(if, true), x0), x1)
   app(app(app(if, false), x0), 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(app(le, app(s, x)), app(s, y)) -> APP(app(le, x), y)

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

   app(p, 0)
   app(p, app(s, x0))
   app(app(le, 0), x0)
   app(app(le, app(s, x0)), 0)
   app(app(le, app(s, x0)), app(s, x1))
   app(app(minus, x0), x1)
   app(app(app(if, true), x0), x1)
   app(app(app(if, false), x0), 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:

   le1(s(x), s(y)) -> le1(x, y)

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

   p(0)
   p(s(x0))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   minus(x0, x1)
   if(true, x0, x1)
   if(false, x0, 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].

   p(0)
   p(s(x0))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   minus(x0, x1)
   if(true, x0, x1)
   if(false, x0, 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:

   le1(s(x), s(y)) -> le1(x, y)

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:
*le1(s(x), s(y)) -> le1(x, y)
The graph contains the following edges 1 > 1, 2 > 2


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

(13)
YES

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

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

   APP(app(minus, x), y) -> APP(app(app(if, app(app(le, x), y)), x), y)
   APP(app(app(if, false), x), y) -> APP(app(minus, app(p, x)), y)

The TRS R consists of the following rules:

   app(p, 0) -> 0
   app(p, app(s, x)) -> x
   app(app(le, 0), y) -> true
   app(app(le, app(s, x)), 0) -> false
   app(app(le, app(s, x)), app(s, y)) -> app(app(le, x), y)
   app(app(minus, x), y) -> app(app(app(if, app(app(le, x), y)), x), y)
   app(app(app(if, true), x), y) -> 0
   app(app(app(if, false), x), y) -> app(s, app(app(minus, app(p, 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)

The set Q consists of the following terms:

   app(p, 0)
   app(p, app(s, x0))
   app(app(le, 0), x0)
   app(app(le, app(s, x0)), 0)
   app(app(le, app(s, x0)), app(s, x1))
   app(app(minus, x0), x1)
   app(app(app(if, true), x0), x1)
   app(app(app(if, false), x0), 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(minus, x), y) -> APP(app(app(if, app(app(le, x), y)), x), y)
   APP(app(app(if, false), x), y) -> APP(app(minus, app(p, x)), y)

The TRS R consists of the following rules:

   app(p, 0) -> 0
   app(p, app(s, x)) -> x
   app(app(le, 0), y) -> true
   app(app(le, app(s, x)), 0) -> false
   app(app(le, app(s, x)), app(s, y)) -> app(app(le, x), y)

The set Q consists of the following terms:

   app(p, 0)
   app(p, app(s, x0))
   app(app(le, 0), x0)
   app(app(le, app(s, x0)), 0)
   app(app(le, app(s, x0)), app(s, x1))
   app(app(minus, x0), x1)
   app(app(app(if, true), x0), x1)
   app(app(app(if, false), x0), 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:

   minus1(x, y) -> if1(le(x, y), x, y)
   if1(false, x, y) -> minus1(p(x), y)

The TRS R consists of the following rules:

   p(0) -> 0
   p(s(x)) -> x
   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)

The set Q consists of the following terms:

   p(0)
   p(s(x0))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   minus(x0, x1)
   if(true, x0, x1)
   if(false, x0, 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].

   minus(x0, x1)
   if(true, x0, x1)
   if(false, x0, 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)


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

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

   minus1(x, y) -> if1(le(x, y), x, y)
   if1(false, x, y) -> minus1(p(x), y)

The TRS R consists of the following rules:

   p(0) -> 0
   p(s(x)) -> x
   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)

The set Q consists of the following terms:

   p(0)
   p(s(x0))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(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.

   if1(false, x, y) -> minus1(p(x), y)
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO,RATPOLO]:

   POL(0) = 0
   POL(false) = [1/4]
   POL(if1(x_1, x_2, x_3)) = [4]x_1 + [1/2]x_2
   POL(le(x_1, x_2)) = [1/4]x_1
   POL(minus1(x_1, x_2)) = [2]x_1
   POL(p(x_1)) = [1/4]x_1
   POL(s(x_1)) = [4] + [4]x_1
   POL(true) = 0
The value of delta used in the strict ordering is 1.
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   p(0) -> 0
   p(s(x)) -> x


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

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

   minus1(x, y) -> if1(le(x, y), x, y)

The TRS R consists of the following rules:

   p(0) -> 0
   p(s(x)) -> x
   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)

The set Q consists of the following terms:

   p(0)
   p(s(x0))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))

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

(23) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
----------------------------------------

(24)
TRUE

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

(25)
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(p, 0) -> 0
   app(p, app(s, x)) -> x
   app(app(le, 0), y) -> true
   app(app(le, app(s, x)), 0) -> false
   app(app(le, app(s, x)), app(s, y)) -> app(app(le, x), y)
   app(app(minus, x), y) -> app(app(app(if, app(app(le, x), y)), x), y)
   app(app(app(if, true), x), y) -> 0
   app(app(app(if, false), x), y) -> app(s, app(app(minus, app(p, 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)

The set Q consists of the following terms:

   app(p, 0)
   app(p, app(s, x0))
   app(app(le, 0), x0)
   app(app(le, app(s, x0)), 0)
   app(app(le, app(s, x0)), app(s, x1))
   app(app(minus, x0), x1)
   app(app(app(if, true), x0), x1)
   app(app(app(if, false), x0), 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, 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


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(27)
YES