/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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


YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


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

(0) QTRS
(1) Overlay + Local Confluence [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 40 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) ATransformationProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) QReductionProof [EQUIVALENT, 0 ms]
        (13) QDP
        (14) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (15) YES
    (16) QDP
        (17) UsableRulesProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) ATransformationProof [EQUIVALENT, 0 ms]
        (20) QDP
        (21) QReductionProof [EQUIVALENT, 0 ms]
        (22) QDP
        (23) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (24) YES
    (25) QDP
        (26) UsableRulesProof [EQUIVALENT, 0 ms]
        (27) QDP
        (28) QDPOrderProof [EQUIVALENT, 12 ms]
        (29) QDP
        (30) UsableRulesProof [EQUIVALENT, 0 ms]
        (31) QDP
        (32) ATransformationProof [EQUIVALENT, 0 ms]
        (33) QDP
        (34) QReductionProof [EQUIVALENT, 0 ms]
        (35) QDP
        (36) TransformationProof [EQUIVALENT, 0 ms]
        (37) QDP
        (38) DependencyGraphProof [EQUIVALENT, 0 ms]
        (39) QDP
        (40) UsableRulesProof [EQUIVALENT, 0 ms]
        (41) QDP
        (42) QReductionProof [EQUIVALENT, 0 ms]
        (43) QDP
        (44) MRRProof [EQUIVALENT, 7 ms]
        (45) QDP
        (46) QDPOrderProof [EQUIVALENT, 7 ms]
        (47) QDP
        (48) DependencyGraphProof [EQUIVALENT, 0 ms]
        (49) TRUE


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

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

   ap(ap(map, f), xs) -> ap(ap(ap(if, ap(isEmpty, xs)), f), xs)
   ap(ap(ap(if, true), f), xs) -> nil
   ap(ap(ap(if, false), f), xs) -> ap(ap(cons, ap(f, ap(last, xs))), ap(ap(map, f), ap(dropLast, xs)))
   ap(isEmpty, nil) -> true
   ap(isEmpty, ap(ap(cons, x), xs)) -> false
   ap(last, ap(ap(cons, x), nil)) -> x
   ap(last, ap(ap(cons, x), ap(ap(cons, y), ys))) -> ap(last, ap(ap(cons, y), ys))
   ap(dropLast, nil) -> nil
   ap(dropLast, ap(ap(cons, x), nil)) -> nil
   ap(dropLast, ap(ap(cons, x), ap(ap(cons, y), ys))) -> ap(ap(cons, x), ap(dropLast, ap(ap(cons, y), ys)))

Q is empty.

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

(1) Overlay + Local Confluence (EQUIVALENT)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
----------------------------------------

(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   ap(ap(map, f), xs) -> ap(ap(ap(if, ap(isEmpty, xs)), f), xs)
   ap(ap(ap(if, true), f), xs) -> nil
   ap(ap(ap(if, false), f), xs) -> ap(ap(cons, ap(f, ap(last, xs))), ap(ap(map, f), ap(dropLast, xs)))
   ap(isEmpty, nil) -> true
   ap(isEmpty, ap(ap(cons, x), xs)) -> false
   ap(last, ap(ap(cons, x), nil)) -> x
   ap(last, ap(ap(cons, x), ap(ap(cons, y), ys))) -> ap(last, ap(ap(cons, y), ys))
   ap(dropLast, nil) -> nil
   ap(dropLast, ap(ap(cons, x), nil)) -> nil
   ap(dropLast, ap(ap(cons, x), ap(ap(cons, y), ys))) -> ap(ap(cons, x), ap(dropLast, ap(ap(cons, y), ys)))

The set Q consists of the following terms:

   ap(ap(map, x0), x1)
   ap(ap(ap(if, true), x0), x1)
   ap(ap(ap(if, false), x0), x1)
   ap(isEmpty, nil)
   ap(isEmpty, ap(ap(cons, x0), x1))
   ap(last, ap(ap(cons, x0), nil))
   ap(last, ap(ap(cons, x0), ap(ap(cons, x1), x2)))
   ap(dropLast, nil)
   ap(dropLast, ap(ap(cons, x0), nil))
   ap(dropLast, ap(ap(cons, x0), ap(ap(cons, x1), x2)))


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

(3) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   AP(ap(map, f), xs) -> AP(ap(ap(if, ap(isEmpty, xs)), f), xs)
   AP(ap(map, f), xs) -> AP(ap(if, ap(isEmpty, xs)), f)
   AP(ap(map, f), xs) -> AP(if, ap(isEmpty, xs))
   AP(ap(map, f), xs) -> AP(isEmpty, xs)
   AP(ap(ap(if, false), f), xs) -> AP(ap(cons, ap(f, ap(last, xs))), ap(ap(map, f), ap(dropLast, xs)))
   AP(ap(ap(if, false), f), xs) -> AP(cons, ap(f, ap(last, xs)))
   AP(ap(ap(if, false), f), xs) -> AP(f, ap(last, xs))
   AP(ap(ap(if, false), f), xs) -> AP(last, xs)
   AP(ap(ap(if, false), f), xs) -> AP(ap(map, f), ap(dropLast, xs))
   AP(ap(ap(if, false), f), xs) -> AP(map, f)
   AP(ap(ap(if, false), f), xs) -> AP(dropLast, xs)
   AP(last, ap(ap(cons, x), ap(ap(cons, y), ys))) -> AP(last, ap(ap(cons, y), ys))
   AP(dropLast, ap(ap(cons, x), ap(ap(cons, y), ys))) -> AP(ap(cons, x), ap(dropLast, ap(ap(cons, y), ys)))
   AP(dropLast, ap(ap(cons, x), ap(ap(cons, y), ys))) -> AP(dropLast, ap(ap(cons, y), ys))

The TRS R consists of the following rules:

   ap(ap(map, f), xs) -> ap(ap(ap(if, ap(isEmpty, xs)), f), xs)
   ap(ap(ap(if, true), f), xs) -> nil
   ap(ap(ap(if, false), f), xs) -> ap(ap(cons, ap(f, ap(last, xs))), ap(ap(map, f), ap(dropLast, xs)))
   ap(isEmpty, nil) -> true
   ap(isEmpty, ap(ap(cons, x), xs)) -> false
   ap(last, ap(ap(cons, x), nil)) -> x
   ap(last, ap(ap(cons, x), ap(ap(cons, y), ys))) -> ap(last, ap(ap(cons, y), ys))
   ap(dropLast, nil) -> nil
   ap(dropLast, ap(ap(cons, x), nil)) -> nil
   ap(dropLast, ap(ap(cons, x), ap(ap(cons, y), ys))) -> ap(ap(cons, x), ap(dropLast, ap(ap(cons, y), ys)))

The set Q consists of the following terms:

   ap(ap(map, x0), x1)
   ap(ap(ap(if, true), x0), x1)
   ap(ap(ap(if, false), x0), x1)
   ap(isEmpty, nil)
   ap(isEmpty, ap(ap(cons, x0), x1))
   ap(last, ap(ap(cons, x0), nil))
   ap(last, ap(ap(cons, x0), ap(ap(cons, x1), x2)))
   ap(dropLast, nil)
   ap(dropLast, ap(ap(cons, x0), nil))
   ap(dropLast, ap(ap(cons, x0), ap(ap(cons, x1), x2)))

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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 9 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

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

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

   AP(dropLast, ap(ap(cons, x), ap(ap(cons, y), ys))) -> AP(dropLast, ap(ap(cons, y), ys))

The TRS R consists of the following rules:

   ap(ap(map, f), xs) -> ap(ap(ap(if, ap(isEmpty, xs)), f), xs)
   ap(ap(ap(if, true), f), xs) -> nil
   ap(ap(ap(if, false), f), xs) -> ap(ap(cons, ap(f, ap(last, xs))), ap(ap(map, f), ap(dropLast, xs)))
   ap(isEmpty, nil) -> true
   ap(isEmpty, ap(ap(cons, x), xs)) -> false
   ap(last, ap(ap(cons, x), nil)) -> x
   ap(last, ap(ap(cons, x), ap(ap(cons, y), ys))) -> ap(last, ap(ap(cons, y), ys))
   ap(dropLast, nil) -> nil
   ap(dropLast, ap(ap(cons, x), nil)) -> nil
   ap(dropLast, ap(ap(cons, x), ap(ap(cons, y), ys))) -> ap(ap(cons, x), ap(dropLast, ap(ap(cons, y), ys)))

The set Q consists of the following terms:

   ap(ap(map, x0), x1)
   ap(ap(ap(if, true), x0), x1)
   ap(ap(ap(if, false), x0), x1)
   ap(isEmpty, nil)
   ap(isEmpty, ap(ap(cons, x0), x1))
   ap(last, ap(ap(cons, x0), nil))
   ap(last, ap(ap(cons, x0), ap(ap(cons, x1), x2)))
   ap(dropLast, nil)
   ap(dropLast, ap(ap(cons, x0), nil))
   ap(dropLast, ap(ap(cons, x0), ap(ap(cons, x1), x2)))

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

(8) 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.
----------------------------------------

(9)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   AP(dropLast, ap(ap(cons, x), ap(ap(cons, y), ys))) -> AP(dropLast, ap(ap(cons, y), ys))

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

   ap(ap(map, x0), x1)
   ap(ap(ap(if, true), x0), x1)
   ap(ap(ap(if, false), x0), x1)
   ap(isEmpty, nil)
   ap(isEmpty, ap(ap(cons, x0), x1))
   ap(last, ap(ap(cons, x0), nil))
   ap(last, ap(ap(cons, x0), ap(ap(cons, x1), x2)))
   ap(dropLast, nil)
   ap(dropLast, ap(ap(cons, x0), nil))
   ap(dropLast, ap(ap(cons, x0), ap(ap(cons, x1), x2)))

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

(10) ATransformationProof (EQUIVALENT)
We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. 
----------------------------------------

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

   dropLast1(cons(x, cons(y, ys))) -> dropLast1(cons(y, ys))

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

   map(x0, x1)
   if(true, x0, x1)
   if(false, x0, x1)
   isEmpty(nil)
   isEmpty(cons(x0, x1))
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   dropLast(nil)
   dropLast(cons(x0, nil))
   dropLast(cons(x0, cons(x1, x2)))

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

(12) 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, x1)
   if(true, x0, x1)
   if(false, x0, x1)
   isEmpty(nil)
   isEmpty(cons(x0, x1))
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   dropLast(nil)
   dropLast(cons(x0, nil))
   dropLast(cons(x0, cons(x1, x2)))


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

(13)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   dropLast1(cons(x, cons(y, ys))) -> dropLast1(cons(y, ys))

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

(14) 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:
*dropLast1(cons(x, cons(y, ys))) -> dropLast1(cons(y, ys))
The graph contains the following edges 1 > 1


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

(15)
YES

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

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

   AP(last, ap(ap(cons, x), ap(ap(cons, y), ys))) -> AP(last, ap(ap(cons, y), ys))

The TRS R consists of the following rules:

   ap(ap(map, f), xs) -> ap(ap(ap(if, ap(isEmpty, xs)), f), xs)
   ap(ap(ap(if, true), f), xs) -> nil
   ap(ap(ap(if, false), f), xs) -> ap(ap(cons, ap(f, ap(last, xs))), ap(ap(map, f), ap(dropLast, xs)))
   ap(isEmpty, nil) -> true
   ap(isEmpty, ap(ap(cons, x), xs)) -> false
   ap(last, ap(ap(cons, x), nil)) -> x
   ap(last, ap(ap(cons, x), ap(ap(cons, y), ys))) -> ap(last, ap(ap(cons, y), ys))
   ap(dropLast, nil) -> nil
   ap(dropLast, ap(ap(cons, x), nil)) -> nil
   ap(dropLast, ap(ap(cons, x), ap(ap(cons, y), ys))) -> ap(ap(cons, x), ap(dropLast, ap(ap(cons, y), ys)))

The set Q consists of the following terms:

   ap(ap(map, x0), x1)
   ap(ap(ap(if, true), x0), x1)
   ap(ap(ap(if, false), x0), x1)
   ap(isEmpty, nil)
   ap(isEmpty, ap(ap(cons, x0), x1))
   ap(last, ap(ap(cons, x0), nil))
   ap(last, ap(ap(cons, x0), ap(ap(cons, x1), x2)))
   ap(dropLast, nil)
   ap(dropLast, ap(ap(cons, x0), nil))
   ap(dropLast, ap(ap(cons, x0), ap(ap(cons, x1), x2)))

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

(17) 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.
----------------------------------------

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

   AP(last, ap(ap(cons, x), ap(ap(cons, y), ys))) -> AP(last, ap(ap(cons, y), ys))

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

   ap(ap(map, x0), x1)
   ap(ap(ap(if, true), x0), x1)
   ap(ap(ap(if, false), x0), x1)
   ap(isEmpty, nil)
   ap(isEmpty, ap(ap(cons, x0), x1))
   ap(last, ap(ap(cons, x0), nil))
   ap(last, ap(ap(cons, x0), ap(ap(cons, x1), x2)))
   ap(dropLast, nil)
   ap(dropLast, ap(ap(cons, x0), nil))
   ap(dropLast, ap(ap(cons, x0), ap(ap(cons, x1), x2)))

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

(19) ATransformationProof (EQUIVALENT)
We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. 
----------------------------------------

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

   last1(cons(x, cons(y, ys))) -> last1(cons(y, ys))

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

   map(x0, x1)
   if(true, x0, x1)
   if(false, x0, x1)
   isEmpty(nil)
   isEmpty(cons(x0, x1))
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   dropLast(nil)
   dropLast(cons(x0, nil))
   dropLast(cons(x0, cons(x1, x2)))

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

(21) 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, x1)
   if(true, x0, x1)
   if(false, x0, x1)
   isEmpty(nil)
   isEmpty(cons(x0, x1))
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   dropLast(nil)
   dropLast(cons(x0, nil))
   dropLast(cons(x0, cons(x1, x2)))


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

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

   last1(cons(x, cons(y, ys))) -> last1(cons(y, ys))

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

(23) 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:
*last1(cons(x, cons(y, ys))) -> last1(cons(y, ys))
The graph contains the following edges 1 > 1


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

(24)
YES

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

(25)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   AP(ap(ap(if, false), f), xs) -> AP(f, ap(last, xs))
   AP(ap(map, f), xs) -> AP(ap(ap(if, ap(isEmpty, xs)), f), xs)
   AP(ap(ap(if, false), f), xs) -> AP(ap(map, f), ap(dropLast, xs))

The TRS R consists of the following rules:

   ap(ap(map, f), xs) -> ap(ap(ap(if, ap(isEmpty, xs)), f), xs)
   ap(ap(ap(if, true), f), xs) -> nil
   ap(ap(ap(if, false), f), xs) -> ap(ap(cons, ap(f, ap(last, xs))), ap(ap(map, f), ap(dropLast, xs)))
   ap(isEmpty, nil) -> true
   ap(isEmpty, ap(ap(cons, x), xs)) -> false
   ap(last, ap(ap(cons, x), nil)) -> x
   ap(last, ap(ap(cons, x), ap(ap(cons, y), ys))) -> ap(last, ap(ap(cons, y), ys))
   ap(dropLast, nil) -> nil
   ap(dropLast, ap(ap(cons, x), nil)) -> nil
   ap(dropLast, ap(ap(cons, x), ap(ap(cons, y), ys))) -> ap(ap(cons, x), ap(dropLast, ap(ap(cons, y), ys)))

The set Q consists of the following terms:

   ap(ap(map, x0), x1)
   ap(ap(ap(if, true), x0), x1)
   ap(ap(ap(if, false), x0), x1)
   ap(isEmpty, nil)
   ap(isEmpty, ap(ap(cons, x0), x1))
   ap(last, ap(ap(cons, x0), nil))
   ap(last, ap(ap(cons, x0), ap(ap(cons, x1), x2)))
   ap(dropLast, nil)
   ap(dropLast, ap(ap(cons, x0), nil))
   ap(dropLast, ap(ap(cons, x0), ap(ap(cons, x1), x2)))

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

(26) 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.
----------------------------------------

(27)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   AP(ap(ap(if, false), f), xs) -> AP(f, ap(last, xs))
   AP(ap(map, f), xs) -> AP(ap(ap(if, ap(isEmpty, xs)), f), xs)
   AP(ap(ap(if, false), f), xs) -> AP(ap(map, f), ap(dropLast, xs))

The TRS R consists of the following rules:

   ap(dropLast, nil) -> nil
   ap(dropLast, ap(ap(cons, x), nil)) -> nil
   ap(dropLast, ap(ap(cons, x), ap(ap(cons, y), ys))) -> ap(ap(cons, x), ap(dropLast, ap(ap(cons, y), ys)))
   ap(isEmpty, nil) -> true
   ap(isEmpty, ap(ap(cons, x), xs)) -> false
   ap(last, ap(ap(cons, x), nil)) -> x
   ap(last, ap(ap(cons, x), ap(ap(cons, y), ys))) -> ap(last, ap(ap(cons, y), ys))

The set Q consists of the following terms:

   ap(ap(map, x0), x1)
   ap(ap(ap(if, true), x0), x1)
   ap(ap(ap(if, false), x0), x1)
   ap(isEmpty, nil)
   ap(isEmpty, ap(ap(cons, x0), x1))
   ap(last, ap(ap(cons, x0), nil))
   ap(last, ap(ap(cons, x0), ap(ap(cons, x1), x2)))
   ap(dropLast, nil)
   ap(dropLast, ap(ap(cons, x0), nil))
   ap(dropLast, ap(ap(cons, x0), ap(ap(cons, x1), x2)))

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

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


The following pairs can be oriented strictly and are deleted.

   AP(ap(ap(if, false), f), xs) -> AP(f, ap(last, xs))
The remaining pairs can at least be oriented weakly.
Used ordering:  Combined order from the following AFS and order.
AP(x1, x2)  =  x1

ap(x1, x2)  =  ap(x2)


Knuth-Bendix order [KBO] with precedence:trivial

and weight map:

   dummyConstant=1
   ap_1=1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none


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

(29)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   AP(ap(map, f), xs) -> AP(ap(ap(if, ap(isEmpty, xs)), f), xs)
   AP(ap(ap(if, false), f), xs) -> AP(ap(map, f), ap(dropLast, xs))

The TRS R consists of the following rules:

   ap(dropLast, nil) -> nil
   ap(dropLast, ap(ap(cons, x), nil)) -> nil
   ap(dropLast, ap(ap(cons, x), ap(ap(cons, y), ys))) -> ap(ap(cons, x), ap(dropLast, ap(ap(cons, y), ys)))
   ap(isEmpty, nil) -> true
   ap(isEmpty, ap(ap(cons, x), xs)) -> false
   ap(last, ap(ap(cons, x), nil)) -> x
   ap(last, ap(ap(cons, x), ap(ap(cons, y), ys))) -> ap(last, ap(ap(cons, y), ys))

The set Q consists of the following terms:

   ap(ap(map, x0), x1)
   ap(ap(ap(if, true), x0), x1)
   ap(ap(ap(if, false), x0), x1)
   ap(isEmpty, nil)
   ap(isEmpty, ap(ap(cons, x0), x1))
   ap(last, ap(ap(cons, x0), nil))
   ap(last, ap(ap(cons, x0), ap(ap(cons, x1), x2)))
   ap(dropLast, nil)
   ap(dropLast, ap(ap(cons, x0), nil))
   ap(dropLast, ap(ap(cons, x0), ap(ap(cons, x1), x2)))

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

(30) 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.
----------------------------------------

(31)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   AP(ap(map, f), xs) -> AP(ap(ap(if, ap(isEmpty, xs)), f), xs)
   AP(ap(ap(if, false), f), xs) -> AP(ap(map, f), ap(dropLast, xs))

The TRS R consists of the following rules:

   ap(dropLast, nil) -> nil
   ap(dropLast, ap(ap(cons, x), nil)) -> nil
   ap(dropLast, ap(ap(cons, x), ap(ap(cons, y), ys))) -> ap(ap(cons, x), ap(dropLast, ap(ap(cons, y), ys)))
   ap(isEmpty, nil) -> true
   ap(isEmpty, ap(ap(cons, x), xs)) -> false

The set Q consists of the following terms:

   ap(ap(map, x0), x1)
   ap(ap(ap(if, true), x0), x1)
   ap(ap(ap(if, false), x0), x1)
   ap(isEmpty, nil)
   ap(isEmpty, ap(ap(cons, x0), x1))
   ap(last, ap(ap(cons, x0), nil))
   ap(last, ap(ap(cons, x0), ap(ap(cons, x1), x2)))
   ap(dropLast, nil)
   ap(dropLast, ap(ap(cons, x0), nil))
   ap(dropLast, ap(ap(cons, x0), ap(ap(cons, x1), x2)))

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

(32) ATransformationProof (EQUIVALENT)
We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. 
----------------------------------------

(33)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   map1(f, xs) -> if1(isEmpty(xs), f, xs)
   if1(false, f, xs) -> map1(f, dropLast(xs))

The TRS R consists of the following rules:

   dropLast(nil) -> nil
   dropLast(cons(x, nil)) -> nil
   dropLast(cons(x, cons(y, ys))) -> cons(x, dropLast(cons(y, ys)))
   isEmpty(nil) -> true
   isEmpty(cons(x, xs)) -> false

The set Q consists of the following terms:

   map(x0, x1)
   if(true, x0, x1)
   if(false, x0, x1)
   isEmpty(nil)
   isEmpty(cons(x0, x1))
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   dropLast(nil)
   dropLast(cons(x0, nil))
   dropLast(cons(x0, cons(x1, x2)))

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

(34) 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, x1)
   if(true, x0, x1)
   if(false, x0, x1)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))


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

(35)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   map1(f, xs) -> if1(isEmpty(xs), f, xs)
   if1(false, f, xs) -> map1(f, dropLast(xs))

The TRS R consists of the following rules:

   dropLast(nil) -> nil
   dropLast(cons(x, nil)) -> nil
   dropLast(cons(x, cons(y, ys))) -> cons(x, dropLast(cons(y, ys)))
   isEmpty(nil) -> true
   isEmpty(cons(x, xs)) -> false

The set Q consists of the following terms:

   isEmpty(nil)
   isEmpty(cons(x0, x1))
   dropLast(nil)
   dropLast(cons(x0, nil))
   dropLast(cons(x0, cons(x1, x2)))

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

(36) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule map1(f, xs) -> if1(isEmpty(xs), f, xs) at position [0] we obtained the following new rules [LPAR04]:

   (map1(y0, nil) -> if1(true, y0, nil),map1(y0, nil) -> if1(true, y0, nil))
   (map1(y0, cons(x0, x1)) -> if1(false, y0, cons(x0, x1)),map1(y0, cons(x0, x1)) -> if1(false, y0, cons(x0, x1)))


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

(37)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   if1(false, f, xs) -> map1(f, dropLast(xs))
   map1(y0, nil) -> if1(true, y0, nil)
   map1(y0, cons(x0, x1)) -> if1(false, y0, cons(x0, x1))

The TRS R consists of the following rules:

   dropLast(nil) -> nil
   dropLast(cons(x, nil)) -> nil
   dropLast(cons(x, cons(y, ys))) -> cons(x, dropLast(cons(y, ys)))
   isEmpty(nil) -> true
   isEmpty(cons(x, xs)) -> false

The set Q consists of the following terms:

   isEmpty(nil)
   isEmpty(cons(x0, x1))
   dropLast(nil)
   dropLast(cons(x0, nil))
   dropLast(cons(x0, cons(x1, x2)))

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

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

(39)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   map1(y0, cons(x0, x1)) -> if1(false, y0, cons(x0, x1))
   if1(false, f, xs) -> map1(f, dropLast(xs))

The TRS R consists of the following rules:

   dropLast(nil) -> nil
   dropLast(cons(x, nil)) -> nil
   dropLast(cons(x, cons(y, ys))) -> cons(x, dropLast(cons(y, ys)))
   isEmpty(nil) -> true
   isEmpty(cons(x, xs)) -> false

The set Q consists of the following terms:

   isEmpty(nil)
   isEmpty(cons(x0, x1))
   dropLast(nil)
   dropLast(cons(x0, nil))
   dropLast(cons(x0, cons(x1, x2)))

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

(40) 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.
----------------------------------------

(41)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   map1(y0, cons(x0, x1)) -> if1(false, y0, cons(x0, x1))
   if1(false, f, xs) -> map1(f, dropLast(xs))

The TRS R consists of the following rules:

   dropLast(nil) -> nil
   dropLast(cons(x, nil)) -> nil
   dropLast(cons(x, cons(y, ys))) -> cons(x, dropLast(cons(y, ys)))

The set Q consists of the following terms:

   isEmpty(nil)
   isEmpty(cons(x0, x1))
   dropLast(nil)
   dropLast(cons(x0, nil))
   dropLast(cons(x0, cons(x1, x2)))

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

(42) 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].

   isEmpty(nil)
   isEmpty(cons(x0, x1))


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

(43)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   map1(y0, cons(x0, x1)) -> if1(false, y0, cons(x0, x1))
   if1(false, f, xs) -> map1(f, dropLast(xs))

The TRS R consists of the following rules:

   dropLast(nil) -> nil
   dropLast(cons(x, nil)) -> nil
   dropLast(cons(x, cons(y, ys))) -> cons(x, dropLast(cons(y, ys)))

The set Q consists of the following terms:

   dropLast(nil)
   dropLast(cons(x0, nil))
   dropLast(cons(x0, cons(x1, x2)))

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

(44) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.


Strictly oriented rules of the TRS R:

   dropLast(cons(x, nil)) -> nil

Used ordering: Polynomial interpretation [POLO]:

   POL(cons(x_1, x_2)) = x_1 + 2*x_2
   POL(dropLast(x_1)) = x_1
   POL(false) = 0
   POL(if1(x_1, x_2, x_3)) = 2*x_1 + x_2 + x_3
   POL(map1(x_1, x_2)) = x_1 + x_2
   POL(nil) = 1


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

(45)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   map1(y0, cons(x0, x1)) -> if1(false, y0, cons(x0, x1))
   if1(false, f, xs) -> map1(f, dropLast(xs))

The TRS R consists of the following rules:

   dropLast(nil) -> nil
   dropLast(cons(x, cons(y, ys))) -> cons(x, dropLast(cons(y, ys)))

The set Q consists of the following terms:

   dropLast(nil)
   dropLast(cons(x0, nil))
   dropLast(cons(x0, cons(x1, x2)))

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

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


The following pairs can be oriented strictly and are deleted.

   map1(y0, cons(x0, x1)) -> if1(false, y0, cons(x0, x1))
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] to (N^2, +, *, >=, >) :

   <<<
 POL(map1(x_1, x_2)) =  	[[0]] 	 +  	[[1, 1]] 	* 	x_1 	 +  	[[1, 0]] 	* 	x_2
>>>

   <<<
 POL(cons(x_1, x_2)) =  	[[1], [0]] 	 +  	[[0, 0], [0, 0]] 	* 	x_1 	 +  	[[1, 0], [1, 0]] 	* 	x_2
>>>

   <<<
 POL(if1(x_1, x_2, x_3)) =  	[[0]] 	 +  	[[1, 1]] 	* 	x_1 	 +  	[[1, 1]] 	* 	x_2 	 +  	[[0, 1]] 	* 	x_3
>>>

   <<<
 POL(false) =  	[[0], [0]]
>>>

   <<<
 POL(dropLast(x_1)) =  	[[0], [0]] 	 +  	[[0, 1], [1, 0]] 	* 	x_1
>>>

   <<<
 POL(nil) =  	[[1], [1]]
>>>


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

   dropLast(nil) -> nil
   dropLast(cons(x, cons(y, ys))) -> cons(x, dropLast(cons(y, ys)))


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

(47)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   if1(false, f, xs) -> map1(f, dropLast(xs))

The TRS R consists of the following rules:

   dropLast(nil) -> nil
   dropLast(cons(x, cons(y, ys))) -> cons(x, dropLast(cons(y, ys)))

The set Q consists of the following terms:

   dropLast(nil)
   dropLast(cons(x0, nil))
   dropLast(cons(x0, cons(x1, x2)))

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

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

(49)
TRUE