/export/starexec/sandbox2/solver/bin/starexec_run_complexity /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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WORST_CASE(Omega(n^1), O(n^1))
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, n^1).

(0) CpxTRS
(1) NestedDefinedSymbolProof [UPPER BOUND(ID), 0 ms]
(2) CpxTRS
    (3) RelTrsToTrsProof [UPPER BOUND(ID), 0 ms]
    (4) CpxTRS
    (5) CpxTrsMatchBoundsTAProof [FINISHED, 68 ms]
    (6) BOUNDS(1, n^1)
(7) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms]
(8) TRS for Loop Detection
    (9) DecreasingLoopProof [LOWER BOUND(ID), 0 ms]
    (10) BEST
        (11) proven lower bound
            (12) LowerBoundPropagationProof [FINISHED, 0 ms]
            (13) BOUNDS(n^1, INF)
        (14) TRS for Loop Detection


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

(0)
Obligation:
The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, n^1).


The TRS R consists of the following rules:

   active(fst(0, Z)) -> mark(nil)
   active(fst(s(X), cons(Y, Z))) -> mark(cons(Y, fst(X, Z)))
   active(from(X)) -> mark(cons(X, from(s(X))))
   active(add(0, X)) -> mark(X)
   active(add(s(X), Y)) -> mark(s(add(X, Y)))
   active(len(nil)) -> mark(0)
   active(len(cons(X, Z))) -> mark(s(len(Z)))
   active(cons(X1, X2)) -> cons(active(X1), X2)
   active(fst(X1, X2)) -> fst(active(X1), X2)
   active(fst(X1, X2)) -> fst(X1, active(X2))
   active(from(X)) -> from(active(X))
   active(add(X1, X2)) -> add(active(X1), X2)
   active(add(X1, X2)) -> add(X1, active(X2))
   active(len(X)) -> len(active(X))
   cons(mark(X1), X2) -> mark(cons(X1, X2))
   fst(mark(X1), X2) -> mark(fst(X1, X2))
   fst(X1, mark(X2)) -> mark(fst(X1, X2))
   from(mark(X)) -> mark(from(X))
   add(mark(X1), X2) -> mark(add(X1, X2))
   add(X1, mark(X2)) -> mark(add(X1, X2))
   len(mark(X)) -> mark(len(X))
   proper(0) -> ok(0)
   proper(s(X)) -> s(proper(X))
   proper(nil) -> ok(nil)
   proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
   proper(fst(X1, X2)) -> fst(proper(X1), proper(X2))
   proper(from(X)) -> from(proper(X))
   proper(add(X1, X2)) -> add(proper(X1), proper(X2))
   proper(len(X)) -> len(proper(X))
   s(ok(X)) -> ok(s(X))
   cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
   fst(ok(X1), ok(X2)) -> ok(fst(X1, X2))
   from(ok(X)) -> ok(from(X))
   add(ok(X1), ok(X2)) -> ok(add(X1, X2))
   len(ok(X)) -> ok(len(X))
   top(mark(X)) -> top(proper(X))
   top(ok(X)) -> top(active(X))

S is empty.
Rewrite Strategy: FULL
----------------------------------------

(1) NestedDefinedSymbolProof (UPPER BOUND(ID))
The following defined symbols can occur below the 0th argument of top: proper, active
The following defined symbols can occur below the 0th argument of proper: proper, active
The following defined symbols can occur below the 0th argument of active: proper, active

Hence, the left-hand sides of the following rules are not basic-reachable and can be removed:
active(fst(0, Z)) -> mark(nil)
active(fst(s(X), cons(Y, Z))) -> mark(cons(Y, fst(X, Z)))
active(from(X)) -> mark(cons(X, from(s(X))))
active(add(0, X)) -> mark(X)
active(add(s(X), Y)) -> mark(s(add(X, Y)))
active(len(nil)) -> mark(0)
active(len(cons(X, Z))) -> mark(s(len(Z)))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(fst(X1, X2)) -> fst(active(X1), X2)
active(fst(X1, X2)) -> fst(X1, active(X2))
active(from(X)) -> from(active(X))
active(add(X1, X2)) -> add(active(X1), X2)
active(add(X1, X2)) -> add(X1, active(X2))
active(len(X)) -> len(active(X))
proper(s(X)) -> s(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(fst(X1, X2)) -> fst(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(add(X1, X2)) -> add(proper(X1), proper(X2))
proper(len(X)) -> len(proper(X))

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

(2)
Obligation:
The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(1, n^1).


The TRS R consists of the following rules:

   cons(mark(X1), X2) -> mark(cons(X1, X2))
   fst(mark(X1), X2) -> mark(fst(X1, X2))
   fst(X1, mark(X2)) -> mark(fst(X1, X2))
   from(mark(X)) -> mark(from(X))
   add(mark(X1), X2) -> mark(add(X1, X2))
   add(X1, mark(X2)) -> mark(add(X1, X2))
   len(mark(X)) -> mark(len(X))
   proper(0) -> ok(0)
   proper(nil) -> ok(nil)
   s(ok(X)) -> ok(s(X))
   cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
   fst(ok(X1), ok(X2)) -> ok(fst(X1, X2))
   from(ok(X)) -> ok(from(X))
   add(ok(X1), ok(X2)) -> ok(add(X1, X2))
   len(ok(X)) -> ok(len(X))
   top(mark(X)) -> top(proper(X))
   top(ok(X)) -> top(active(X))

S is empty.
Rewrite Strategy: FULL
----------------------------------------

(3) RelTrsToTrsProof (UPPER BOUND(ID))
transformed relative TRS to TRS
----------------------------------------

(4)
Obligation:
The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(1, n^1).


The TRS R consists of the following rules:

   cons(mark(X1), X2) -> mark(cons(X1, X2))
   fst(mark(X1), X2) -> mark(fst(X1, X2))
   fst(X1, mark(X2)) -> mark(fst(X1, X2))
   from(mark(X)) -> mark(from(X))
   add(mark(X1), X2) -> mark(add(X1, X2))
   add(X1, mark(X2)) -> mark(add(X1, X2))
   len(mark(X)) -> mark(len(X))
   proper(0) -> ok(0)
   proper(nil) -> ok(nil)
   s(ok(X)) -> ok(s(X))
   cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
   fst(ok(X1), ok(X2)) -> ok(fst(X1, X2))
   from(ok(X)) -> ok(from(X))
   add(ok(X1), ok(X2)) -> ok(add(X1, X2))
   len(ok(X)) -> ok(len(X))
   top(mark(X)) -> top(proper(X))
   top(ok(X)) -> top(active(X))

S is empty.
Rewrite Strategy: FULL
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(5) CpxTrsMatchBoundsTAProof (FINISHED)
A linear upper bound on the runtime complexity of the TRS R could be shown with a Match-Bound[TAB_LEFTLINEAR,TAB_NONLEFTLINEAR] (for contructor-based start-terms) of 2. 

The compatible tree automaton used to show the Match-Boundedness (for constructor-based start-terms) is represented by: 
final states : [1, 2, 3, 4, 5, 6, 7, 8]
transitions: 
mark0(0) -> 0
00() -> 0
ok0(0) -> 0
nil0() -> 0
active0(0) -> 0
cons0(0, 0) -> 1
fst0(0, 0) -> 2
from0(0) -> 3
add0(0, 0) -> 4
len0(0) -> 5
proper0(0) -> 6
s0(0) -> 7
top0(0) -> 8
cons1(0, 0) -> 9
mark1(9) -> 1
fst1(0, 0) -> 10
mark1(10) -> 2
from1(0) -> 11
mark1(11) -> 3
add1(0, 0) -> 12
mark1(12) -> 4
len1(0) -> 13
mark1(13) -> 5
01() -> 14
ok1(14) -> 6
nil1() -> 15
ok1(15) -> 6
s1(0) -> 16
ok1(16) -> 7
cons1(0, 0) -> 17
ok1(17) -> 1
fst1(0, 0) -> 18
ok1(18) -> 2
from1(0) -> 19
ok1(19) -> 3
add1(0, 0) -> 20
ok1(20) -> 4
len1(0) -> 21
ok1(21) -> 5
proper1(0) -> 22
top1(22) -> 8
active1(0) -> 23
top1(23) -> 8
mark1(9) -> 9
mark1(9) -> 17
mark1(10) -> 10
mark1(10) -> 18
mark1(11) -> 11
mark1(11) -> 19
mark1(12) -> 12
mark1(12) -> 20
mark1(13) -> 13
mark1(13) -> 21
ok1(14) -> 22
ok1(15) -> 22
ok1(16) -> 16
ok1(17) -> 9
ok1(17) -> 17
ok1(18) -> 10
ok1(18) -> 18
ok1(19) -> 11
ok1(19) -> 19
ok1(20) -> 12
ok1(20) -> 20
ok1(21) -> 13
ok1(21) -> 21
active2(14) -> 24
top2(24) -> 8
active2(15) -> 24

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(6)
BOUNDS(1, n^1)

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(7) RelTrsToDecreasingLoopProblemProof (LOWER BOUND(ID))
Transformed a relative TRS into a decreasing-loop problem.
----------------------------------------

(8)
Obligation:
Analyzing the following TRS for decreasing loops:

The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, n^1).


The TRS R consists of the following rules:

   active(fst(0, Z)) -> mark(nil)
   active(fst(s(X), cons(Y, Z))) -> mark(cons(Y, fst(X, Z)))
   active(from(X)) -> mark(cons(X, from(s(X))))
   active(add(0, X)) -> mark(X)
   active(add(s(X), Y)) -> mark(s(add(X, Y)))
   active(len(nil)) -> mark(0)
   active(len(cons(X, Z))) -> mark(s(len(Z)))
   active(cons(X1, X2)) -> cons(active(X1), X2)
   active(fst(X1, X2)) -> fst(active(X1), X2)
   active(fst(X1, X2)) -> fst(X1, active(X2))
   active(from(X)) -> from(active(X))
   active(add(X1, X2)) -> add(active(X1), X2)
   active(add(X1, X2)) -> add(X1, active(X2))
   active(len(X)) -> len(active(X))
   cons(mark(X1), X2) -> mark(cons(X1, X2))
   fst(mark(X1), X2) -> mark(fst(X1, X2))
   fst(X1, mark(X2)) -> mark(fst(X1, X2))
   from(mark(X)) -> mark(from(X))
   add(mark(X1), X2) -> mark(add(X1, X2))
   add(X1, mark(X2)) -> mark(add(X1, X2))
   len(mark(X)) -> mark(len(X))
   proper(0) -> ok(0)
   proper(s(X)) -> s(proper(X))
   proper(nil) -> ok(nil)
   proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
   proper(fst(X1, X2)) -> fst(proper(X1), proper(X2))
   proper(from(X)) -> from(proper(X))
   proper(add(X1, X2)) -> add(proper(X1), proper(X2))
   proper(len(X)) -> len(proper(X))
   s(ok(X)) -> ok(s(X))
   cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
   fst(ok(X1), ok(X2)) -> ok(fst(X1, X2))
   from(ok(X)) -> ok(from(X))
   add(ok(X1), ok(X2)) -> ok(add(X1, X2))
   len(ok(X)) -> ok(len(X))
   top(mark(X)) -> top(proper(X))
   top(ok(X)) -> top(active(X))

S is empty.
Rewrite Strategy: FULL
----------------------------------------

(9) DecreasingLoopProof (LOWER BOUND(ID))
The following loop(s) give(s) rise to the lower bound Omega(n^1):

The rewrite sequence

from(ok(X)) ->^+ ok(from(X))

gives rise to a decreasing loop by considering the right hand sides subterm at position [0].

The pumping substitution is [X / ok(X)].

The result substitution is [ ].




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(10)
Complex Obligation (BEST)

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

(11)
Obligation:
Proved the lower bound n^1 for the following obligation:

The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, n^1).


The TRS R consists of the following rules:

   active(fst(0, Z)) -> mark(nil)
   active(fst(s(X), cons(Y, Z))) -> mark(cons(Y, fst(X, Z)))
   active(from(X)) -> mark(cons(X, from(s(X))))
   active(add(0, X)) -> mark(X)
   active(add(s(X), Y)) -> mark(s(add(X, Y)))
   active(len(nil)) -> mark(0)
   active(len(cons(X, Z))) -> mark(s(len(Z)))
   active(cons(X1, X2)) -> cons(active(X1), X2)
   active(fst(X1, X2)) -> fst(active(X1), X2)
   active(fst(X1, X2)) -> fst(X1, active(X2))
   active(from(X)) -> from(active(X))
   active(add(X1, X2)) -> add(active(X1), X2)
   active(add(X1, X2)) -> add(X1, active(X2))
   active(len(X)) -> len(active(X))
   cons(mark(X1), X2) -> mark(cons(X1, X2))
   fst(mark(X1), X2) -> mark(fst(X1, X2))
   fst(X1, mark(X2)) -> mark(fst(X1, X2))
   from(mark(X)) -> mark(from(X))
   add(mark(X1), X2) -> mark(add(X1, X2))
   add(X1, mark(X2)) -> mark(add(X1, X2))
   len(mark(X)) -> mark(len(X))
   proper(0) -> ok(0)
   proper(s(X)) -> s(proper(X))
   proper(nil) -> ok(nil)
   proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
   proper(fst(X1, X2)) -> fst(proper(X1), proper(X2))
   proper(from(X)) -> from(proper(X))
   proper(add(X1, X2)) -> add(proper(X1), proper(X2))
   proper(len(X)) -> len(proper(X))
   s(ok(X)) -> ok(s(X))
   cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
   fst(ok(X1), ok(X2)) -> ok(fst(X1, X2))
   from(ok(X)) -> ok(from(X))
   add(ok(X1), ok(X2)) -> ok(add(X1, X2))
   len(ok(X)) -> ok(len(X))
   top(mark(X)) -> top(proper(X))
   top(ok(X)) -> top(active(X))

S is empty.
Rewrite Strategy: FULL
----------------------------------------

(12) LowerBoundPropagationProof (FINISHED)
Propagated lower bound.
----------------------------------------

(13)
BOUNDS(n^1, INF)

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

(14)
Obligation:
Analyzing the following TRS for decreasing loops:

The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, n^1).


The TRS R consists of the following rules:

   active(fst(0, Z)) -> mark(nil)
   active(fst(s(X), cons(Y, Z))) -> mark(cons(Y, fst(X, Z)))
   active(from(X)) -> mark(cons(X, from(s(X))))
   active(add(0, X)) -> mark(X)
   active(add(s(X), Y)) -> mark(s(add(X, Y)))
   active(len(nil)) -> mark(0)
   active(len(cons(X, Z))) -> mark(s(len(Z)))
   active(cons(X1, X2)) -> cons(active(X1), X2)
   active(fst(X1, X2)) -> fst(active(X1), X2)
   active(fst(X1, X2)) -> fst(X1, active(X2))
   active(from(X)) -> from(active(X))
   active(add(X1, X2)) -> add(active(X1), X2)
   active(add(X1, X2)) -> add(X1, active(X2))
   active(len(X)) -> len(active(X))
   cons(mark(X1), X2) -> mark(cons(X1, X2))
   fst(mark(X1), X2) -> mark(fst(X1, X2))
   fst(X1, mark(X2)) -> mark(fst(X1, X2))
   from(mark(X)) -> mark(from(X))
   add(mark(X1), X2) -> mark(add(X1, X2))
   add(X1, mark(X2)) -> mark(add(X1, X2))
   len(mark(X)) -> mark(len(X))
   proper(0) -> ok(0)
   proper(s(X)) -> s(proper(X))
   proper(nil) -> ok(nil)
   proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
   proper(fst(X1, X2)) -> fst(proper(X1), proper(X2))
   proper(from(X)) -> from(proper(X))
   proper(add(X1, X2)) -> add(proper(X1), proper(X2))
   proper(len(X)) -> len(proper(X))
   s(ok(X)) -> ok(s(X))
   cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
   fst(ok(X1), ok(X2)) -> ok(fst(X1, X2))
   from(ok(X)) -> ok(from(X))
   add(ok(X1), ok(X2)) -> ok(add(X1, X2))
   len(ok(X)) -> ok(len(X))
   top(mark(X)) -> top(proper(X))
   top(ok(X)) -> top(active(X))

S is empty.
Rewrite Strategy: FULL