/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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), 7 ms]
(2) CpxTRS
    (3) RelTrsToTrsProof [UPPER BOUND(ID), 0 ms]
    (4) CpxTRS
    (5) CpxTrsMatchBoundsTAProof [FINISHED, 364 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


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(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(zeros) -> mark(cons(0, zeros))
   active(tail(cons(X, XS))) -> mark(XS)
   active(cons(X1, X2)) -> cons(active(X1), X2)
   active(tail(X)) -> tail(active(X))
   cons(mark(X1), X2) -> mark(cons(X1, X2))
   tail(mark(X)) -> mark(tail(X))
   proper(zeros) -> ok(zeros)
   proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
   proper(0) -> ok(0)
   proper(tail(X)) -> tail(proper(X))
   cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
   tail(ok(X)) -> ok(tail(X))
   top(mark(X)) -> top(proper(X))
   top(ok(X)) -> top(active(X))

S is empty.
Rewrite Strategy: FULL
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(1) NestedDefinedSymbolProof (UPPER BOUND(ID))
The following defined symbols can occur below the 0th argument of cons: active, proper, cons
The following defined symbols can occur below the 1th argument of cons: active, proper, cons
The following defined symbols can occur below the 0th argument of top: active, proper, cons
The following defined symbols can occur below the 0th argument of proper: active, proper, cons
The following defined symbols can occur below the 0th argument of active: active, proper, cons

Hence, the left-hand sides of the following rules are not basic-reachable and can be removed:
active(tail(cons(X, XS))) -> mark(XS)
active(tail(X)) -> tail(active(X))
proper(tail(X)) -> tail(proper(X))

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(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:

   active(zeros) -> mark(cons(0, zeros))
   active(cons(X1, X2)) -> cons(active(X1), X2)
   cons(mark(X1), X2) -> mark(cons(X1, X2))
   tail(mark(X)) -> mark(tail(X))
   proper(zeros) -> ok(zeros)
   proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
   proper(0) -> ok(0)
   cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
   tail(ok(X)) -> ok(tail(X))
   top(mark(X)) -> top(proper(X))
   top(ok(X)) -> top(active(X))

S is empty.
Rewrite Strategy: FULL
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(3) RelTrsToTrsProof (UPPER BOUND(ID))
transformed relative TRS to TRS
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(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:

   active(zeros) -> mark(cons(0, zeros))
   active(cons(X1, X2)) -> cons(active(X1), X2)
   cons(mark(X1), X2) -> mark(cons(X1, X2))
   tail(mark(X)) -> mark(tail(X))
   proper(zeros) -> ok(zeros)
   proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
   proper(0) -> ok(0)
   cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
   tail(ok(X)) -> ok(tail(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 5. 

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]
transitions: 
zeros0() -> 0
mark0(0) -> 0
00() -> 0
ok0(0) -> 0
active0(0) -> 1
cons0(0, 0) -> 2
tail0(0) -> 3
proper0(0) -> 4
top0(0) -> 5
01() -> 7
zeros1() -> 8
cons1(7, 8) -> 6
mark1(6) -> 1
cons1(0, 0) -> 9
mark1(9) -> 2
tail1(0) -> 10
mark1(10) -> 3
zeros1() -> 11
ok1(11) -> 4
01() -> 12
ok1(12) -> 4
cons1(0, 0) -> 13
ok1(13) -> 2
tail1(0) -> 14
ok1(14) -> 3
proper1(0) -> 15
top1(15) -> 5
active1(0) -> 16
top1(16) -> 5
mark1(6) -> 16
mark1(9) -> 9
mark1(9) -> 13
mark1(10) -> 10
mark1(10) -> 14
ok1(11) -> 15
ok1(12) -> 15
ok1(13) -> 9
ok1(13) -> 13
ok1(14) -> 10
ok1(14) -> 14
proper2(6) -> 17
top2(17) -> 5
active2(11) -> 18
top2(18) -> 5
active2(12) -> 18
02() -> 20
zeros2() -> 21
cons2(20, 21) -> 19
mark2(19) -> 18
proper2(7) -> 22
proper2(8) -> 23
cons2(22, 23) -> 17
zeros2() -> 24
ok2(24) -> 23
02() -> 25
ok2(25) -> 22
proper3(19) -> 26
top3(26) -> 5
proper3(20) -> 27
proper3(21) -> 28
cons3(27, 28) -> 26
cons3(25, 24) -> 29
ok3(29) -> 17
zeros3() -> 30
ok3(30) -> 28
03() -> 31
ok3(31) -> 27
active3(29) -> 32
top3(32) -> 5
cons4(31, 30) -> 33
ok4(33) -> 26
active4(25) -> 34
cons4(34, 24) -> 32
active4(33) -> 35
top4(35) -> 5
active5(31) -> 36
cons5(36, 30) -> 35

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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.
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(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(zeros) -> mark(cons(0, zeros))
   active(tail(cons(X, XS))) -> mark(XS)
   active(cons(X1, X2)) -> cons(active(X1), X2)
   active(tail(X)) -> tail(active(X))
   cons(mark(X1), X2) -> mark(cons(X1, X2))
   tail(mark(X)) -> mark(tail(X))
   proper(zeros) -> ok(zeros)
   proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
   proper(0) -> ok(0)
   proper(tail(X)) -> tail(proper(X))
   cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
   tail(ok(X)) -> ok(tail(X))
   top(mark(X)) -> top(proper(X))
   top(ok(X)) -> top(active(X))

S is empty.
Rewrite Strategy: FULL
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(9) DecreasingLoopProof (LOWER BOUND(ID))
The following loop(s) give(s) rise to the lower bound Omega(n^1):

The rewrite sequence

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

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(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(zeros) -> mark(cons(0, zeros))
   active(tail(cons(X, XS))) -> mark(XS)
   active(cons(X1, X2)) -> cons(active(X1), X2)
   active(tail(X)) -> tail(active(X))
   cons(mark(X1), X2) -> mark(cons(X1, X2))
   tail(mark(X)) -> mark(tail(X))
   proper(zeros) -> ok(zeros)
   proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
   proper(0) -> ok(0)
   proper(tail(X)) -> tail(proper(X))
   cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
   tail(ok(X)) -> ok(tail(X))
   top(mark(X)) -> top(proper(X))
   top(ok(X)) -> top(active(X))

S is empty.
Rewrite Strategy: FULL
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(12) LowerBoundPropagationProof (FINISHED)
Propagated lower bound.
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(13)
BOUNDS(n^1, INF)

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(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(zeros) -> mark(cons(0, zeros))
   active(tail(cons(X, XS))) -> mark(XS)
   active(cons(X1, X2)) -> cons(active(X1), X2)
   active(tail(X)) -> tail(active(X))
   cons(mark(X1), X2) -> mark(cons(X1, X2))
   tail(mark(X)) -> mark(tail(X))
   proper(zeros) -> ok(zeros)
   proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
   proper(0) -> ok(0)
   proper(tail(X)) -> tail(proper(X))
   cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
   tail(ok(X)) -> ok(tail(X))
   top(mark(X)) -> top(proper(X))
   top(ok(X)) -> top(active(X))

S is empty.
Rewrite Strategy: FULL