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


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WORST_CASE(NON_POLY, ?)
proof of /export/starexec/sandbox/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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(EXP, INF).

(0) CpxTRS
(1) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms]
(2) TRS for Loop Detection
(3) DecreasingLoopProof [LOWER BOUND(ID), 0 ms]
(4) BEST
    (5) proven lower bound
        (6) LowerBoundPropagationProof [FINISHED, 0 ms]
        (7) BOUNDS(n^1, INF)
    (8) TRS for Loop Detection
        (9) DecreasingLoopProof [FINISHED, 0 ms]
        (10) BOUNDS(EXP, INF)


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


The TRS R consists of the following rules:

   sort(l) -> st(0, l)
   st(n, l) -> cond1(member(n, l), n, l)
   cond1(true, n, l) -> cons(n, st(s(n), l))
   cond1(false, n, l) -> cond2(gt(n, max(l)), n, l)
   cond2(true, n, l) -> nil
   cond2(false, n, l) -> st(s(n), l)
   member(n, nil) -> false
   member(n, cons(m, l)) -> or(equal(n, m), member(n, l))
   or(x, true) -> true
   or(x, false) -> x
   equal(0, 0) -> true
   equal(s(x), 0) -> false
   equal(0, s(y)) -> false
   equal(s(x), s(y)) -> equal(x, y)
   gt(0, v) -> false
   gt(s(u), 0) -> true
   gt(s(u), s(v)) -> gt(u, v)
   max(nil) -> 0
   max(cons(u, l)) -> if(gt(u, max(l)), u, max(l))
   if(true, u, v) -> u
   if(false, u, v) -> v

S is empty.
Rewrite Strategy: FULL
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(1) RelTrsToDecreasingLoopProblemProof (LOWER BOUND(ID))
Transformed a relative TRS into a decreasing-loop problem.
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(2)
Obligation:
Analyzing the following TRS for decreasing loops:

The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF).


The TRS R consists of the following rules:

   sort(l) -> st(0, l)
   st(n, l) -> cond1(member(n, l), n, l)
   cond1(true, n, l) -> cons(n, st(s(n), l))
   cond1(false, n, l) -> cond2(gt(n, max(l)), n, l)
   cond2(true, n, l) -> nil
   cond2(false, n, l) -> st(s(n), l)
   member(n, nil) -> false
   member(n, cons(m, l)) -> or(equal(n, m), member(n, l))
   or(x, true) -> true
   or(x, false) -> x
   equal(0, 0) -> true
   equal(s(x), 0) -> false
   equal(0, s(y)) -> false
   equal(s(x), s(y)) -> equal(x, y)
   gt(0, v) -> false
   gt(s(u), 0) -> true
   gt(s(u), s(v)) -> gt(u, v)
   max(nil) -> 0
   max(cons(u, l)) -> if(gt(u, max(l)), u, max(l))
   if(true, u, v) -> u
   if(false, u, v) -> v

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

The rewrite sequence

gt(s(u), s(v)) ->^+ gt(u, v)

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

The pumping substitution is [u / s(u), v / s(v)].

The result substitution is [ ].




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(4)
Complex Obligation (BEST)

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(5)
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(EXP, INF).


The TRS R consists of the following rules:

   sort(l) -> st(0, l)
   st(n, l) -> cond1(member(n, l), n, l)
   cond1(true, n, l) -> cons(n, st(s(n), l))
   cond1(false, n, l) -> cond2(gt(n, max(l)), n, l)
   cond2(true, n, l) -> nil
   cond2(false, n, l) -> st(s(n), l)
   member(n, nil) -> false
   member(n, cons(m, l)) -> or(equal(n, m), member(n, l))
   or(x, true) -> true
   or(x, false) -> x
   equal(0, 0) -> true
   equal(s(x), 0) -> false
   equal(0, s(y)) -> false
   equal(s(x), s(y)) -> equal(x, y)
   gt(0, v) -> false
   gt(s(u), 0) -> true
   gt(s(u), s(v)) -> gt(u, v)
   max(nil) -> 0
   max(cons(u, l)) -> if(gt(u, max(l)), u, max(l))
   if(true, u, v) -> u
   if(false, u, v) -> v

S is empty.
Rewrite Strategy: FULL
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(6) LowerBoundPropagationProof (FINISHED)
Propagated lower bound.
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(7)
BOUNDS(n^1, INF)

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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(EXP, INF).


The TRS R consists of the following rules:

   sort(l) -> st(0, l)
   st(n, l) -> cond1(member(n, l), n, l)
   cond1(true, n, l) -> cons(n, st(s(n), l))
   cond1(false, n, l) -> cond2(gt(n, max(l)), n, l)
   cond2(true, n, l) -> nil
   cond2(false, n, l) -> st(s(n), l)
   member(n, nil) -> false
   member(n, cons(m, l)) -> or(equal(n, m), member(n, l))
   or(x, true) -> true
   or(x, false) -> x
   equal(0, 0) -> true
   equal(s(x), 0) -> false
   equal(0, s(y)) -> false
   equal(s(x), s(y)) -> equal(x, y)
   gt(0, v) -> false
   gt(s(u), 0) -> true
   gt(s(u), s(v)) -> gt(u, v)
   max(nil) -> 0
   max(cons(u, l)) -> if(gt(u, max(l)), u, max(l))
   if(true, u, v) -> u
   if(false, u, v) -> v

S is empty.
Rewrite Strategy: FULL
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(9) DecreasingLoopProof (FINISHED)
The following loop(s) give(s) rise to the lower bound EXP:

The rewrite sequence

max(cons(u, l)) ->^+ if(gt(u, max(l)), u, max(l))

gives rise to a decreasing loop by considering the right hand sides subterm at position [0,1].

The pumping substitution is [l / cons(u, l)].

The result substitution is [ ].



The rewrite sequence

max(cons(u, l)) ->^+ if(gt(u, max(l)), u, max(l))

gives rise to a decreasing loop by considering the right hand sides subterm at position [2].

The pumping substitution is [l / cons(u, l)].

The result substitution is [ ].




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(10)
BOUNDS(EXP, INF)