/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), ?)
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 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


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


The TRS R consists of the following rules:

   half(0) -> 0
   half(s(0)) -> 0
   half(s(s(x))) -> s(half(x))
   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   inc(0) -> 0
   inc(s(x)) -> s(inc(x))
   log(x) -> log2(x, 0)
   log2(x, y) -> if(le(x, s(0)), x, inc(y))
   if(true, x, s(y)) -> y
   if(false, x, y) -> log2(half(x), y)

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(n^1, INF).


The TRS R consists of the following rules:

   half(0) -> 0
   half(s(0)) -> 0
   half(s(s(x))) -> s(half(x))
   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   inc(0) -> 0
   inc(s(x)) -> s(inc(x))
   log(x) -> log2(x, 0)
   log2(x, y) -> if(le(x, s(0)), x, inc(y))
   if(true, x, s(y)) -> y
   if(false, x, y) -> log2(half(x), y)

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

half(s(s(x))) ->^+ s(half(x))

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

The pumping substitution is [x / s(s(x))].

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(n^1, INF).


The TRS R consists of the following rules:

   half(0) -> 0
   half(s(0)) -> 0
   half(s(s(x))) -> s(half(x))
   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   inc(0) -> 0
   inc(s(x)) -> s(inc(x))
   log(x) -> log2(x, 0)
   log2(x, y) -> if(le(x, s(0)), x, inc(y))
   if(true, x, s(y)) -> y
   if(false, x, y) -> log2(half(x), y)

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(n^1, INF).


The TRS R consists of the following rules:

   half(0) -> 0
   half(s(0)) -> 0
   half(s(s(x))) -> s(half(x))
   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   inc(0) -> 0
   inc(s(x)) -> s(inc(x))
   log(x) -> log2(x, 0)
   log2(x, y) -> if(le(x, s(0)), x, inc(y))
   if(true, x, s(y)) -> y
   if(false, x, y) -> log2(half(x), y)

S is empty.
Rewrite Strategy: FULL