/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(Omega(n^1), ?)
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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:

   eq(0, 0) -> true
   eq(0, s(m)) -> false
   eq(s(n), 0) -> false
   eq(s(n), s(m)) -> eq(n, m)
   le(0, m) -> true
   le(s(n), 0) -> false
   le(s(n), s(m)) -> le(n, m)
   min(cons(x, nil)) -> x
   min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
   if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
   if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
   replace(n, m, nil) -> nil
   replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
   if_replace(true, n, m, cons(k, x)) -> cons(m, x)
   if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
   empty(nil) -> true
   empty(cons(n, x)) -> false
   head(cons(n, x)) -> n
   tail(nil) -> nil
   tail(cons(n, x)) -> x
   sort(x) -> sortIter(x, nil)
   sortIter(x, y) -> if(empty(x), x, y, append(y, cons(min(x), nil)))
   if(true, x, y, z) -> y
   if(false, x, y, z) -> sortIter(replace(min(x), head(x), tail(x)), z)

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:

   eq(0, 0) -> true
   eq(0, s(m)) -> false
   eq(s(n), 0) -> false
   eq(s(n), s(m)) -> eq(n, m)
   le(0, m) -> true
   le(s(n), 0) -> false
   le(s(n), s(m)) -> le(n, m)
   min(cons(x, nil)) -> x
   min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
   if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
   if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
   replace(n, m, nil) -> nil
   replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
   if_replace(true, n, m, cons(k, x)) -> cons(m, x)
   if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
   empty(nil) -> true
   empty(cons(n, x)) -> false
   head(cons(n, x)) -> n
   tail(nil) -> nil
   tail(cons(n, x)) -> x
   sort(x) -> sortIter(x, nil)
   sortIter(x, y) -> if(empty(x), x, y, append(y, cons(min(x), nil)))
   if(true, x, y, z) -> y
   if(false, x, y, z) -> sortIter(replace(min(x), head(x), tail(x)), z)

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

eq(s(n), s(m)) ->^+ eq(n, m)

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

The pumping substitution is [n / s(n), m / s(m)].

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:

   eq(0, 0) -> true
   eq(0, s(m)) -> false
   eq(s(n), 0) -> false
   eq(s(n), s(m)) -> eq(n, m)
   le(0, m) -> true
   le(s(n), 0) -> false
   le(s(n), s(m)) -> le(n, m)
   min(cons(x, nil)) -> x
   min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
   if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
   if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
   replace(n, m, nil) -> nil
   replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
   if_replace(true, n, m, cons(k, x)) -> cons(m, x)
   if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
   empty(nil) -> true
   empty(cons(n, x)) -> false
   head(cons(n, x)) -> n
   tail(nil) -> nil
   tail(cons(n, x)) -> x
   sort(x) -> sortIter(x, nil)
   sortIter(x, y) -> if(empty(x), x, y, append(y, cons(min(x), nil)))
   if(true, x, y, z) -> y
   if(false, x, y, z) -> sortIter(replace(min(x), head(x), tail(x)), z)

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:

   eq(0, 0) -> true
   eq(0, s(m)) -> false
   eq(s(n), 0) -> false
   eq(s(n), s(m)) -> eq(n, m)
   le(0, m) -> true
   le(s(n), 0) -> false
   le(s(n), s(m)) -> le(n, m)
   min(cons(x, nil)) -> x
   min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
   if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
   if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
   replace(n, m, nil) -> nil
   replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
   if_replace(true, n, m, cons(k, x)) -> cons(m, x)
   if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
   empty(nil) -> true
   empty(cons(n, x)) -> false
   head(cons(n, x)) -> n
   tail(nil) -> nil
   tail(cons(n, x)) -> x
   sort(x) -> sortIter(x, nil)
   sortIter(x, y) -> if(empty(x), x, y, append(y, cons(min(x), nil)))
   if(true, x, y, z) -> y
   if(false, x, y, z) -> sortIter(replace(min(x), head(x), tail(x)), z)

S is empty.
Rewrite Strategy: FULL