/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) RcToIrcProof [BOTH BOUNDS(ID, ID), 0 ms]
(2) CpxTRS
    (3) CpxTrsToCdtProof [UPPER BOUND(ID), 0 ms]
    (4) CdtProblem
    (5) CdtLeafRemovalProof [BOTH BOUNDS(ID, ID), 0 ms]
    (6) CdtProblem
    (7) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 37 ms]
    (8) CdtProblem
    (9) SIsEmptyProof [BOTH BOUNDS(ID, ID), 0 ms]
    (10) BOUNDS(1, 1)
(11) RenamingProof [BOTH BOUNDS(ID, ID), 0 ms]
(12) CpxTRS
    (13) SlicingProof [LOWER BOUND(ID), 0 ms]
    (14) CpxTRS
    (15) TypeInferenceProof [BOTH BOUNDS(ID, ID), 0 ms]
    (16) typed CpxTrs
    (17) OrderProof [LOWER BOUND(ID), 0 ms]
    (18) typed CpxTrs
    (19) RewriteLemmaProof [LOWER BOUND(ID), 194 ms]
    (20) proven lower bound
    (21) LowerBoundPropagationProof [FINISHED, 0 ms]
    (22) BOUNDS(n^1, INF)


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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:

   f(0, y) -> 0
   f(s(x), y) -> f(f(x, y), y)

S is empty.
Rewrite Strategy: FULL
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(1) RcToIrcProof (BOTH BOUNDS(ID, ID))
Converted rc-obligation to irc-obligation.

The duplicating contexts are:
f(s(x), [])


The defined contexts are:
f([], x1)


As the TRS is an overlay system and the defined contexts and the duplicating contexts don't overlap, we have rc = irc.
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(2)
Obligation:
The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(1, n^1).


The TRS R consists of the following rules:

   f(0, y) -> 0
   f(s(x), y) -> f(f(x, y), y)

S is empty.
Rewrite Strategy: INNERMOST
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(3) CpxTrsToCdtProof (UPPER BOUND(ID))
Converted Cpx (relative) TRS to CDT
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(4)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   f(0, z0) -> 0
   f(s(z0), z1) -> f(f(z0, z1), z1)
Tuples:
   F(0, z0) -> c
   F(s(z0), z1) -> c1(F(f(z0, z1), z1), F(z0, z1))
S tuples:
   F(0, z0) -> c
   F(s(z0), z1) -> c1(F(f(z0, z1), z1), F(z0, z1))
K tuples:none
Defined Rule Symbols:   f_2

Defined Pair Symbols:   F_2

Compound Symbols:   c, c1_2


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(5) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
Removed 1 trailing nodes:
   F(0, z0) -> c

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(6)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   f(0, z0) -> 0
   f(s(z0), z1) -> f(f(z0, z1), z1)
Tuples:
   F(s(z0), z1) -> c1(F(f(z0, z1), z1), F(z0, z1))
S tuples:
   F(s(z0), z1) -> c1(F(f(z0, z1), z1), F(z0, z1))
K tuples:none
Defined Rule Symbols:   f_2

Defined Pair Symbols:   F_2

Compound Symbols:   c1_2


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(7) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   F(s(z0), z1) -> c1(F(f(z0, z1), z1), F(z0, z1))
We considered the (Usable) Rules:
   f(0, z0) -> 0
   f(s(z0), z1) -> f(f(z0, z1), z1)
And the Tuples:
   F(s(z0), z1) -> c1(F(f(z0, z1), z1), F(z0, z1))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0) = 0
   POL(F(x_1, x_2)) = x_1
   POL(c1(x_1, x_2)) = x_1 + x_2
   POL(f(x_1, x_2)) = 0
   POL(s(x_1)) = [1] + x_1

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(8)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   f(0, z0) -> 0
   f(s(z0), z1) -> f(f(z0, z1), z1)
Tuples:
   F(s(z0), z1) -> c1(F(f(z0, z1), z1), F(z0, z1))
S tuples:none
K tuples:
   F(s(z0), z1) -> c1(F(f(z0, z1), z1), F(z0, z1))
Defined Rule Symbols:   f_2

Defined Pair Symbols:   F_2

Compound Symbols:   c1_2


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(9) SIsEmptyProof (BOTH BOUNDS(ID, ID))
The set S is empty
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(10)
BOUNDS(1, 1)

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(11) RenamingProof (BOTH BOUNDS(ID, ID))
Renamed function symbols to avoid clashes with predefined symbol.
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(12)
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:

   f(0', y) -> 0'
   f(s(x), y) -> f(f(x, y), y)

S is empty.
Rewrite Strategy: FULL
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(13) SlicingProof (LOWER BOUND(ID))
Sliced the following arguments:
f/1

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(14)
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:

   f(0') -> 0'
   f(s(x)) -> f(f(x))

S is empty.
Rewrite Strategy: FULL
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(15) TypeInferenceProof (BOTH BOUNDS(ID, ID))
Infered types.
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(16)
Obligation:
TRS:
Rules:
f(0') -> 0'
f(s(x)) -> f(f(x))

Types:
f :: 0':s -> 0':s
0' :: 0':s
s :: 0':s -> 0':s
hole_0':s1_0 :: 0':s
gen_0':s2_0 :: Nat -> 0':s

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(17) OrderProof (LOWER BOUND(ID))
Heuristically decided to analyse the following defined symbols:
f
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(18)
Obligation:
TRS:
Rules:
f(0') -> 0'
f(s(x)) -> f(f(x))

Types:
f :: 0':s -> 0':s
0' :: 0':s
s :: 0':s -> 0':s
hole_0':s1_0 :: 0':s
gen_0':s2_0 :: Nat -> 0':s


Generator Equations:
gen_0':s2_0(0) <=> 0'
gen_0':s2_0(+(x, 1)) <=> s(gen_0':s2_0(x))


The following defined symbols remain to be analysed:
f
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(19) RewriteLemmaProof (LOWER BOUND(ID))
Proved the following rewrite lemma:
f(gen_0':s2_0(n4_0)) -> gen_0':s2_0(0), rt in Omega(1 + n4_0)

Induction Base:
f(gen_0':s2_0(0)) ->_R^Omega(1)
0'

Induction Step:
f(gen_0':s2_0(+(n4_0, 1))) ->_R^Omega(1)
f(f(gen_0':s2_0(n4_0))) ->_IH
f(gen_0':s2_0(0)) ->_R^Omega(1)
0'

We have rt in Omega(n^1) and sz in O(n). Thus, we have irc_R in Omega(n).
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(20)
Obligation:
Proved the lower bound n^1 for the following obligation:

TRS:
Rules:
f(0') -> 0'
f(s(x)) -> f(f(x))

Types:
f :: 0':s -> 0':s
0' :: 0':s
s :: 0':s -> 0':s
hole_0':s1_0 :: 0':s
gen_0':s2_0 :: Nat -> 0':s


Generator Equations:
gen_0':s2_0(0) <=> 0'
gen_0':s2_0(+(x, 1)) <=> s(gen_0':s2_0(x))


The following defined symbols remain to be analysed:
f
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(21) LowerBoundPropagationProof (FINISHED)
Propagated lower bound.
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(22)
BOUNDS(n^1, INF)