/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), O(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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 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) CdtUsableRulesProof [BOTH BOUNDS(ID, ID), 2 ms]
    (8) CdtProblem
    (9) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 13 ms]
    (10) CdtProblem
    (11) CdtNarrowingProof [BOTH BOUNDS(ID, ID), 0 ms]
    (12) CdtProblem
    (13) CdtLeafRemovalProof [BOTH BOUNDS(ID, ID), 0 ms]
    (14) CdtProblem
    (15) CdtRhsSimplificationProcessorProof [BOTH BOUNDS(ID, ID), 0 ms]
    (16) CdtProblem
    (17) CdtLeafRemovalProof [ComplexityIfPolyImplication, 0 ms]
    (18) CdtProblem
    (19) CdtUsableRulesProof [BOTH BOUNDS(ID, ID), 0 ms]
    (20) CdtProblem
    (21) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 0 ms]
    (22) CdtProblem
    (23) SIsEmptyProof [BOTH BOUNDS(ID, ID), 0 ms]
    (24) BOUNDS(1, 1)
(25) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms]
(26) TRS for Loop Detection
    (27) DecreasingLoopProof [LOWER BOUND(ID), 0 ms]
    (28) BEST
        (29) proven lower bound
            (30) LowerBoundPropagationProof [FINISHED, 0 ms]
            (31) BOUNDS(n^1, INF)
        (32) 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:

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

S is empty.
Rewrite Strategy: FULL
----------------------------------------

(1) RcToIrcProof (BOTH BOUNDS(ID, ID))
Converted rc-obligation to irc-obligation.

The duplicating contexts are:
g(0, [])


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


As the TRS is an overlay system and the defined contexts and the duplicating contexts don't overlap, we have rc = irc.
----------------------------------------

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

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(z0, 0) -> s(0)
   f(s(z0), s(z1)) -> s(f(z0, z1))
   g(0, z0) -> g(f(z0, z0), z0)
Tuples:
   F(z0, 0) -> c
   F(s(z0), s(z1)) -> c1(F(z0, z1))
   G(0, z0) -> c2(G(f(z0, z0), z0), F(z0, z0))
S tuples:
   F(z0, 0) -> c
   F(s(z0), s(z1)) -> c1(F(z0, z1))
   G(0, z0) -> c2(G(f(z0, z0), z0), F(z0, z0))
K tuples:none
Defined Rule Symbols:   f_2, g_2

Defined Pair Symbols:   F_2, G_2

Compound Symbols:   c, c1_1, c2_2


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

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(6)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   f(z0, 0) -> s(0)
   f(s(z0), s(z1)) -> s(f(z0, z1))
   g(0, z0) -> g(f(z0, z0), z0)
Tuples:
   F(s(z0), s(z1)) -> c1(F(z0, z1))
   G(0, z0) -> c2(G(f(z0, z0), z0), F(z0, z0))
S tuples:
   F(s(z0), s(z1)) -> c1(F(z0, z1))
   G(0, z0) -> c2(G(f(z0, z0), z0), F(z0, z0))
K tuples:none
Defined Rule Symbols:   f_2, g_2

Defined Pair Symbols:   F_2, G_2

Compound Symbols:   c1_1, c2_2


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(7) CdtUsableRulesProof (BOTH BOUNDS(ID, ID))
The following rules are not usable and were removed:
   g(0, z0) -> g(f(z0, z0), z0)

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(8)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   f(z0, 0) -> s(0)
   f(s(z0), s(z1)) -> s(f(z0, z1))
Tuples:
   F(s(z0), s(z1)) -> c1(F(z0, z1))
   G(0, z0) -> c2(G(f(z0, z0), z0), F(z0, z0))
S tuples:
   F(s(z0), s(z1)) -> c1(F(z0, z1))
   G(0, z0) -> c2(G(f(z0, z0), z0), F(z0, z0))
K tuples:none
Defined Rule Symbols:   f_2

Defined Pair Symbols:   F_2, G_2

Compound Symbols:   c1_1, c2_2


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(9) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   G(0, z0) -> c2(G(f(z0, z0), z0), F(z0, z0))
We considered the (Usable) Rules:
   f(s(z0), s(z1)) -> s(f(z0, z1))
   f(z0, 0) -> s(0)
And the Tuples:
   F(s(z0), s(z1)) -> c1(F(z0, z1))
   G(0, z0) -> c2(G(f(z0, z0), z0), F(z0, z0))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0) = [1]
   POL(F(x_1, x_2)) = 0
   POL(G(x_1, x_2)) = x_1
   POL(c1(x_1)) = x_1
   POL(c2(x_1, x_2)) = x_1 + x_2
   POL(f(x_1, x_2)) = 0
   POL(s(x_1)) = 0

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(10)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   f(z0, 0) -> s(0)
   f(s(z0), s(z1)) -> s(f(z0, z1))
Tuples:
   F(s(z0), s(z1)) -> c1(F(z0, z1))
   G(0, z0) -> c2(G(f(z0, z0), z0), F(z0, z0))
S tuples:
   F(s(z0), s(z1)) -> c1(F(z0, z1))
K tuples:
   G(0, z0) -> c2(G(f(z0, z0), z0), F(z0, z0))
Defined Rule Symbols:   f_2

Defined Pair Symbols:   F_2, G_2

Compound Symbols:   c1_1, c2_2


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(11) CdtNarrowingProof (BOTH BOUNDS(ID, ID))
Use narrowing to replace G(0, z0) -> c2(G(f(z0, z0), z0), F(z0, z0)) by 
   G(0, 0) -> c2(G(s(0), 0), F(0, 0))
   G(0, s(z0)) -> c2(G(s(f(z0, z0)), s(z0)), F(s(z0), s(z0)))

----------------------------------------

(12)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   f(z0, 0) -> s(0)
   f(s(z0), s(z1)) -> s(f(z0, z1))
Tuples:
   F(s(z0), s(z1)) -> c1(F(z0, z1))
   G(0, 0) -> c2(G(s(0), 0), F(0, 0))
   G(0, s(z0)) -> c2(G(s(f(z0, z0)), s(z0)), F(s(z0), s(z0)))
S tuples:
   F(s(z0), s(z1)) -> c1(F(z0, z1))
K tuples:
   G(0, z0) -> c2(G(f(z0, z0), z0), F(z0, z0))
Defined Rule Symbols:   f_2

Defined Pair Symbols:   F_2, G_2

Compound Symbols:   c1_1, c2_2


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(13) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
Removed 1 trailing nodes:
   G(0, 0) -> c2(G(s(0), 0), F(0, 0))

----------------------------------------

(14)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   f(z0, 0) -> s(0)
   f(s(z0), s(z1)) -> s(f(z0, z1))
Tuples:
   F(s(z0), s(z1)) -> c1(F(z0, z1))
   G(0, s(z0)) -> c2(G(s(f(z0, z0)), s(z0)), F(s(z0), s(z0)))
S tuples:
   F(s(z0), s(z1)) -> c1(F(z0, z1))
K tuples:none
Defined Rule Symbols:   f_2

Defined Pair Symbols:   F_2, G_2

Compound Symbols:   c1_1, c2_2


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(15) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID))
Removed 1 trailing tuple parts
----------------------------------------

(16)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   f(z0, 0) -> s(0)
   f(s(z0), s(z1)) -> s(f(z0, z1))
Tuples:
   F(s(z0), s(z1)) -> c1(F(z0, z1))
   G(0, s(z0)) -> c2(F(s(z0), s(z0)))
S tuples:
   F(s(z0), s(z1)) -> c1(F(z0, z1))
K tuples:none
Defined Rule Symbols:   f_2

Defined Pair Symbols:   F_2, G_2

Compound Symbols:   c1_1, c2_1


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(17) CdtLeafRemovalProof (ComplexityIfPolyImplication)
Removed 1 leading nodes:
   G(0, s(z0)) -> c2(F(s(z0), s(z0)))

----------------------------------------

(18)
Obligation:
Complexity Dependency Tuples Problem

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

Defined Pair Symbols:   F_2

Compound Symbols:   c1_1


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(19) CdtUsableRulesProof (BOTH BOUNDS(ID, ID))
The following rules are not usable and were removed:
   f(z0, 0) -> s(0)
   f(s(z0), s(z1)) -> s(f(z0, z1))

----------------------------------------

(20)
Obligation:
Complexity Dependency Tuples Problem

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

Defined Pair Symbols:   F_2

Compound Symbols:   c1_1


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(21) 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), s(z1)) -> c1(F(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:
   F(s(z0), s(z1)) -> c1(F(z0, z1))
The order we found is given by the following interpretation:

Polynomial interpretation :

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

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(22)
Obligation:
Complexity Dependency Tuples Problem

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

Defined Pair Symbols:   F_2

Compound Symbols:   c1_1


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(23) SIsEmptyProof (BOTH BOUNDS(ID, ID))
The set S is empty
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(24)
BOUNDS(1, 1)

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(25) RelTrsToDecreasingLoopProblemProof (LOWER BOUND(ID))
Transformed a relative TRS into a decreasing-loop problem.
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(26)
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:

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

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

The rewrite sequence

f(s(x), s(y)) ->^+ s(f(x, y))

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

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

The result substitution is [ ].




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(28)
Complex Obligation (BEST)

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(29)
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:

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

S is empty.
Rewrite Strategy: FULL
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(30) LowerBoundPropagationProof (FINISHED)
Propagated lower bound.
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(31)
BOUNDS(n^1, INF)

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(32)
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:

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

S is empty.
Rewrite Strategy: FULL