/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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 (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, n^1).

(0) CpxTRS
(1) CpxTrsToCdtProof [UPPER BOUND(ID), 0 ms]
(2) CdtProblem
    (3) CdtLeafRemovalProof [BOTH BOUNDS(ID, ID), 0 ms]
    (4) CdtProblem
    (5) CdtUsableRulesProof [BOTH BOUNDS(ID, ID), 0 ms]
    (6) CdtProblem
    (7) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 40 ms]
    (8) CdtProblem
    (9) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 0 ms]
    (10) CdtProblem
    (11) SIsEmptyProof [BOTH BOUNDS(ID, ID), 0 ms]
    (12) BOUNDS(1, 1)
(13) RenamingProof [BOTH BOUNDS(ID, ID), 0 ms]
(14) CpxTRS
    (15) SlicingProof [LOWER BOUND(ID), 0 ms]
    (16) CpxTRS
    (17) TypeInferenceProof [BOTH BOUNDS(ID, ID), 0 ms]
    (18) typed CpxTrs
    (19) OrderProof [LOWER BOUND(ID), 0 ms]
    (20) typed CpxTrs
    (21) RewriteLemmaProof [LOWER BOUND(ID), 2229 ms]
    (22) BEST
        (23) proven lower bound
            (24) LowerBoundPropagationProof [FINISHED, 0 ms]
            (25) BOUNDS(n^1, INF)
        (26) typed CpxTrs


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


The TRS R consists of the following rules:

   sum(0) -> 0
   sum(s(x)) -> +(sum(x), s(x))
   sum1(0) -> 0
   sum1(s(x)) -> s(+(sum1(x), +(x, x)))

S is empty.
Rewrite Strategy: INNERMOST
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(1) CpxTrsToCdtProof (UPPER BOUND(ID))
Converted Cpx (relative) TRS to CDT
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(2)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   sum(0) -> 0
   sum(s(z0)) -> +(sum(z0), s(z0))
   sum1(0) -> 0
   sum1(s(z0)) -> s(+(sum1(z0), +(z0, z0)))
Tuples:
   SUM(0) -> c
   SUM(s(z0)) -> c1(SUM(z0))
   SUM1(0) -> c2
   SUM1(s(z0)) -> c3(SUM1(z0))
S tuples:
   SUM(0) -> c
   SUM(s(z0)) -> c1(SUM(z0))
   SUM1(0) -> c2
   SUM1(s(z0)) -> c3(SUM1(z0))
K tuples:none
Defined Rule Symbols:   sum_1, sum1_1

Defined Pair Symbols:   SUM_1, SUM1_1

Compound Symbols:   c, c1_1, c2, c3_1


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(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
Removed 2 trailing nodes:
   SUM1(0) -> c2
   SUM(0) -> c

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(4)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   sum(0) -> 0
   sum(s(z0)) -> +(sum(z0), s(z0))
   sum1(0) -> 0
   sum1(s(z0)) -> s(+(sum1(z0), +(z0, z0)))
Tuples:
   SUM(s(z0)) -> c1(SUM(z0))
   SUM1(s(z0)) -> c3(SUM1(z0))
S tuples:
   SUM(s(z0)) -> c1(SUM(z0))
   SUM1(s(z0)) -> c3(SUM1(z0))
K tuples:none
Defined Rule Symbols:   sum_1, sum1_1

Defined Pair Symbols:   SUM_1, SUM1_1

Compound Symbols:   c1_1, c3_1


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(5) CdtUsableRulesProof (BOTH BOUNDS(ID, ID))
The following rules are not usable and were removed:
   sum(0) -> 0
   sum(s(z0)) -> +(sum(z0), s(z0))
   sum1(0) -> 0
   sum1(s(z0)) -> s(+(sum1(z0), +(z0, z0)))

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(6)
Obligation:
Complexity Dependency Tuples Problem

Rules:none
Tuples:
   SUM(s(z0)) -> c1(SUM(z0))
   SUM1(s(z0)) -> c3(SUM1(z0))
S tuples:
   SUM(s(z0)) -> c1(SUM(z0))
   SUM1(s(z0)) -> c3(SUM1(z0))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:   SUM_1, SUM1_1

Compound Symbols:   c1_1, c3_1


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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.
   SUM1(s(z0)) -> c3(SUM1(z0))
We considered the (Usable) Rules:none
And the Tuples:
   SUM(s(z0)) -> c1(SUM(z0))
   SUM1(s(z0)) -> c3(SUM1(z0))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(SUM(x_1)) = 0
   POL(SUM1(x_1)) = x_1
   POL(c1(x_1)) = x_1
   POL(c3(x_1)) = x_1
   POL(s(x_1)) = [1] + x_1

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(8)
Obligation:
Complexity Dependency Tuples Problem

Rules:none
Tuples:
   SUM(s(z0)) -> c1(SUM(z0))
   SUM1(s(z0)) -> c3(SUM1(z0))
S tuples:
   SUM(s(z0)) -> c1(SUM(z0))
K tuples:
   SUM1(s(z0)) -> c3(SUM1(z0))
Defined Rule Symbols:none

Defined Pair Symbols:   SUM_1, SUM1_1

Compound Symbols:   c1_1, c3_1


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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.
   SUM(s(z0)) -> c1(SUM(z0))
We considered the (Usable) Rules:none
And the Tuples:
   SUM(s(z0)) -> c1(SUM(z0))
   SUM1(s(z0)) -> c3(SUM1(z0))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(SUM(x_1)) = x_1
   POL(SUM1(x_1)) = 0
   POL(c1(x_1)) = x_1
   POL(c3(x_1)) = x_1
   POL(s(x_1)) = [1] + x_1

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(10)
Obligation:
Complexity Dependency Tuples Problem

Rules:none
Tuples:
   SUM(s(z0)) -> c1(SUM(z0))
   SUM1(s(z0)) -> c3(SUM1(z0))
S tuples:none
K tuples:
   SUM1(s(z0)) -> c3(SUM1(z0))
   SUM(s(z0)) -> c1(SUM(z0))
Defined Rule Symbols:none

Defined Pair Symbols:   SUM_1, SUM1_1

Compound Symbols:   c1_1, c3_1


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(11) SIsEmptyProof (BOTH BOUNDS(ID, ID))
The set S is empty
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(12)
BOUNDS(1, 1)

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(13) RenamingProof (BOTH BOUNDS(ID, ID))
Renamed function symbols to avoid clashes with predefined symbol.
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(14)
Obligation:
The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).


The TRS R consists of the following rules:

   sum(0') -> 0'
   sum(s(x)) -> +'(sum(x), s(x))
   sum1(0') -> 0'
   sum1(s(x)) -> s(+'(sum1(x), +'(x, x)))

S is empty.
Rewrite Strategy: INNERMOST
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(15) SlicingProof (LOWER BOUND(ID))
Sliced the following arguments:
+'/1

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(16)
Obligation:
The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).


The TRS R consists of the following rules:

   sum(0') -> 0'
   sum(s(x)) -> +'(sum(x))
   sum1(0') -> 0'
   sum1(s(x)) -> s(+'(sum1(x)))

S is empty.
Rewrite Strategy: INNERMOST
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(17) TypeInferenceProof (BOTH BOUNDS(ID, ID))
Infered types.
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(18)
Obligation:
Innermost TRS:
Rules:
sum(0') -> 0'
sum(s(x)) -> +'(sum(x))
sum1(0') -> 0'
sum1(s(x)) -> s(+'(sum1(x)))

Types:
sum :: 0':s:+' -> 0':s:+'
0' :: 0':s:+'
s :: 0':s:+' -> 0':s:+'
+' :: 0':s:+' -> 0':s:+'
sum1 :: 0':s:+' -> 0':s:+'
hole_0':s:+'1_0 :: 0':s:+'
gen_0':s:+'2_0 :: Nat -> 0':s:+'

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(19) OrderProof (LOWER BOUND(ID))
Heuristically decided to analyse the following defined symbols:
sum, sum1
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(20)
Obligation:
Innermost TRS:
Rules:
sum(0') -> 0'
sum(s(x)) -> +'(sum(x))
sum1(0') -> 0'
sum1(s(x)) -> s(+'(sum1(x)))

Types:
sum :: 0':s:+' -> 0':s:+'
0' :: 0':s:+'
s :: 0':s:+' -> 0':s:+'
+' :: 0':s:+' -> 0':s:+'
sum1 :: 0':s:+' -> 0':s:+'
hole_0':s:+'1_0 :: 0':s:+'
gen_0':s:+'2_0 :: Nat -> 0':s:+'


Generator Equations:
gen_0':s:+'2_0(0) <=> 0'
gen_0':s:+'2_0(+(x, 1)) <=> s(gen_0':s:+'2_0(x))


The following defined symbols remain to be analysed:
sum, sum1
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(21) RewriteLemmaProof (LOWER BOUND(ID))
Proved the following rewrite lemma:
sum(gen_0':s:+'2_0(+(1, n4_0))) -> *3_0, rt in Omega(n4_0)

Induction Base:
sum(gen_0':s:+'2_0(+(1, 0)))

Induction Step:
sum(gen_0':s:+'2_0(+(1, +(n4_0, 1)))) ->_R^Omega(1)
+'(sum(gen_0':s:+'2_0(+(1, n4_0)))) ->_IH
+'(*3_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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(22)
Complex Obligation (BEST)

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(23)
Obligation:
Proved the lower bound n^1 for the following obligation:

Innermost TRS:
Rules:
sum(0') -> 0'
sum(s(x)) -> +'(sum(x))
sum1(0') -> 0'
sum1(s(x)) -> s(+'(sum1(x)))

Types:
sum :: 0':s:+' -> 0':s:+'
0' :: 0':s:+'
s :: 0':s:+' -> 0':s:+'
+' :: 0':s:+' -> 0':s:+'
sum1 :: 0':s:+' -> 0':s:+'
hole_0':s:+'1_0 :: 0':s:+'
gen_0':s:+'2_0 :: Nat -> 0':s:+'


Generator Equations:
gen_0':s:+'2_0(0) <=> 0'
gen_0':s:+'2_0(+(x, 1)) <=> s(gen_0':s:+'2_0(x))


The following defined symbols remain to be analysed:
sum, sum1
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(24) LowerBoundPropagationProof (FINISHED)
Propagated lower bound.
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(25)
BOUNDS(n^1, INF)

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(26)
Obligation:
Innermost TRS:
Rules:
sum(0') -> 0'
sum(s(x)) -> +'(sum(x))
sum1(0') -> 0'
sum1(s(x)) -> s(+'(sum1(x)))

Types:
sum :: 0':s:+' -> 0':s:+'
0' :: 0':s:+'
s :: 0':s:+' -> 0':s:+'
+' :: 0':s:+' -> 0':s:+'
sum1 :: 0':s:+' -> 0':s:+'
hole_0':s:+'1_0 :: 0':s:+'
gen_0':s:+'2_0 :: Nat -> 0':s:+'


Lemmas:
sum(gen_0':s:+'2_0(+(1, n4_0))) -> *3_0, rt in Omega(n4_0)


Generator Equations:
gen_0':s:+'2_0(0) <=> 0'
gen_0':s:+'2_0(+(x, 1)) <=> s(gen_0':s:+'2_0(x))


The following defined symbols remain to be analysed:
sum1