/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) CdtRhsSimplificationProcessorProof [BOTH BOUNDS(ID, ID), 0 ms]
    (6) CdtProblem
    (7) CdtUsableRulesProof [BOTH BOUNDS(ID, ID), 0 ms]
    (8) CdtProblem
    (9) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 69 ms]
    (10) CdtProblem
    (11) CdtKnowledgeProof [FINISHED, 0 ms]
    (12) BOUNDS(1, 1)
(13) RenamingProof [BOTH BOUNDS(ID, 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), 414 ms]
    (20) proven lower bound
    (21) LowerBoundPropagationProof [FINISHED, 0 ms]
    (22) BOUNDS(n^1, INF)


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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:

   del(.(x, .(y, z))) -> f(=(x, y), x, y, z)
   f(true, x, y, z) -> del(.(y, z))
   f(false, x, y, z) -> .(x, del(.(y, z)))
   =(nil, nil) -> true
   =(.(x, y), nil) -> false
   =(nil, .(y, z)) -> false
   =(.(x, y), .(u, v)) -> and(=(x, u), =(y, v))

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

(1) CpxTrsToCdtProof (UPPER BOUND(ID))
Converted Cpx (relative) TRS to CDT
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(2)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   del(.(z0, .(z1, z2))) -> f(=(z0, z1), z0, z1, z2)
   f(true, z0, z1, z2) -> del(.(z1, z2))
   f(false, z0, z1, z2) -> .(z0, del(.(z1, z2)))
   =(nil, nil) -> true
   =(.(z0, z1), nil) -> false
   =(nil, .(z0, z1)) -> false
   =(.(z0, z1), .(u, v)) -> and(=(z0, u), =(z1, v))
Tuples:
   DEL(.(z0, .(z1, z2))) -> c(F(=(z0, z1), z0, z1, z2), ='(z0, z1))
   F(true, z0, z1, z2) -> c1(DEL(.(z1, z2)))
   F(false, z0, z1, z2) -> c2(DEL(.(z1, z2)))
   ='(nil, nil) -> c3
   ='(.(z0, z1), nil) -> c4
   ='(nil, .(z0, z1)) -> c5
   ='(.(z0, z1), .(u, v)) -> c6(='(z0, u), ='(z1, v))
S tuples:
   DEL(.(z0, .(z1, z2))) -> c(F(=(z0, z1), z0, z1, z2), ='(z0, z1))
   F(true, z0, z1, z2) -> c1(DEL(.(z1, z2)))
   F(false, z0, z1, z2) -> c2(DEL(.(z1, z2)))
   ='(nil, nil) -> c3
   ='(.(z0, z1), nil) -> c4
   ='(nil, .(z0, z1)) -> c5
   ='(.(z0, z1), .(u, v)) -> c6(='(z0, u), ='(z1, v))
K tuples:none
Defined Rule Symbols:   del_1, f_4, =_2

Defined Pair Symbols:   DEL_1, F_4, ='_2

Compound Symbols:   c_2, c1_1, c2_1, c3, c4, c5, c6_2


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(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
Removed 4 trailing nodes:
   ='(.(z0, z1), .(u, v)) -> c6(='(z0, u), ='(z1, v))
   ='(nil, .(z0, z1)) -> c5
   ='(.(z0, z1), nil) -> c4
   ='(nil, nil) -> c3

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

(4)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   del(.(z0, .(z1, z2))) -> f(=(z0, z1), z0, z1, z2)
   f(true, z0, z1, z2) -> del(.(z1, z2))
   f(false, z0, z1, z2) -> .(z0, del(.(z1, z2)))
   =(nil, nil) -> true
   =(.(z0, z1), nil) -> false
   =(nil, .(z0, z1)) -> false
   =(.(z0, z1), .(u, v)) -> and(=(z0, u), =(z1, v))
Tuples:
   DEL(.(z0, .(z1, z2))) -> c(F(=(z0, z1), z0, z1, z2), ='(z0, z1))
   F(true, z0, z1, z2) -> c1(DEL(.(z1, z2)))
   F(false, z0, z1, z2) -> c2(DEL(.(z1, z2)))
S tuples:
   DEL(.(z0, .(z1, z2))) -> c(F(=(z0, z1), z0, z1, z2), ='(z0, z1))
   F(true, z0, z1, z2) -> c1(DEL(.(z1, z2)))
   F(false, z0, z1, z2) -> c2(DEL(.(z1, z2)))
K tuples:none
Defined Rule Symbols:   del_1, f_4, =_2

Defined Pair Symbols:   DEL_1, F_4

Compound Symbols:   c_2, c1_1, c2_1


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

(6)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   del(.(z0, .(z1, z2))) -> f(=(z0, z1), z0, z1, z2)
   f(true, z0, z1, z2) -> del(.(z1, z2))
   f(false, z0, z1, z2) -> .(z0, del(.(z1, z2)))
   =(nil, nil) -> true
   =(.(z0, z1), nil) -> false
   =(nil, .(z0, z1)) -> false
   =(.(z0, z1), .(u, v)) -> and(=(z0, u), =(z1, v))
Tuples:
   F(true, z0, z1, z2) -> c1(DEL(.(z1, z2)))
   F(false, z0, z1, z2) -> c2(DEL(.(z1, z2)))
   DEL(.(z0, .(z1, z2))) -> c(F(=(z0, z1), z0, z1, z2))
S tuples:
   F(true, z0, z1, z2) -> c1(DEL(.(z1, z2)))
   F(false, z0, z1, z2) -> c2(DEL(.(z1, z2)))
   DEL(.(z0, .(z1, z2))) -> c(F(=(z0, z1), z0, z1, z2))
K tuples:none
Defined Rule Symbols:   del_1, f_4, =_2

Defined Pair Symbols:   F_4, DEL_1

Compound Symbols:   c1_1, c2_1, c_1


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(7) CdtUsableRulesProof (BOTH BOUNDS(ID, ID))
The following rules are not usable and were removed:
   del(.(z0, .(z1, z2))) -> f(=(z0, z1), z0, z1, z2)
   f(true, z0, z1, z2) -> del(.(z1, z2))
   f(false, z0, z1, z2) -> .(z0, del(.(z1, z2)))

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

(8)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   =(nil, nil) -> true
   =(.(z0, z1), nil) -> false
   =(nil, .(z0, z1)) -> false
   =(.(z0, z1), .(u, v)) -> and(=(z0, u), =(z1, v))
Tuples:
   F(true, z0, z1, z2) -> c1(DEL(.(z1, z2)))
   F(false, z0, z1, z2) -> c2(DEL(.(z1, z2)))
   DEL(.(z0, .(z1, z2))) -> c(F(=(z0, z1), z0, z1, z2))
S tuples:
   F(true, z0, z1, z2) -> c1(DEL(.(z1, z2)))
   F(false, z0, z1, z2) -> c2(DEL(.(z1, z2)))
   DEL(.(z0, .(z1, z2))) -> c(F(=(z0, z1), z0, z1, z2))
K tuples:none
Defined Rule Symbols:   =_2

Defined Pair Symbols:   F_4, DEL_1

Compound Symbols:   c1_1, c2_1, c_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.
   DEL(.(z0, .(z1, z2))) -> c(F(=(z0, z1), z0, z1, z2))
We considered the (Usable) Rules:none
And the Tuples:
   F(true, z0, z1, z2) -> c1(DEL(.(z1, z2)))
   F(false, z0, z1, z2) -> c2(DEL(.(z1, z2)))
   DEL(.(z0, .(z1, z2))) -> c(F(=(z0, z1), z0, z1, z2))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(.(x_1, x_2)) = [3] + x_2
   POL(=(x_1, x_2)) = [3] + x_1 + [2]x_2
   POL(DEL(x_1)) = x_1
   POL(F(x_1, x_2, x_3, x_4)) = [3] + x_4
   POL(and(x_1, x_2)) = [3] + x_2
   POL(c(x_1)) = x_1
   POL(c1(x_1)) = x_1
   POL(c2(x_1)) = x_1
   POL(false) = 0
   POL(nil) = 0
   POL(true) = [2]
   POL(u) = [3]
   POL(v) = [3]

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

(10)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   =(nil, nil) -> true
   =(.(z0, z1), nil) -> false
   =(nil, .(z0, z1)) -> false
   =(.(z0, z1), .(u, v)) -> and(=(z0, u), =(z1, v))
Tuples:
   F(true, z0, z1, z2) -> c1(DEL(.(z1, z2)))
   F(false, z0, z1, z2) -> c2(DEL(.(z1, z2)))
   DEL(.(z0, .(z1, z2))) -> c(F(=(z0, z1), z0, z1, z2))
S tuples:
   F(true, z0, z1, z2) -> c1(DEL(.(z1, z2)))
   F(false, z0, z1, z2) -> c2(DEL(.(z1, z2)))
K tuples:
   DEL(.(z0, .(z1, z2))) -> c(F(=(z0, z1), z0, z1, z2))
Defined Rule Symbols:   =_2

Defined Pair Symbols:   F_4, DEL_1

Compound Symbols:   c1_1, c2_1, c_1


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(11) CdtKnowledgeProof (FINISHED)
The following tuples could be moved from S to K by knowledge propagation:
   F(true, z0, z1, z2) -> c1(DEL(.(z1, z2)))
   F(false, z0, z1, z2) -> c2(DEL(.(z1, z2)))
   DEL(.(z0, .(z1, z2))) -> c(F(=(z0, z1), z0, z1, z2))
   DEL(.(z0, .(z1, z2))) -> c(F(=(z0, z1), z0, z1, z2))
Now 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.
----------------------------------------

(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:

   del(.(x, .(y, z))) -> f(='(x, y), x, y, z)
   f(true, x, y, z) -> del(.(y, z))
   f(false, x, y, z) -> .(x, del(.(y, z)))
   ='(nil, nil) -> true
   ='(.(x, y), nil) -> false
   ='(nil, .(y, z)) -> false
   ='(.(x, y), .(u, v)) -> and(='(x, u), ='(y, v))

S is empty.
Rewrite Strategy: INNERMOST
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(15) TypeInferenceProof (BOTH BOUNDS(ID, ID))
Infered types.
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(16)
Obligation:
Innermost TRS:
Rules:
del(.(x, .(y, z))) -> f(='(x, y), x, y, z)
f(true, x, y, z) -> del(.(y, z))
f(false, x, y, z) -> .(x, del(.(y, z)))
='(nil, nil) -> true
='(.(x, y), nil) -> false
='(nil, .(y, z)) -> false
='(.(x, y), .(u, v)) -> and(='(x, u), ='(y, v))

Types:
del :: .:nil:u:v -> .:nil:u:v
. :: .:nil:u:v -> .:nil:u:v -> .:nil:u:v
f :: true:false:and -> .:nil:u:v -> .:nil:u:v -> .:nil:u:v -> .:nil:u:v
=' :: .:nil:u:v -> .:nil:u:v -> true:false:and
true :: true:false:and
false :: true:false:and
nil :: .:nil:u:v
u :: .:nil:u:v
v :: .:nil:u:v
and :: true:false:and -> true:false:and -> true:false:and
hole_.:nil:u:v1_0 :: .:nil:u:v
hole_true:false:and2_0 :: true:false:and
gen_.:nil:u:v3_0 :: Nat -> .:nil:u:v
gen_true:false:and4_0 :: Nat -> true:false:and

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(17) OrderProof (LOWER BOUND(ID))
Heuristically decided to analyse the following defined symbols:
del, ='

They will be analysed ascendingly in the following order:
=' < del

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

(18)
Obligation:
Innermost TRS:
Rules:
del(.(x, .(y, z))) -> f(='(x, y), x, y, z)
f(true, x, y, z) -> del(.(y, z))
f(false, x, y, z) -> .(x, del(.(y, z)))
='(nil, nil) -> true
='(.(x, y), nil) -> false
='(nil, .(y, z)) -> false
='(.(x, y), .(u, v)) -> and(='(x, u), ='(y, v))

Types:
del :: .:nil:u:v -> .:nil:u:v
. :: .:nil:u:v -> .:nil:u:v -> .:nil:u:v
f :: true:false:and -> .:nil:u:v -> .:nil:u:v -> .:nil:u:v -> .:nil:u:v
=' :: .:nil:u:v -> .:nil:u:v -> true:false:and
true :: true:false:and
false :: true:false:and
nil :: .:nil:u:v
u :: .:nil:u:v
v :: .:nil:u:v
and :: true:false:and -> true:false:and -> true:false:and
hole_.:nil:u:v1_0 :: .:nil:u:v
hole_true:false:and2_0 :: true:false:and
gen_.:nil:u:v3_0 :: Nat -> .:nil:u:v
gen_true:false:and4_0 :: Nat -> true:false:and


Generator Equations:
gen_.:nil:u:v3_0(0) <=> nil
gen_.:nil:u:v3_0(+(x, 1)) <=> .(nil, gen_.:nil:u:v3_0(x))
gen_true:false:and4_0(0) <=> false
gen_true:false:and4_0(+(x, 1)) <=> and(false, gen_true:false:and4_0(x))


The following defined symbols remain to be analysed:
=', del

They will be analysed ascendingly in the following order:
=' < del

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(19) RewriteLemmaProof (LOWER BOUND(ID))
Proved the following rewrite lemma:
del(gen_.:nil:u:v3_0(+(2, n56_0))) -> *5_0, rt in Omega(n56_0)

Induction Base:
del(gen_.:nil:u:v3_0(+(2, 0)))

Induction Step:
del(gen_.:nil:u:v3_0(+(2, +(n56_0, 1)))) ->_R^Omega(1)
f(='(nil, nil), nil, nil, gen_.:nil:u:v3_0(+(1, n56_0))) ->_R^Omega(1)
f(true, nil, nil, gen_.:nil:u:v3_0(+(1, n56_0))) ->_R^Omega(1)
del(.(nil, gen_.:nil:u:v3_0(+(1, n56_0)))) ->_IH
*5_0

We have rt in Omega(n^1) and sz in O(n). Thus, we have irc_R in Omega(n).
----------------------------------------

(20)
Obligation:
Proved the lower bound n^1 for the following obligation:

Innermost TRS:
Rules:
del(.(x, .(y, z))) -> f(='(x, y), x, y, z)
f(true, x, y, z) -> del(.(y, z))
f(false, x, y, z) -> .(x, del(.(y, z)))
='(nil, nil) -> true
='(.(x, y), nil) -> false
='(nil, .(y, z)) -> false
='(.(x, y), .(u, v)) -> and(='(x, u), ='(y, v))

Types:
del :: .:nil:u:v -> .:nil:u:v
. :: .:nil:u:v -> .:nil:u:v -> .:nil:u:v
f :: true:false:and -> .:nil:u:v -> .:nil:u:v -> .:nil:u:v -> .:nil:u:v
=' :: .:nil:u:v -> .:nil:u:v -> true:false:and
true :: true:false:and
false :: true:false:and
nil :: .:nil:u:v
u :: .:nil:u:v
v :: .:nil:u:v
and :: true:false:and -> true:false:and -> true:false:and
hole_.:nil:u:v1_0 :: .:nil:u:v
hole_true:false:and2_0 :: true:false:and
gen_.:nil:u:v3_0 :: Nat -> .:nil:u:v
gen_true:false:and4_0 :: Nat -> true:false:and


Generator Equations:
gen_.:nil:u:v3_0(0) <=> nil
gen_.:nil:u:v3_0(+(x, 1)) <=> .(nil, gen_.:nil:u:v3_0(x))
gen_true:false:and4_0(0) <=> false
gen_true:false:and4_0(+(x, 1)) <=> and(false, gen_true:false:and4_0(x))


The following defined symbols remain to be analysed:
del
----------------------------------------

(21) LowerBoundPropagationProof (FINISHED)
Propagated lower bound.
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(22)
BOUNDS(n^1, INF)