/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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), 8 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), 1 ms]
    (8) CdtProblem
    (9) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 52 ms]
    (10) CdtProblem
    (11) CdtKnowledgeProof [FINISHED, 0 ms]
    (12) BOUNDS(1, 1)
(13) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms]
(14) TRS for Loop Detection
    (15) DecreasingLoopProof [LOWER BOUND(ID), 0 ms]
    (16) BEST
        (17) proven lower bound
            (18) LowerBoundPropagationProof [FINISHED, 0 ms]
            (19) BOUNDS(n^1, INF)
        (20) TRS for Loop Detection


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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
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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:
   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:
   ='(nil, .(z0, z1)) -> c5
   ='(.(z0, z1), nil) -> c4
   ='(nil, nil) -> c3
   ='(.(z0, z1), .(u, v)) -> c6(='(z0, u), ='(z1, v))

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(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
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(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)))

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(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)) = [1] + x_1 + x_2
   POL(=(x_1, x_2)) = [1]
   POL(DEL(x_1)) = x_1
   POL(F(x_1, x_2, x_3, x_4)) = [1] + x_2 + x_3 + x_4
   POL(and(x_1, x_2)) = [1] + x_1 + x_2
   POL(c(x_1)) = x_1
   POL(c1(x_1)) = x_1
   POL(c2(x_1)) = x_1
   POL(false) = [1]
   POL(nil) = [1]
   POL(true) = [1]
   POL(u) = [1]
   POL(v) = [1]

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(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) RelTrsToDecreasingLoopProblemProof (LOWER BOUND(ID))
Transformed a relative TRS into a decreasing-loop problem.
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(14)
Obligation:
Analyzing the following TRS for decreasing loops:

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
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(15) DecreasingLoopProof (LOWER BOUND(ID))
The following loop(s) give(s) rise to the lower bound Omega(n^1):

The rewrite sequence

f(true, x, nil, .(nil, z3_0)) ->^+ f(true, nil, nil, z3_0)

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

The pumping substitution is [z3_0 / .(nil, z3_0)].

The result substitution is [x / nil].




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(16)
Complex Obligation (BEST)

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(17)
Obligation:
Proved the lower bound n^1 for the following 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
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(18) LowerBoundPropagationProof (FINISHED)
Propagated lower bound.
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(19)
BOUNDS(n^1, INF)

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(20)
Obligation:
Analyzing the following TRS for decreasing loops:

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