/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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 [ComplexityIfPolyImplication, 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)), 54 ms]
    (10) CdtProblem
    (11) SIsEmptyProof [BOTH BOUNDS(ID, ID), 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:

   foldl(x, Cons(S(0), xs)) -> foldl(S(x), xs)
   foldl(S(0), Cons(x, xs)) -> foldl(S(x), xs)
   foldr(a, Cons(x, xs)) -> op(x, foldr(a, xs))
   foldr(a, Nil) -> a
   foldl(a, Nil) -> a
   notEmpty(Cons(x, xs)) -> True
   notEmpty(Nil) -> False
   op(x, S(0)) -> S(x)
   op(S(0), y) -> S(y)
   fold(a, xs) -> Cons(foldl(a, xs), Cons(foldr(a, xs), Nil))

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

(1) CpxTrsToCdtProof (UPPER BOUND(ID))
Converted Cpx (relative) TRS to CDT
----------------------------------------

(2)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   foldl(z0, Cons(S(0), z1)) -> foldl(S(z0), z1)
   foldl(S(0), Cons(z0, z1)) -> foldl(S(z0), z1)
   foldl(z0, Nil) -> z0
   foldr(z0, Cons(z1, z2)) -> op(z1, foldr(z0, z2))
   foldr(z0, Nil) -> z0
   notEmpty(Cons(z0, z1)) -> True
   notEmpty(Nil) -> False
   op(z0, S(0)) -> S(z0)
   op(S(0), z0) -> S(z0)
   fold(z0, z1) -> Cons(foldl(z0, z1), Cons(foldr(z0, z1), Nil))
Tuples:
   FOLDL(z0, Cons(S(0), z1)) -> c(FOLDL(S(z0), z1))
   FOLDL(S(0), Cons(z0, z1)) -> c1(FOLDL(S(z0), z1))
   FOLDL(z0, Nil) -> c2
   FOLDR(z0, Cons(z1, z2)) -> c3(OP(z1, foldr(z0, z2)), FOLDR(z0, z2))
   FOLDR(z0, Nil) -> c4
   NOTEMPTY(Cons(z0, z1)) -> c5
   NOTEMPTY(Nil) -> c6
   OP(z0, S(0)) -> c7
   OP(S(0), z0) -> c8
   FOLD(z0, z1) -> c9(FOLDL(z0, z1), FOLDR(z0, z1))
S tuples:
   FOLDL(z0, Cons(S(0), z1)) -> c(FOLDL(S(z0), z1))
   FOLDL(S(0), Cons(z0, z1)) -> c1(FOLDL(S(z0), z1))
   FOLDL(z0, Nil) -> c2
   FOLDR(z0, Cons(z1, z2)) -> c3(OP(z1, foldr(z0, z2)), FOLDR(z0, z2))
   FOLDR(z0, Nil) -> c4
   NOTEMPTY(Cons(z0, z1)) -> c5
   NOTEMPTY(Nil) -> c6
   OP(z0, S(0)) -> c7
   OP(S(0), z0) -> c8
   FOLD(z0, z1) -> c9(FOLDL(z0, z1), FOLDR(z0, z1))
K tuples:none
Defined Rule Symbols:   foldl_2, foldr_2, notEmpty_1, op_2, fold_2

Defined Pair Symbols:   FOLDL_2, FOLDR_2, NOTEMPTY_1, OP_2, FOLD_2

Compound Symbols:   c_1, c1_1, c2, c3_2, c4, c5, c6, c7, c8, c9_2


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(3) CdtLeafRemovalProof (ComplexityIfPolyImplication)
Removed 1 leading nodes:
   FOLD(z0, z1) -> c9(FOLDL(z0, z1), FOLDR(z0, z1))
Removed 6 trailing nodes:
   NOTEMPTY(Cons(z0, z1)) -> c5
   FOLDR(z0, Nil) -> c4
   OP(S(0), z0) -> c8
   OP(z0, S(0)) -> c7
   NOTEMPTY(Nil) -> c6
   FOLDL(z0, Nil) -> c2

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

(4)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   foldl(z0, Cons(S(0), z1)) -> foldl(S(z0), z1)
   foldl(S(0), Cons(z0, z1)) -> foldl(S(z0), z1)
   foldl(z0, Nil) -> z0
   foldr(z0, Cons(z1, z2)) -> op(z1, foldr(z0, z2))
   foldr(z0, Nil) -> z0
   notEmpty(Cons(z0, z1)) -> True
   notEmpty(Nil) -> False
   op(z0, S(0)) -> S(z0)
   op(S(0), z0) -> S(z0)
   fold(z0, z1) -> Cons(foldl(z0, z1), Cons(foldr(z0, z1), Nil))
Tuples:
   FOLDL(z0, Cons(S(0), z1)) -> c(FOLDL(S(z0), z1))
   FOLDL(S(0), Cons(z0, z1)) -> c1(FOLDL(S(z0), z1))
   FOLDR(z0, Cons(z1, z2)) -> c3(OP(z1, foldr(z0, z2)), FOLDR(z0, z2))
S tuples:
   FOLDL(z0, Cons(S(0), z1)) -> c(FOLDL(S(z0), z1))
   FOLDL(S(0), Cons(z0, z1)) -> c1(FOLDL(S(z0), z1))
   FOLDR(z0, Cons(z1, z2)) -> c3(OP(z1, foldr(z0, z2)), FOLDR(z0, z2))
K tuples:none
Defined Rule Symbols:   foldl_2, foldr_2, notEmpty_1, op_2, fold_2

Defined Pair Symbols:   FOLDL_2, FOLDR_2

Compound Symbols:   c_1, c1_1, c3_2


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

(6)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   foldl(z0, Cons(S(0), z1)) -> foldl(S(z0), z1)
   foldl(S(0), Cons(z0, z1)) -> foldl(S(z0), z1)
   foldl(z0, Nil) -> z0
   foldr(z0, Cons(z1, z2)) -> op(z1, foldr(z0, z2))
   foldr(z0, Nil) -> z0
   notEmpty(Cons(z0, z1)) -> True
   notEmpty(Nil) -> False
   op(z0, S(0)) -> S(z0)
   op(S(0), z0) -> S(z0)
   fold(z0, z1) -> Cons(foldl(z0, z1), Cons(foldr(z0, z1), Nil))
Tuples:
   FOLDL(z0, Cons(S(0), z1)) -> c(FOLDL(S(z0), z1))
   FOLDL(S(0), Cons(z0, z1)) -> c1(FOLDL(S(z0), z1))
   FOLDR(z0, Cons(z1, z2)) -> c3(FOLDR(z0, z2))
S tuples:
   FOLDL(z0, Cons(S(0), z1)) -> c(FOLDL(S(z0), z1))
   FOLDL(S(0), Cons(z0, z1)) -> c1(FOLDL(S(z0), z1))
   FOLDR(z0, Cons(z1, z2)) -> c3(FOLDR(z0, z2))
K tuples:none
Defined Rule Symbols:   foldl_2, foldr_2, notEmpty_1, op_2, fold_2

Defined Pair Symbols:   FOLDL_2, FOLDR_2

Compound Symbols:   c_1, c1_1, c3_1


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(7) CdtUsableRulesProof (BOTH BOUNDS(ID, ID))
The following rules are not usable and were removed:
   foldl(z0, Cons(S(0), z1)) -> foldl(S(z0), z1)
   foldl(S(0), Cons(z0, z1)) -> foldl(S(z0), z1)
   foldl(z0, Nil) -> z0
   foldr(z0, Cons(z1, z2)) -> op(z1, foldr(z0, z2))
   foldr(z0, Nil) -> z0
   notEmpty(Cons(z0, z1)) -> True
   notEmpty(Nil) -> False
   op(z0, S(0)) -> S(z0)
   op(S(0), z0) -> S(z0)
   fold(z0, z1) -> Cons(foldl(z0, z1), Cons(foldr(z0, z1), Nil))

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

(8)
Obligation:
Complexity Dependency Tuples Problem

Rules:none
Tuples:
   FOLDL(z0, Cons(S(0), z1)) -> c(FOLDL(S(z0), z1))
   FOLDL(S(0), Cons(z0, z1)) -> c1(FOLDL(S(z0), z1))
   FOLDR(z0, Cons(z1, z2)) -> c3(FOLDR(z0, z2))
S tuples:
   FOLDL(z0, Cons(S(0), z1)) -> c(FOLDL(S(z0), z1))
   FOLDL(S(0), Cons(z0, z1)) -> c1(FOLDL(S(z0), z1))
   FOLDR(z0, Cons(z1, z2)) -> c3(FOLDR(z0, z2))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:   FOLDL_2, FOLDR_2

Compound Symbols:   c_1, 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.
   FOLDL(z0, Cons(S(0), z1)) -> c(FOLDL(S(z0), z1))
   FOLDL(S(0), Cons(z0, z1)) -> c1(FOLDL(S(z0), z1))
   FOLDR(z0, Cons(z1, z2)) -> c3(FOLDR(z0, z2))
We considered the (Usable) Rules:none
And the Tuples:
   FOLDL(z0, Cons(S(0), z1)) -> c(FOLDL(S(z0), z1))
   FOLDL(S(0), Cons(z0, z1)) -> c1(FOLDL(S(z0), z1))
   FOLDR(z0, Cons(z1, z2)) -> c3(FOLDR(z0, z2))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0) = 0
   POL(Cons(x_1, x_2)) = [1] + x_1 + x_2
   POL(FOLDL(x_1, x_2)) = x_2
   POL(FOLDR(x_1, x_2)) = [3]x_2
   POL(S(x_1)) = [1] + x_1
   POL(c(x_1)) = x_1
   POL(c1(x_1)) = x_1
   POL(c3(x_1)) = x_1

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

(10)
Obligation:
Complexity Dependency Tuples Problem

Rules:none
Tuples:
   FOLDL(z0, Cons(S(0), z1)) -> c(FOLDL(S(z0), z1))
   FOLDL(S(0), Cons(z0, z1)) -> c1(FOLDL(S(z0), z1))
   FOLDR(z0, Cons(z1, z2)) -> c3(FOLDR(z0, z2))
S tuples:none
K tuples:
   FOLDL(z0, Cons(S(0), z1)) -> c(FOLDL(S(z0), z1))
   FOLDL(S(0), Cons(z0, z1)) -> c1(FOLDL(S(z0), z1))
   FOLDR(z0, Cons(z1, z2)) -> c3(FOLDR(z0, z2))
Defined Rule Symbols:none

Defined Pair Symbols:   FOLDL_2, FOLDR_2

Compound Symbols:   c_1, c1_1, c3_1


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

(12)
BOUNDS(1, 1)

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

(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:

   foldl(x, Cons(S(0), xs)) -> foldl(S(x), xs)
   foldl(S(0), Cons(x, xs)) -> foldl(S(x), xs)
   foldr(a, Cons(x, xs)) -> op(x, foldr(a, xs))
   foldr(a, Nil) -> a
   foldl(a, Nil) -> a
   notEmpty(Cons(x, xs)) -> True
   notEmpty(Nil) -> False
   op(x, S(0)) -> S(x)
   op(S(0), y) -> S(y)
   fold(a, xs) -> Cons(foldl(a, xs), Cons(foldr(a, xs), Nil))

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

foldl(x, Cons(S(0), xs)) ->^+ foldl(S(x), xs)

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

The pumping substitution is [xs / Cons(S(0), xs)].

The result substitution is [x / S(x)].




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(16)
Complex Obligation (BEST)

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

(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:

   foldl(x, Cons(S(0), xs)) -> foldl(S(x), xs)
   foldl(S(0), Cons(x, xs)) -> foldl(S(x), xs)
   foldr(a, Cons(x, xs)) -> op(x, foldr(a, xs))
   foldr(a, Nil) -> a
   foldl(a, Nil) -> a
   notEmpty(Cons(x, xs)) -> True
   notEmpty(Nil) -> False
   op(x, S(0)) -> S(x)
   op(S(0), y) -> S(y)
   fold(a, xs) -> Cons(foldl(a, xs), Cons(foldr(a, xs), Nil))

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

(18) LowerBoundPropagationProof (FINISHED)
Propagated lower bound.
----------------------------------------

(19)
BOUNDS(n^1, INF)

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

(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:

   foldl(x, Cons(S(0), xs)) -> foldl(S(x), xs)
   foldl(S(0), Cons(x, xs)) -> foldl(S(x), xs)
   foldr(a, Cons(x, xs)) -> op(x, foldr(a, xs))
   foldr(a, Nil) -> a
   foldl(a, Nil) -> a
   notEmpty(Cons(x, xs)) -> True
   notEmpty(Nil) -> False
   op(x, S(0)) -> S(x)
   op(S(0), y) -> S(y)
   fold(a, xs) -> Cons(foldl(a, xs), Cons(foldr(a, xs), Nil))

S is empty.
Rewrite Strategy: INNERMOST