/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) CdtUsableRulesProof [BOTH BOUNDS(ID, ID), 0 ms]
    (6) CdtProblem
    (7) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 91 ms]
    (8) CdtProblem
    (9) CdtKnowledgeProof [FINISHED, 0 ms]
    (10) BOUNDS(1, 1)
(11) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms]
(12) TRS for Loop Detection
    (13) DecreasingLoopProof [LOWER BOUND(ID), 0 ms]
    (14) BEST
        (15) proven lower bound
            (16) LowerBoundPropagationProof [FINISHED, 0 ms]
            (17) BOUNDS(n^1, INF)
        (18) TRS for Loop Detection


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

(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:

   fst(0, Z) -> nil
   fst(s(X), cons(Y, Z)) -> cons(Y, n__fst(activate(X), activate(Z)))
   from(X) -> cons(X, n__from(s(X)))
   add(0, X) -> X
   add(s(X), Y) -> s(n__add(activate(X), Y))
   len(nil) -> 0
   len(cons(X, Z)) -> s(n__len(activate(Z)))
   fst(X1, X2) -> n__fst(X1, X2)
   from(X) -> n__from(X)
   add(X1, X2) -> n__add(X1, X2)
   len(X) -> n__len(X)
   activate(n__fst(X1, X2)) -> fst(X1, X2)
   activate(n__from(X)) -> from(X)
   activate(n__add(X1, X2)) -> add(X1, X2)
   activate(n__len(X)) -> len(X)
   activate(X) -> X

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

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

(2)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   fst(0, z0) -> nil
   fst(s(z0), cons(z1, z2)) -> cons(z1, n__fst(activate(z0), activate(z2)))
   fst(z0, z1) -> n__fst(z0, z1)
   from(z0) -> cons(z0, n__from(s(z0)))
   from(z0) -> n__from(z0)
   add(0, z0) -> z0
   add(s(z0), z1) -> s(n__add(activate(z0), z1))
   add(z0, z1) -> n__add(z0, z1)
   len(nil) -> 0
   len(cons(z0, z1)) -> s(n__len(activate(z1)))
   len(z0) -> n__len(z0)
   activate(n__fst(z0, z1)) -> fst(z0, z1)
   activate(n__from(z0)) -> from(z0)
   activate(n__add(z0, z1)) -> add(z0, z1)
   activate(n__len(z0)) -> len(z0)
   activate(z0) -> z0
Tuples:
   FST(0, z0) -> c
   FST(s(z0), cons(z1, z2)) -> c1(ACTIVATE(z0), ACTIVATE(z2))
   FST(z0, z1) -> c2
   FROM(z0) -> c3
   FROM(z0) -> c4
   ADD(0, z0) -> c5
   ADD(s(z0), z1) -> c6(ACTIVATE(z0))
   ADD(z0, z1) -> c7
   LEN(nil) -> c8
   LEN(cons(z0, z1)) -> c9(ACTIVATE(z1))
   LEN(z0) -> c10
   ACTIVATE(n__fst(z0, z1)) -> c11(FST(z0, z1))
   ACTIVATE(n__from(z0)) -> c12(FROM(z0))
   ACTIVATE(n__add(z0, z1)) -> c13(ADD(z0, z1))
   ACTIVATE(n__len(z0)) -> c14(LEN(z0))
   ACTIVATE(z0) -> c15
S tuples:
   FST(0, z0) -> c
   FST(s(z0), cons(z1, z2)) -> c1(ACTIVATE(z0), ACTIVATE(z2))
   FST(z0, z1) -> c2
   FROM(z0) -> c3
   FROM(z0) -> c4
   ADD(0, z0) -> c5
   ADD(s(z0), z1) -> c6(ACTIVATE(z0))
   ADD(z0, z1) -> c7
   LEN(nil) -> c8
   LEN(cons(z0, z1)) -> c9(ACTIVATE(z1))
   LEN(z0) -> c10
   ACTIVATE(n__fst(z0, z1)) -> c11(FST(z0, z1))
   ACTIVATE(n__from(z0)) -> c12(FROM(z0))
   ACTIVATE(n__add(z0, z1)) -> c13(ADD(z0, z1))
   ACTIVATE(n__len(z0)) -> c14(LEN(z0))
   ACTIVATE(z0) -> c15
K tuples:none
Defined Rule Symbols:   fst_2, from_1, add_2, len_1, activate_1

Defined Pair Symbols:   FST_2, FROM_1, ADD_2, LEN_1, ACTIVATE_1

Compound Symbols:   c, c1_2, c2, c3, c4, c5, c6_1, c7, c8, c9_1, c10, c11_1, c12_1, c13_1, c14_1, c15


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(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
Removed 10 trailing nodes:
   LEN(z0) -> c10
   ACTIVATE(n__from(z0)) -> c12(FROM(z0))
   ACTIVATE(z0) -> c15
   FST(0, z0) -> c
   LEN(nil) -> c8
   FROM(z0) -> c4
   FST(z0, z1) -> c2
   ADD(z0, z1) -> c7
   ADD(0, z0) -> c5
   FROM(z0) -> c3

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

(4)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   fst(0, z0) -> nil
   fst(s(z0), cons(z1, z2)) -> cons(z1, n__fst(activate(z0), activate(z2)))
   fst(z0, z1) -> n__fst(z0, z1)
   from(z0) -> cons(z0, n__from(s(z0)))
   from(z0) -> n__from(z0)
   add(0, z0) -> z0
   add(s(z0), z1) -> s(n__add(activate(z0), z1))
   add(z0, z1) -> n__add(z0, z1)
   len(nil) -> 0
   len(cons(z0, z1)) -> s(n__len(activate(z1)))
   len(z0) -> n__len(z0)
   activate(n__fst(z0, z1)) -> fst(z0, z1)
   activate(n__from(z0)) -> from(z0)
   activate(n__add(z0, z1)) -> add(z0, z1)
   activate(n__len(z0)) -> len(z0)
   activate(z0) -> z0
Tuples:
   FST(s(z0), cons(z1, z2)) -> c1(ACTIVATE(z0), ACTIVATE(z2))
   ADD(s(z0), z1) -> c6(ACTIVATE(z0))
   LEN(cons(z0, z1)) -> c9(ACTIVATE(z1))
   ACTIVATE(n__fst(z0, z1)) -> c11(FST(z0, z1))
   ACTIVATE(n__add(z0, z1)) -> c13(ADD(z0, z1))
   ACTIVATE(n__len(z0)) -> c14(LEN(z0))
S tuples:
   FST(s(z0), cons(z1, z2)) -> c1(ACTIVATE(z0), ACTIVATE(z2))
   ADD(s(z0), z1) -> c6(ACTIVATE(z0))
   LEN(cons(z0, z1)) -> c9(ACTIVATE(z1))
   ACTIVATE(n__fst(z0, z1)) -> c11(FST(z0, z1))
   ACTIVATE(n__add(z0, z1)) -> c13(ADD(z0, z1))
   ACTIVATE(n__len(z0)) -> c14(LEN(z0))
K tuples:none
Defined Rule Symbols:   fst_2, from_1, add_2, len_1, activate_1

Defined Pair Symbols:   FST_2, ADD_2, LEN_1, ACTIVATE_1

Compound Symbols:   c1_2, c6_1, c9_1, c11_1, c13_1, c14_1


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

(5) CdtUsableRulesProof (BOTH BOUNDS(ID, ID))
The following rules are not usable and were removed:
   fst(0, z0) -> nil
   fst(s(z0), cons(z1, z2)) -> cons(z1, n__fst(activate(z0), activate(z2)))
   fst(z0, z1) -> n__fst(z0, z1)
   from(z0) -> cons(z0, n__from(s(z0)))
   from(z0) -> n__from(z0)
   add(0, z0) -> z0
   add(s(z0), z1) -> s(n__add(activate(z0), z1))
   add(z0, z1) -> n__add(z0, z1)
   len(nil) -> 0
   len(cons(z0, z1)) -> s(n__len(activate(z1)))
   len(z0) -> n__len(z0)
   activate(n__fst(z0, z1)) -> fst(z0, z1)
   activate(n__from(z0)) -> from(z0)
   activate(n__add(z0, z1)) -> add(z0, z1)
   activate(n__len(z0)) -> len(z0)
   activate(z0) -> z0

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

(6)
Obligation:
Complexity Dependency Tuples Problem

Rules:none
Tuples:
   FST(s(z0), cons(z1, z2)) -> c1(ACTIVATE(z0), ACTIVATE(z2))
   ADD(s(z0), z1) -> c6(ACTIVATE(z0))
   LEN(cons(z0, z1)) -> c9(ACTIVATE(z1))
   ACTIVATE(n__fst(z0, z1)) -> c11(FST(z0, z1))
   ACTIVATE(n__add(z0, z1)) -> c13(ADD(z0, z1))
   ACTIVATE(n__len(z0)) -> c14(LEN(z0))
S tuples:
   FST(s(z0), cons(z1, z2)) -> c1(ACTIVATE(z0), ACTIVATE(z2))
   ADD(s(z0), z1) -> c6(ACTIVATE(z0))
   LEN(cons(z0, z1)) -> c9(ACTIVATE(z1))
   ACTIVATE(n__fst(z0, z1)) -> c11(FST(z0, z1))
   ACTIVATE(n__add(z0, z1)) -> c13(ADD(z0, z1))
   ACTIVATE(n__len(z0)) -> c14(LEN(z0))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:   FST_2, ADD_2, LEN_1, ACTIVATE_1

Compound Symbols:   c1_2, c6_1, c9_1, c11_1, c13_1, c14_1


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

(7) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   ADD(s(z0), z1) -> c6(ACTIVATE(z0))
   LEN(cons(z0, z1)) -> c9(ACTIVATE(z1))
   ACTIVATE(n__fst(z0, z1)) -> c11(FST(z0, z1))
   ACTIVATE(n__add(z0, z1)) -> c13(ADD(z0, z1))
   ACTIVATE(n__len(z0)) -> c14(LEN(z0))
We considered the (Usable) Rules:none
And the Tuples:
   FST(s(z0), cons(z1, z2)) -> c1(ACTIVATE(z0), ACTIVATE(z2))
   ADD(s(z0), z1) -> c6(ACTIVATE(z0))
   LEN(cons(z0, z1)) -> c9(ACTIVATE(z1))
   ACTIVATE(n__fst(z0, z1)) -> c11(FST(z0, z1))
   ACTIVATE(n__add(z0, z1)) -> c13(ADD(z0, z1))
   ACTIVATE(n__len(z0)) -> c14(LEN(z0))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(ACTIVATE(x_1)) = [1] + x_1
   POL(ADD(x_1, x_2)) = [1] + x_1 + x_2
   POL(FST(x_1, x_2)) = x_1 + x_2
   POL(LEN(x_1)) = [1] + x_1
   POL(c1(x_1, x_2)) = x_1 + x_2
   POL(c11(x_1)) = x_1
   POL(c13(x_1)) = x_1
   POL(c14(x_1)) = x_1
   POL(c6(x_1)) = x_1
   POL(c9(x_1)) = x_1
   POL(cons(x_1, x_2)) = [1] + x_2
   POL(n__add(x_1, x_2)) = [1] + x_1 + x_2
   POL(n__fst(x_1, x_2)) = [1] + x_1 + x_2
   POL(n__len(x_1)) = [1] + x_1
   POL(s(x_1)) = [1] + x_1

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

(8)
Obligation:
Complexity Dependency Tuples Problem

Rules:none
Tuples:
   FST(s(z0), cons(z1, z2)) -> c1(ACTIVATE(z0), ACTIVATE(z2))
   ADD(s(z0), z1) -> c6(ACTIVATE(z0))
   LEN(cons(z0, z1)) -> c9(ACTIVATE(z1))
   ACTIVATE(n__fst(z0, z1)) -> c11(FST(z0, z1))
   ACTIVATE(n__add(z0, z1)) -> c13(ADD(z0, z1))
   ACTIVATE(n__len(z0)) -> c14(LEN(z0))
S tuples:
   FST(s(z0), cons(z1, z2)) -> c1(ACTIVATE(z0), ACTIVATE(z2))
K tuples:
   ADD(s(z0), z1) -> c6(ACTIVATE(z0))
   LEN(cons(z0, z1)) -> c9(ACTIVATE(z1))
   ACTIVATE(n__fst(z0, z1)) -> c11(FST(z0, z1))
   ACTIVATE(n__add(z0, z1)) -> c13(ADD(z0, z1))
   ACTIVATE(n__len(z0)) -> c14(LEN(z0))
Defined Rule Symbols:none

Defined Pair Symbols:   FST_2, ADD_2, LEN_1, ACTIVATE_1

Compound Symbols:   c1_2, c6_1, c9_1, c11_1, c13_1, c14_1


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

(9) CdtKnowledgeProof (FINISHED)
The following tuples could be moved from S to K by knowledge propagation:
   FST(s(z0), cons(z1, z2)) -> c1(ACTIVATE(z0), ACTIVATE(z2))
   ACTIVATE(n__fst(z0, z1)) -> c11(FST(z0, z1))
   ACTIVATE(n__add(z0, z1)) -> c13(ADD(z0, z1))
   ACTIVATE(n__len(z0)) -> c14(LEN(z0))
Now S is empty
----------------------------------------

(10)
BOUNDS(1, 1)

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

(12)
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:

   fst(0, Z) -> nil
   fst(s(X), cons(Y, Z)) -> cons(Y, n__fst(activate(X), activate(Z)))
   from(X) -> cons(X, n__from(s(X)))
   add(0, X) -> X
   add(s(X), Y) -> s(n__add(activate(X), Y))
   len(nil) -> 0
   len(cons(X, Z)) -> s(n__len(activate(Z)))
   fst(X1, X2) -> n__fst(X1, X2)
   from(X) -> n__from(X)
   add(X1, X2) -> n__add(X1, X2)
   len(X) -> n__len(X)
   activate(n__fst(X1, X2)) -> fst(X1, X2)
   activate(n__from(X)) -> from(X)
   activate(n__add(X1, X2)) -> add(X1, X2)
   activate(n__len(X)) -> len(X)
   activate(X) -> X

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

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

The rewrite sequence

activate(n__fst(s(X1_0), cons(Y2_0, Z3_0))) ->^+ cons(Y2_0, n__fst(activate(X1_0), activate(Z3_0)))

gives rise to a decreasing loop by considering the right hand sides subterm at position [1,0].

The pumping substitution is [X1_0 / n__fst(s(X1_0), cons(Y2_0, Z3_0))].

The result substitution is [ ].




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

(14)
Complex Obligation (BEST)

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

(15)
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:

   fst(0, Z) -> nil
   fst(s(X), cons(Y, Z)) -> cons(Y, n__fst(activate(X), activate(Z)))
   from(X) -> cons(X, n__from(s(X)))
   add(0, X) -> X
   add(s(X), Y) -> s(n__add(activate(X), Y))
   len(nil) -> 0
   len(cons(X, Z)) -> s(n__len(activate(Z)))
   fst(X1, X2) -> n__fst(X1, X2)
   from(X) -> n__from(X)
   add(X1, X2) -> n__add(X1, X2)
   len(X) -> n__len(X)
   activate(n__fst(X1, X2)) -> fst(X1, X2)
   activate(n__from(X)) -> from(X)
   activate(n__add(X1, X2)) -> add(X1, X2)
   activate(n__len(X)) -> len(X)
   activate(X) -> X

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

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

(17)
BOUNDS(n^1, INF)

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

(18)
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:

   fst(0, Z) -> nil
   fst(s(X), cons(Y, Z)) -> cons(Y, n__fst(activate(X), activate(Z)))
   from(X) -> cons(X, n__from(s(X)))
   add(0, X) -> X
   add(s(X), Y) -> s(n__add(activate(X), Y))
   len(nil) -> 0
   len(cons(X, Z)) -> s(n__len(activate(Z)))
   fst(X1, X2) -> n__fst(X1, X2)
   from(X) -> n__from(X)
   add(X1, X2) -> n__add(X1, X2)
   len(X) -> n__len(X)
   activate(n__fst(X1, X2)) -> fst(X1, X2)
   activate(n__from(X)) -> from(X)
   activate(n__add(X1, X2)) -> add(X1, X2)
   activate(n__len(X)) -> len(X)
   activate(X) -> X

S is empty.
Rewrite Strategy: INNERMOST