/export/starexec/sandbox2/solver/bin/starexec_run_complexity /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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


WORST_CASE(Omega(n^1), O(n^1))
proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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), 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)), 25 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:

   g(x, y) -> x
   g(x, y) -> y
   f(0, 1, x) -> f(s(x), x, x)
   f(x, y, s(z)) -> s(f(0, 1, z))

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

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

(2)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   g(z0, z1) -> z0
   g(z0, z1) -> z1
   f(0, 1, z0) -> f(s(z0), z0, z0)
   f(z0, z1, s(z2)) -> s(f(0, 1, z2))
Tuples:
   G(z0, z1) -> c
   G(z0, z1) -> c1
   F(0, 1, z0) -> c2(F(s(z0), z0, z0))
   F(z0, z1, s(z2)) -> c3(F(0, 1, z2))
S tuples:
   G(z0, z1) -> c
   G(z0, z1) -> c1
   F(0, 1, z0) -> c2(F(s(z0), z0, z0))
   F(z0, z1, s(z2)) -> c3(F(0, 1, z2))
K tuples:none
Defined Rule Symbols:   g_2, f_3

Defined Pair Symbols:   G_2, F_3

Compound Symbols:   c, c1, c2_1, c3_1


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

(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
Removed 2 trailing nodes:
   G(z0, z1) -> c
   G(z0, z1) -> c1

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

(4)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   g(z0, z1) -> z0
   g(z0, z1) -> z1
   f(0, 1, z0) -> f(s(z0), z0, z0)
   f(z0, z1, s(z2)) -> s(f(0, 1, z2))
Tuples:
   F(0, 1, z0) -> c2(F(s(z0), z0, z0))
   F(z0, z1, s(z2)) -> c3(F(0, 1, z2))
S tuples:
   F(0, 1, z0) -> c2(F(s(z0), z0, z0))
   F(z0, z1, s(z2)) -> c3(F(0, 1, z2))
K tuples:none
Defined Rule Symbols:   g_2, f_3

Defined Pair Symbols:   F_3

Compound Symbols:   c2_1, c3_1


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

(5) CdtUsableRulesProof (BOTH BOUNDS(ID, ID))
The following rules are not usable and were removed:
   g(z0, z1) -> z0
   g(z0, z1) -> z1
   f(0, 1, z0) -> f(s(z0), z0, z0)
   f(z0, z1, s(z2)) -> s(f(0, 1, z2))

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

(6)
Obligation:
Complexity Dependency Tuples Problem

Rules:none
Tuples:
   F(0, 1, z0) -> c2(F(s(z0), z0, z0))
   F(z0, z1, s(z2)) -> c3(F(0, 1, z2))
S tuples:
   F(0, 1, z0) -> c2(F(s(z0), z0, z0))
   F(z0, z1, s(z2)) -> c3(F(0, 1, z2))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:   F_3

Compound Symbols:   c2_1, c3_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.
   F(z0, z1, s(z2)) -> c3(F(0, 1, z2))
We considered the (Usable) Rules:none
And the Tuples:
   F(0, 1, z0) -> c2(F(s(z0), z0, z0))
   F(z0, z1, s(z2)) -> c3(F(0, 1, z2))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0) = [3]
   POL(1) = 0
   POL(F(x_1, x_2, x_3)) = [2]x_3
   POL(c2(x_1)) = x_1
   POL(c3(x_1)) = x_1
   POL(s(x_1)) = [1] + x_1

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

(8)
Obligation:
Complexity Dependency Tuples Problem

Rules:none
Tuples:
   F(0, 1, z0) -> c2(F(s(z0), z0, z0))
   F(z0, z1, s(z2)) -> c3(F(0, 1, z2))
S tuples:
   F(0, 1, z0) -> c2(F(s(z0), z0, z0))
K tuples:
   F(z0, z1, s(z2)) -> c3(F(0, 1, z2))
Defined Rule Symbols:none

Defined Pair Symbols:   F_3

Compound Symbols:   c2_1, c3_1


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

(9) CdtKnowledgeProof (FINISHED)
The following tuples could be moved from S to K by knowledge propagation:
   F(0, 1, z0) -> c2(F(s(z0), z0, z0))
   F(z0, z1, s(z2)) -> c3(F(0, 1, z2))
Now S is empty
----------------------------------------

(10)
BOUNDS(1, 1)

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

(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:

   g(x, y) -> x
   g(x, y) -> y
   f(0, 1, x) -> f(s(x), x, x)
   f(x, y, s(z)) -> s(f(0, 1, z))

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

f(x, y, s(z)) ->^+ s(f(0, 1, z))

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

The pumping substitution is [z / s(z)].

The result substitution is [x / 0, y / 1].




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

(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:

   g(x, y) -> x
   g(x, y) -> y
   f(0, 1, x) -> f(s(x), x, x)
   f(x, y, s(z)) -> s(f(0, 1, z))

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:

   g(x, y) -> x
   g(x, y) -> y
   f(0, 1, x) -> f(s(x), x, x)
   f(x, y, s(z)) -> s(f(0, 1, z))

S is empty.
Rewrite Strategy: INNERMOST