/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


The Runtime Complexity (innermost) of the given CpxRelTRS could be proven to be BOUNDS(n^1, n^1).

(0) CpxRelTRS
(1) STerminationProof [BOTH CONCRETE BOUNDS(ID, ID), 175 ms]
(2) CpxRelTRS
(3) CpxTrsToCdtProof [UPPER BOUND(ID), 0 ms]
(4) CdtProblem
    (5) CdtLeafRemovalProof [ComplexityIfPolyImplication, 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) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 13 ms]
    (12) CdtProblem
    (13) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 0 ms]
    (14) CdtProblem
    (15) SIsEmptyProof [BOTH BOUNDS(ID, ID), 0 ms]
    (16) BOUNDS(1, 1)
(17) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms]
(18) TRS for Loop Detection
    (19) DecreasingLoopProof [LOWER BOUND(ID), 0 ms]
    (20) BEST
        (21) proven lower bound
            (22) LowerBoundPropagationProof [FINISHED, 0 ms]
            (23) BOUNDS(n^1, INF)
        (24) TRS for Loop Detection


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

(0)
Obligation:
The Runtime Complexity (innermost) of the given CpxRelTRS could be proven to be BOUNDS(n^1, n^1).


The TRS R consists of the following rules:

   merge(Cons(x, xs), Nil) -> Cons(x, xs)
   merge(Cons(x', xs'), Cons(x, xs)) -> merge[Ite](<=(x', x), Cons(x', xs'), Cons(x, xs))
   merge(Nil, ys) -> ys
   goal(xs, ys) -> merge(xs, ys)

The (relative) TRS S consists of the following rules:

   <=(S(x), S(y)) -> <=(x, y)
   <=(0, y) -> True
   <=(S(x), 0) -> False
   merge[Ite](False, xs', Cons(x, xs)) -> Cons(x, merge(xs', xs))
   merge[Ite](True, Cons(x, xs), ys) -> Cons(x, merge(xs, ys))

Rewrite Strategy: INNERMOST
----------------------------------------

(1) STerminationProof (BOTH CONCRETE BOUNDS(ID, ID))
proved termination of relative rules
----------------------------------------

(2)
Obligation:
The Runtime Complexity (innermost) of the given CpxRelTRS could be proven to be BOUNDS(n^1, n^1).


The TRS R consists of the following rules:

   merge(Cons(x, xs), Nil) -> Cons(x, xs)
   merge(Cons(x', xs'), Cons(x, xs)) -> merge[Ite](<=(x', x), Cons(x', xs'), Cons(x, xs))
   merge(Nil, ys) -> ys
   goal(xs, ys) -> merge(xs, ys)

The (relative) TRS S consists of the following rules:

   <=(S(x), S(y)) -> <=(x, y)
   <=(0, y) -> True
   <=(S(x), 0) -> False
   merge[Ite](False, xs', Cons(x, xs)) -> Cons(x, merge(xs', xs))
   merge[Ite](True, Cons(x, xs), ys) -> Cons(x, merge(xs, ys))

Rewrite Strategy: INNERMOST
----------------------------------------

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

(4)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   <=(S(z0), S(z1)) -> <=(z0, z1)
   <=(0, z0) -> True
   <=(S(z0), 0) -> False
   merge[Ite](False, z0, Cons(z1, z2)) -> Cons(z1, merge(z0, z2))
   merge[Ite](True, Cons(z0, z1), z2) -> Cons(z0, merge(z1, z2))
   merge(Cons(z0, z1), Nil) -> Cons(z0, z1)
   merge(Cons(z0, z1), Cons(z2, z3)) -> merge[Ite](<=(z0, z2), Cons(z0, z1), Cons(z2, z3))
   merge(Nil, z0) -> z0
   goal(z0, z1) -> merge(z0, z1)
Tuples:
   <='(S(z0), S(z1)) -> c(<='(z0, z1))
   <='(0, z0) -> c1
   <='(S(z0), 0) -> c2
   MERGE[ITE](False, z0, Cons(z1, z2)) -> c3(MERGE(z0, z2))
   MERGE[ITE](True, Cons(z0, z1), z2) -> c4(MERGE(z1, z2))
   MERGE(Cons(z0, z1), Nil) -> c5
   MERGE(Cons(z0, z1), Cons(z2, z3)) -> c6(MERGE[ITE](<=(z0, z2), Cons(z0, z1), Cons(z2, z3)), <='(z0, z2))
   MERGE(Nil, z0) -> c7
   GOAL(z0, z1) -> c8(MERGE(z0, z1))
S tuples:
   MERGE(Cons(z0, z1), Nil) -> c5
   MERGE(Cons(z0, z1), Cons(z2, z3)) -> c6(MERGE[ITE](<=(z0, z2), Cons(z0, z1), Cons(z2, z3)), <='(z0, z2))
   MERGE(Nil, z0) -> c7
   GOAL(z0, z1) -> c8(MERGE(z0, z1))
K tuples:none
Defined Rule Symbols:   merge_2, goal_2, <=_2, merge[Ite]_3

Defined Pair Symbols:   <='_2, MERGE[ITE]_3, MERGE_2, GOAL_2

Compound Symbols:   c_1, c1, c2, c3_1, c4_1, c5, c6_2, c7, c8_1


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

(5) CdtLeafRemovalProof (ComplexityIfPolyImplication)
Removed 1 leading nodes:
   GOAL(z0, z1) -> c8(MERGE(z0, z1))
Removed 2 trailing nodes:
   <='(S(z0), 0) -> c2
   <='(0, z0) -> c1

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

(6)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   <=(S(z0), S(z1)) -> <=(z0, z1)
   <=(0, z0) -> True
   <=(S(z0), 0) -> False
   merge[Ite](False, z0, Cons(z1, z2)) -> Cons(z1, merge(z0, z2))
   merge[Ite](True, Cons(z0, z1), z2) -> Cons(z0, merge(z1, z2))
   merge(Cons(z0, z1), Nil) -> Cons(z0, z1)
   merge(Cons(z0, z1), Cons(z2, z3)) -> merge[Ite](<=(z0, z2), Cons(z0, z1), Cons(z2, z3))
   merge(Nil, z0) -> z0
   goal(z0, z1) -> merge(z0, z1)
Tuples:
   <='(S(z0), S(z1)) -> c(<='(z0, z1))
   MERGE[ITE](False, z0, Cons(z1, z2)) -> c3(MERGE(z0, z2))
   MERGE[ITE](True, Cons(z0, z1), z2) -> c4(MERGE(z1, z2))
   MERGE(Cons(z0, z1), Nil) -> c5
   MERGE(Cons(z0, z1), Cons(z2, z3)) -> c6(MERGE[ITE](<=(z0, z2), Cons(z0, z1), Cons(z2, z3)), <='(z0, z2))
   MERGE(Nil, z0) -> c7
S tuples:
   MERGE(Cons(z0, z1), Nil) -> c5
   MERGE(Cons(z0, z1), Cons(z2, z3)) -> c6(MERGE[ITE](<=(z0, z2), Cons(z0, z1), Cons(z2, z3)), <='(z0, z2))
   MERGE(Nil, z0) -> c7
K tuples:none
Defined Rule Symbols:   merge_2, goal_2, <=_2, merge[Ite]_3

Defined Pair Symbols:   <='_2, MERGE[ITE]_3, MERGE_2

Compound Symbols:   c_1, c3_1, c4_1, c5, c6_2, c7


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

(7) CdtUsableRulesProof (BOTH BOUNDS(ID, ID))
The following rules are not usable and were removed:
   merge[Ite](False, z0, Cons(z1, z2)) -> Cons(z1, merge(z0, z2))
   merge[Ite](True, Cons(z0, z1), z2) -> Cons(z0, merge(z1, z2))
   merge(Cons(z0, z1), Nil) -> Cons(z0, z1)
   merge(Cons(z0, z1), Cons(z2, z3)) -> merge[Ite](<=(z0, z2), Cons(z0, z1), Cons(z2, z3))
   merge(Nil, z0) -> z0
   goal(z0, z1) -> merge(z0, z1)

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

(8)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   <=(S(z0), S(z1)) -> <=(z0, z1)
   <=(0, z0) -> True
   <=(S(z0), 0) -> False
Tuples:
   <='(S(z0), S(z1)) -> c(<='(z0, z1))
   MERGE[ITE](False, z0, Cons(z1, z2)) -> c3(MERGE(z0, z2))
   MERGE[ITE](True, Cons(z0, z1), z2) -> c4(MERGE(z1, z2))
   MERGE(Cons(z0, z1), Nil) -> c5
   MERGE(Cons(z0, z1), Cons(z2, z3)) -> c6(MERGE[ITE](<=(z0, z2), Cons(z0, z1), Cons(z2, z3)), <='(z0, z2))
   MERGE(Nil, z0) -> c7
S tuples:
   MERGE(Cons(z0, z1), Nil) -> c5
   MERGE(Cons(z0, z1), Cons(z2, z3)) -> c6(MERGE[ITE](<=(z0, z2), Cons(z0, z1), Cons(z2, z3)), <='(z0, z2))
   MERGE(Nil, z0) -> c7
K tuples:none
Defined Rule Symbols:   <=_2

Defined Pair Symbols:   <='_2, MERGE[ITE]_3, MERGE_2

Compound Symbols:   c_1, c3_1, c4_1, c5, c6_2, c7


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

(9) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   MERGE(Nil, z0) -> c7
We considered the (Usable) Rules:
   <=(0, z0) -> True
   <=(S(z0), S(z1)) -> <=(z0, z1)
   <=(S(z0), 0) -> False
And the Tuples:
   <='(S(z0), S(z1)) -> c(<='(z0, z1))
   MERGE[ITE](False, z0, Cons(z1, z2)) -> c3(MERGE(z0, z2))
   MERGE[ITE](True, Cons(z0, z1), z2) -> c4(MERGE(z1, z2))
   MERGE(Cons(z0, z1), Nil) -> c5
   MERGE(Cons(z0, z1), Cons(z2, z3)) -> c6(MERGE[ITE](<=(z0, z2), Cons(z0, z1), Cons(z2, z3)), <='(z0, z2))
   MERGE(Nil, z0) -> c7
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0) = [1]
   POL(<=(x_1, x_2)) = 0
   POL(<='(x_1, x_2)) = 0
   POL(Cons(x_1, x_2)) = x_1 + x_2
   POL(False) = 0
   POL(MERGE(x_1, x_2)) = x_1
   POL(MERGE[ITE](x_1, x_2, x_3)) = x_1 + x_2
   POL(Nil) = [1]
   POL(S(x_1)) = [1] + x_1
   POL(True) = 0
   POL(c(x_1)) = x_1
   POL(c3(x_1)) = x_1
   POL(c4(x_1)) = x_1
   POL(c5) = 0
   POL(c6(x_1, x_2)) = x_1 + x_2
   POL(c7) = 0

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

(10)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   <=(S(z0), S(z1)) -> <=(z0, z1)
   <=(0, z0) -> True
   <=(S(z0), 0) -> False
Tuples:
   <='(S(z0), S(z1)) -> c(<='(z0, z1))
   MERGE[ITE](False, z0, Cons(z1, z2)) -> c3(MERGE(z0, z2))
   MERGE[ITE](True, Cons(z0, z1), z2) -> c4(MERGE(z1, z2))
   MERGE(Cons(z0, z1), Nil) -> c5
   MERGE(Cons(z0, z1), Cons(z2, z3)) -> c6(MERGE[ITE](<=(z0, z2), Cons(z0, z1), Cons(z2, z3)), <='(z0, z2))
   MERGE(Nil, z0) -> c7
S tuples:
   MERGE(Cons(z0, z1), Nil) -> c5
   MERGE(Cons(z0, z1), Cons(z2, z3)) -> c6(MERGE[ITE](<=(z0, z2), Cons(z0, z1), Cons(z2, z3)), <='(z0, z2))
K tuples:
   MERGE(Nil, z0) -> c7
Defined Rule Symbols:   <=_2

Defined Pair Symbols:   <='_2, MERGE[ITE]_3, MERGE_2

Compound Symbols:   c_1, c3_1, c4_1, c5, c6_2, c7


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

(11) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   MERGE(Cons(z0, z1), Nil) -> c5
We considered the (Usable) Rules:
   <=(0, z0) -> True
   <=(S(z0), S(z1)) -> <=(z0, z1)
   <=(S(z0), 0) -> False
And the Tuples:
   <='(S(z0), S(z1)) -> c(<='(z0, z1))
   MERGE[ITE](False, z0, Cons(z1, z2)) -> c3(MERGE(z0, z2))
   MERGE[ITE](True, Cons(z0, z1), z2) -> c4(MERGE(z1, z2))
   MERGE(Cons(z0, z1), Nil) -> c5
   MERGE(Cons(z0, z1), Cons(z2, z3)) -> c6(MERGE[ITE](<=(z0, z2), Cons(z0, z1), Cons(z2, z3)), <='(z0, z2))
   MERGE(Nil, z0) -> c7
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0) = [1]
   POL(<=(x_1, x_2)) = [1]
   POL(<='(x_1, x_2)) = 0
   POL(Cons(x_1, x_2)) = [1] + x_1 + x_2
   POL(False) = [1]
   POL(MERGE(x_1, x_2)) = [1] + x_1
   POL(MERGE[ITE](x_1, x_2, x_3)) = x_1 + x_2
   POL(Nil) = [1]
   POL(S(x_1)) = [1] + x_1
   POL(True) = [1]
   POL(c(x_1)) = x_1
   POL(c3(x_1)) = x_1
   POL(c4(x_1)) = x_1
   POL(c5) = 0
   POL(c6(x_1, x_2)) = x_1 + x_2
   POL(c7) = 0

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

(12)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   <=(S(z0), S(z1)) -> <=(z0, z1)
   <=(0, z0) -> True
   <=(S(z0), 0) -> False
Tuples:
   <='(S(z0), S(z1)) -> c(<='(z0, z1))
   MERGE[ITE](False, z0, Cons(z1, z2)) -> c3(MERGE(z0, z2))
   MERGE[ITE](True, Cons(z0, z1), z2) -> c4(MERGE(z1, z2))
   MERGE(Cons(z0, z1), Nil) -> c5
   MERGE(Cons(z0, z1), Cons(z2, z3)) -> c6(MERGE[ITE](<=(z0, z2), Cons(z0, z1), Cons(z2, z3)), <='(z0, z2))
   MERGE(Nil, z0) -> c7
S tuples:
   MERGE(Cons(z0, z1), Cons(z2, z3)) -> c6(MERGE[ITE](<=(z0, z2), Cons(z0, z1), Cons(z2, z3)), <='(z0, z2))
K tuples:
   MERGE(Nil, z0) -> c7
   MERGE(Cons(z0, z1), Nil) -> c5
Defined Rule Symbols:   <=_2

Defined Pair Symbols:   <='_2, MERGE[ITE]_3, MERGE_2

Compound Symbols:   c_1, c3_1, c4_1, c5, c6_2, c7


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

(13) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   MERGE(Cons(z0, z1), Cons(z2, z3)) -> c6(MERGE[ITE](<=(z0, z2), Cons(z0, z1), Cons(z2, z3)), <='(z0, z2))
We considered the (Usable) Rules:none
And the Tuples:
   <='(S(z0), S(z1)) -> c(<='(z0, z1))
   MERGE[ITE](False, z0, Cons(z1, z2)) -> c3(MERGE(z0, z2))
   MERGE[ITE](True, Cons(z0, z1), z2) -> c4(MERGE(z1, z2))
   MERGE(Cons(z0, z1), Nil) -> c5
   MERGE(Cons(z0, z1), Cons(z2, z3)) -> c6(MERGE[ITE](<=(z0, z2), Cons(z0, z1), Cons(z2, z3)), <='(z0, z2))
   MERGE(Nil, z0) -> c7
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0) = [3]
   POL(<=(x_1, x_2)) = [3]
   POL(<='(x_1, x_2)) = 0
   POL(Cons(x_1, x_2)) = [2] + x_2
   POL(False) = [3]
   POL(MERGE(x_1, x_2)) = [1] + [2]x_1 + x_2
   POL(MERGE[ITE](x_1, x_2, x_3)) = [2]x_2 + x_3
   POL(Nil) = 0
   POL(S(x_1)) = [3] + x_1
   POL(True) = [3]
   POL(c(x_1)) = x_1
   POL(c3(x_1)) = x_1
   POL(c4(x_1)) = x_1
   POL(c5) = 0
   POL(c6(x_1, x_2)) = x_1 + x_2
   POL(c7) = 0

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

(14)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   <=(S(z0), S(z1)) -> <=(z0, z1)
   <=(0, z0) -> True
   <=(S(z0), 0) -> False
Tuples:
   <='(S(z0), S(z1)) -> c(<='(z0, z1))
   MERGE[ITE](False, z0, Cons(z1, z2)) -> c3(MERGE(z0, z2))
   MERGE[ITE](True, Cons(z0, z1), z2) -> c4(MERGE(z1, z2))
   MERGE(Cons(z0, z1), Nil) -> c5
   MERGE(Cons(z0, z1), Cons(z2, z3)) -> c6(MERGE[ITE](<=(z0, z2), Cons(z0, z1), Cons(z2, z3)), <='(z0, z2))
   MERGE(Nil, z0) -> c7
S tuples:none
K tuples:
   MERGE(Nil, z0) -> c7
   MERGE(Cons(z0, z1), Nil) -> c5
   MERGE(Cons(z0, z1), Cons(z2, z3)) -> c6(MERGE[ITE](<=(z0, z2), Cons(z0, z1), Cons(z2, z3)), <='(z0, z2))
Defined Rule Symbols:   <=_2

Defined Pair Symbols:   <='_2, MERGE[ITE]_3, MERGE_2

Compound Symbols:   c_1, c3_1, c4_1, c5, c6_2, c7


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

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

(16)
BOUNDS(1, 1)

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

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

(18)
Obligation:
Analyzing the following TRS for decreasing loops:

The Runtime Complexity (innermost) of the given CpxRelTRS could be proven to be BOUNDS(n^1, n^1).


The TRS R consists of the following rules:

   merge(Cons(x, xs), Nil) -> Cons(x, xs)
   merge(Cons(x', xs'), Cons(x, xs)) -> merge[Ite](<=(x', x), Cons(x', xs'), Cons(x, xs))
   merge(Nil, ys) -> ys
   goal(xs, ys) -> merge(xs, ys)

The (relative) TRS S consists of the following rules:

   <=(S(x), S(y)) -> <=(x, y)
   <=(0, y) -> True
   <=(S(x), 0) -> False
   merge[Ite](False, xs', Cons(x, xs)) -> Cons(x, merge(xs', xs))
   merge[Ite](True, Cons(x, xs), ys) -> Cons(x, merge(xs, ys))

Rewrite Strategy: INNERMOST
----------------------------------------

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

The rewrite sequence

merge(Cons(0, xs'), Cons(x, xs)) ->^+ Cons(0, merge(xs', Cons(x, xs)))

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

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

The result substitution is [ ].




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

(20)
Complex Obligation (BEST)

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

(21)
Obligation:
Proved the lower bound n^1 for the following obligation:

The Runtime Complexity (innermost) of the given CpxRelTRS could be proven to be BOUNDS(n^1, n^1).


The TRS R consists of the following rules:

   merge(Cons(x, xs), Nil) -> Cons(x, xs)
   merge(Cons(x', xs'), Cons(x, xs)) -> merge[Ite](<=(x', x), Cons(x', xs'), Cons(x, xs))
   merge(Nil, ys) -> ys
   goal(xs, ys) -> merge(xs, ys)

The (relative) TRS S consists of the following rules:

   <=(S(x), S(y)) -> <=(x, y)
   <=(0, y) -> True
   <=(S(x), 0) -> False
   merge[Ite](False, xs', Cons(x, xs)) -> Cons(x, merge(xs', xs))
   merge[Ite](True, Cons(x, xs), ys) -> Cons(x, merge(xs, ys))

Rewrite Strategy: INNERMOST
----------------------------------------

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

(23)
BOUNDS(n^1, INF)

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

(24)
Obligation:
Analyzing the following TRS for decreasing loops:

The Runtime Complexity (innermost) of the given CpxRelTRS could be proven to be BOUNDS(n^1, n^1).


The TRS R consists of the following rules:

   merge(Cons(x, xs), Nil) -> Cons(x, xs)
   merge(Cons(x', xs'), Cons(x, xs)) -> merge[Ite](<=(x', x), Cons(x', xs'), Cons(x, xs))
   merge(Nil, ys) -> ys
   goal(xs, ys) -> merge(xs, ys)

The (relative) TRS S consists of the following rules:

   <=(S(x), S(y)) -> <=(x, y)
   <=(0, y) -> True
   <=(S(x), 0) -> False
   merge[Ite](False, xs', Cons(x, xs)) -> Cons(x, merge(xs', xs))
   merge[Ite](True, Cons(x, xs), ys) -> Cons(x, merge(xs, ys))

Rewrite Strategy: INNERMOST