/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^3))
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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^3).

(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)), 99 ms]
    (8) CdtProblem
    (9) CdtRuleRemovalProof [UPPER BOUND(ADD(n^2)), 184 ms]
    (10) CdtProblem
    (11) CdtKnowledgeProof [BOTH BOUNDS(ID, ID), 0 ms]
    (12) CdtProblem
    (13) CdtRuleRemovalProof [UPPER BOUND(ADD(n^3)), 1237 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 CpxTRS could be proven to be BOUNDS(n^1, n^3).


The TRS R consists of the following rules:

   le(0, Y) -> true
   le(s(X), 0) -> false
   le(s(X), s(Y)) -> le(X, Y)
   minus(0, Y) -> 0
   minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
   ifMinus(true, s(X), Y) -> 0
   ifMinus(false, s(X), Y) -> s(minus(X, Y))
   quot(0, s(Y)) -> 0
   quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

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

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

(2)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   le(0, z0) -> true
   le(s(z0), 0) -> false
   le(s(z0), s(z1)) -> le(z0, z1)
   minus(0, z0) -> 0
   minus(s(z0), z1) -> ifMinus(le(s(z0), z1), s(z0), z1)
   ifMinus(true, s(z0), z1) -> 0
   ifMinus(false, s(z0), z1) -> s(minus(z0, z1))
   quot(0, s(z0)) -> 0
   quot(s(z0), s(z1)) -> s(quot(minus(z0, z1), s(z1)))
Tuples:
   LE(0, z0) -> c
   LE(s(z0), 0) -> c1
   LE(s(z0), s(z1)) -> c2(LE(z0, z1))
   MINUS(0, z0) -> c3
   MINUS(s(z0), z1) -> c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
   IFMINUS(true, s(z0), z1) -> c5
   IFMINUS(false, s(z0), z1) -> c6(MINUS(z0, z1))
   QUOT(0, s(z0)) -> c7
   QUOT(s(z0), s(z1)) -> c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:
   LE(0, z0) -> c
   LE(s(z0), 0) -> c1
   LE(s(z0), s(z1)) -> c2(LE(z0, z1))
   MINUS(0, z0) -> c3
   MINUS(s(z0), z1) -> c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
   IFMINUS(true, s(z0), z1) -> c5
   IFMINUS(false, s(z0), z1) -> c6(MINUS(z0, z1))
   QUOT(0, s(z0)) -> c7
   QUOT(s(z0), s(z1)) -> c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:   le_2, minus_2, ifMinus_3, quot_2

Defined Pair Symbols:   LE_2, MINUS_2, IFMINUS_3, QUOT_2

Compound Symbols:   c, c1, c2_1, c3, c4_2, c5, c6_1, c7, c8_2


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

(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
Removed 5 trailing nodes:
   QUOT(0, s(z0)) -> c7
   IFMINUS(true, s(z0), z1) -> c5
   LE(0, z0) -> c
   LE(s(z0), 0) -> c1
   MINUS(0, z0) -> c3

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

(4)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   le(0, z0) -> true
   le(s(z0), 0) -> false
   le(s(z0), s(z1)) -> le(z0, z1)
   minus(0, z0) -> 0
   minus(s(z0), z1) -> ifMinus(le(s(z0), z1), s(z0), z1)
   ifMinus(true, s(z0), z1) -> 0
   ifMinus(false, s(z0), z1) -> s(minus(z0, z1))
   quot(0, s(z0)) -> 0
   quot(s(z0), s(z1)) -> s(quot(minus(z0, z1), s(z1)))
Tuples:
   LE(s(z0), s(z1)) -> c2(LE(z0, z1))
   MINUS(s(z0), z1) -> c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
   IFMINUS(false, s(z0), z1) -> c6(MINUS(z0, z1))
   QUOT(s(z0), s(z1)) -> c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:
   LE(s(z0), s(z1)) -> c2(LE(z0, z1))
   MINUS(s(z0), z1) -> c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
   IFMINUS(false, s(z0), z1) -> c6(MINUS(z0, z1))
   QUOT(s(z0), s(z1)) -> c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:   le_2, minus_2, ifMinus_3, quot_2

Defined Pair Symbols:   LE_2, MINUS_2, IFMINUS_3, QUOT_2

Compound Symbols:   c2_1, c4_2, c6_1, c8_2


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

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

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

(6)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   le(s(z0), 0) -> false
   le(s(z0), s(z1)) -> le(z0, z1)
   le(0, z0) -> true
   minus(0, z0) -> 0
   minus(s(z0), z1) -> ifMinus(le(s(z0), z1), s(z0), z1)
   ifMinus(true, s(z0), z1) -> 0
   ifMinus(false, s(z0), z1) -> s(minus(z0, z1))
Tuples:
   LE(s(z0), s(z1)) -> c2(LE(z0, z1))
   MINUS(s(z0), z1) -> c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
   IFMINUS(false, s(z0), z1) -> c6(MINUS(z0, z1))
   QUOT(s(z0), s(z1)) -> c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:
   LE(s(z0), s(z1)) -> c2(LE(z0, z1))
   MINUS(s(z0), z1) -> c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
   IFMINUS(false, s(z0), z1) -> c6(MINUS(z0, z1))
   QUOT(s(z0), s(z1)) -> c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:   le_2, minus_2, ifMinus_3

Defined Pair Symbols:   LE_2, MINUS_2, IFMINUS_3, QUOT_2

Compound Symbols:   c2_1, c4_2, c6_1, c8_2


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

(7) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   QUOT(s(z0), s(z1)) -> c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
We considered the (Usable) Rules:
   ifMinus(true, s(z0), z1) -> 0
   ifMinus(false, s(z0), z1) -> s(minus(z0, z1))
   minus(s(z0), z1) -> ifMinus(le(s(z0), z1), s(z0), z1)
   minus(0, z0) -> 0
And the Tuples:
   LE(s(z0), s(z1)) -> c2(LE(z0, z1))
   MINUS(s(z0), z1) -> c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
   IFMINUS(false, s(z0), z1) -> c6(MINUS(z0, z1))
   QUOT(s(z0), s(z1)) -> c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0) = [1]
   POL(IFMINUS(x_1, x_2, x_3)) = 0
   POL(LE(x_1, x_2)) = 0
   POL(MINUS(x_1, x_2)) = 0
   POL(QUOT(x_1, x_2)) = x_1
   POL(c2(x_1)) = x_1
   POL(c4(x_1, x_2)) = x_1 + x_2
   POL(c6(x_1)) = x_1
   POL(c8(x_1, x_2)) = x_1 + x_2
   POL(false) = [1]
   POL(ifMinus(x_1, x_2, x_3)) = x_2
   POL(le(x_1, x_2)) = [1] + x_2
   POL(minus(x_1, x_2)) = x_1
   POL(s(x_1)) = [1] + x_1
   POL(true) = [1]

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

(8)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   le(s(z0), 0) -> false
   le(s(z0), s(z1)) -> le(z0, z1)
   le(0, z0) -> true
   minus(0, z0) -> 0
   minus(s(z0), z1) -> ifMinus(le(s(z0), z1), s(z0), z1)
   ifMinus(true, s(z0), z1) -> 0
   ifMinus(false, s(z0), z1) -> s(minus(z0, z1))
Tuples:
   LE(s(z0), s(z1)) -> c2(LE(z0, z1))
   MINUS(s(z0), z1) -> c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
   IFMINUS(false, s(z0), z1) -> c6(MINUS(z0, z1))
   QUOT(s(z0), s(z1)) -> c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:
   LE(s(z0), s(z1)) -> c2(LE(z0, z1))
   MINUS(s(z0), z1) -> c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
   IFMINUS(false, s(z0), z1) -> c6(MINUS(z0, z1))
K tuples:
   QUOT(s(z0), s(z1)) -> c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
Defined Rule Symbols:   le_2, minus_2, ifMinus_3

Defined Pair Symbols:   LE_2, MINUS_2, IFMINUS_3, QUOT_2

Compound Symbols:   c2_1, c4_2, c6_1, c8_2


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

(9) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   IFMINUS(false, s(z0), z1) -> c6(MINUS(z0, z1))
We considered the (Usable) Rules:
   ifMinus(true, s(z0), z1) -> 0
   ifMinus(false, s(z0), z1) -> s(minus(z0, z1))
   minus(s(z0), z1) -> ifMinus(le(s(z0), z1), s(z0), z1)
   minus(0, z0) -> 0
And the Tuples:
   LE(s(z0), s(z1)) -> c2(LE(z0, z1))
   MINUS(s(z0), z1) -> c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
   IFMINUS(false, s(z0), z1) -> c6(MINUS(z0, z1))
   QUOT(s(z0), s(z1)) -> c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0) = 0
   POL(IFMINUS(x_1, x_2, x_3)) = [2]x_2
   POL(LE(x_1, x_2)) = 0
   POL(MINUS(x_1, x_2)) = [2]x_1
   POL(QUOT(x_1, x_2)) = x_1^2
   POL(c2(x_1)) = x_1
   POL(c4(x_1, x_2)) = x_1 + x_2
   POL(c6(x_1)) = x_1
   POL(c8(x_1, x_2)) = x_1 + x_2
   POL(false) = 0
   POL(ifMinus(x_1, x_2, x_3)) = [1] + x_2
   POL(le(x_1, x_2)) = 0
   POL(minus(x_1, x_2)) = [1] + x_1
   POL(s(x_1)) = [2] + x_1
   POL(true) = 0

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

(10)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   le(s(z0), 0) -> false
   le(s(z0), s(z1)) -> le(z0, z1)
   le(0, z0) -> true
   minus(0, z0) -> 0
   minus(s(z0), z1) -> ifMinus(le(s(z0), z1), s(z0), z1)
   ifMinus(true, s(z0), z1) -> 0
   ifMinus(false, s(z0), z1) -> s(minus(z0, z1))
Tuples:
   LE(s(z0), s(z1)) -> c2(LE(z0, z1))
   MINUS(s(z0), z1) -> c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
   IFMINUS(false, s(z0), z1) -> c6(MINUS(z0, z1))
   QUOT(s(z0), s(z1)) -> c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:
   LE(s(z0), s(z1)) -> c2(LE(z0, z1))
   MINUS(s(z0), z1) -> c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
K tuples:
   QUOT(s(z0), s(z1)) -> c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
   IFMINUS(false, s(z0), z1) -> c6(MINUS(z0, z1))
Defined Rule Symbols:   le_2, minus_2, ifMinus_3

Defined Pair Symbols:   LE_2, MINUS_2, IFMINUS_3, QUOT_2

Compound Symbols:   c2_1, c4_2, c6_1, c8_2


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

(11) CdtKnowledgeProof (BOTH BOUNDS(ID, ID))
The following tuples could be moved from S to K by knowledge propagation:
   MINUS(s(z0), z1) -> c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
   IFMINUS(false, s(z0), z1) -> c6(MINUS(z0, z1))

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

(12)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   le(s(z0), 0) -> false
   le(s(z0), s(z1)) -> le(z0, z1)
   le(0, z0) -> true
   minus(0, z0) -> 0
   minus(s(z0), z1) -> ifMinus(le(s(z0), z1), s(z0), z1)
   ifMinus(true, s(z0), z1) -> 0
   ifMinus(false, s(z0), z1) -> s(minus(z0, z1))
Tuples:
   LE(s(z0), s(z1)) -> c2(LE(z0, z1))
   MINUS(s(z0), z1) -> c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
   IFMINUS(false, s(z0), z1) -> c6(MINUS(z0, z1))
   QUOT(s(z0), s(z1)) -> c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:
   LE(s(z0), s(z1)) -> c2(LE(z0, z1))
K tuples:
   QUOT(s(z0), s(z1)) -> c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
   IFMINUS(false, s(z0), z1) -> c6(MINUS(z0, z1))
   MINUS(s(z0), z1) -> c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
Defined Rule Symbols:   le_2, minus_2, ifMinus_3

Defined Pair Symbols:   LE_2, MINUS_2, IFMINUS_3, QUOT_2

Compound Symbols:   c2_1, c4_2, c6_1, c8_2


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

(13) CdtRuleRemovalProof (UPPER BOUND(ADD(n^3)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   LE(s(z0), s(z1)) -> c2(LE(z0, z1))
We considered the (Usable) Rules:
   ifMinus(true, s(z0), z1) -> 0
   ifMinus(false, s(z0), z1) -> s(minus(z0, z1))
   minus(s(z0), z1) -> ifMinus(le(s(z0), z1), s(z0), z1)
   minus(0, z0) -> 0
And the Tuples:
   LE(s(z0), s(z1)) -> c2(LE(z0, z1))
   MINUS(s(z0), z1) -> c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
   IFMINUS(false, s(z0), z1) -> c6(MINUS(z0, z1))
   QUOT(s(z0), s(z1)) -> c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0) = 0
   POL(IFMINUS(x_1, x_2, x_3)) = x_2^2
   POL(LE(x_1, x_2)) = [1] + x_1
   POL(MINUS(x_1, x_2)) = [1] + x_1 + x_1^2
   POL(QUOT(x_1, x_2)) = x_1^3
   POL(c2(x_1)) = x_1
   POL(c4(x_1, x_2)) = x_1 + x_2
   POL(c6(x_1)) = x_1
   POL(c8(x_1, x_2)) = x_1 + x_2
   POL(false) = 0
   POL(ifMinus(x_1, x_2, x_3)) = x_2
   POL(le(x_1, x_2)) = 0
   POL(minus(x_1, x_2)) = x_1
   POL(s(x_1)) = [1] + x_1
   POL(true) = 0

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

(14)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   le(s(z0), 0) -> false
   le(s(z0), s(z1)) -> le(z0, z1)
   le(0, z0) -> true
   minus(0, z0) -> 0
   minus(s(z0), z1) -> ifMinus(le(s(z0), z1), s(z0), z1)
   ifMinus(true, s(z0), z1) -> 0
   ifMinus(false, s(z0), z1) -> s(minus(z0, z1))
Tuples:
   LE(s(z0), s(z1)) -> c2(LE(z0, z1))
   MINUS(s(z0), z1) -> c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
   IFMINUS(false, s(z0), z1) -> c6(MINUS(z0, z1))
   QUOT(s(z0), s(z1)) -> c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:none
K tuples:
   QUOT(s(z0), s(z1)) -> c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
   IFMINUS(false, s(z0), z1) -> c6(MINUS(z0, z1))
   MINUS(s(z0), z1) -> c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
   LE(s(z0), s(z1)) -> c2(LE(z0, z1))
Defined Rule Symbols:   le_2, minus_2, ifMinus_3

Defined Pair Symbols:   LE_2, MINUS_2, IFMINUS_3, QUOT_2

Compound Symbols:   c2_1, c4_2, c6_1, c8_2


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

(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 CpxTRS could be proven to be BOUNDS(n^1, n^3).


The TRS R consists of the following rules:

   le(0, Y) -> true
   le(s(X), 0) -> false
   le(s(X), s(Y)) -> le(X, Y)
   minus(0, Y) -> 0
   minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
   ifMinus(true, s(X), Y) -> 0
   ifMinus(false, s(X), Y) -> s(minus(X, Y))
   quot(0, s(Y)) -> 0
   quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

S is empty.
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

le(s(X), s(Y)) ->^+ le(X, Y)

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

The pumping substitution is [X / s(X), Y / s(Y)].

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 CpxTRS could be proven to be BOUNDS(n^1, n^3).


The TRS R consists of the following rules:

   le(0, Y) -> true
   le(s(X), 0) -> false
   le(s(X), s(Y)) -> le(X, Y)
   minus(0, Y) -> 0
   minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
   ifMinus(true, s(X), Y) -> 0
   ifMinus(false, s(X), Y) -> s(minus(X, Y))
   quot(0, s(Y)) -> 0
   quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

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

(22) LowerBoundPropagationProof (FINISHED)
Propagated lower bound.
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(23)
BOUNDS(n^1, INF)

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(24)
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^3).


The TRS R consists of the following rules:

   le(0, Y) -> true
   le(s(X), 0) -> false
   le(s(X), s(Y)) -> le(X, Y)
   minus(0, Y) -> 0
   minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
   ifMinus(true, s(X), Y) -> 0
   ifMinus(false, s(X), Y) -> s(minus(X, Y))
   quot(0, s(Y)) -> 0
   quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

S is empty.
Rewrite Strategy: INNERMOST