WORST_CASE(?, 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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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 CpxRelTRS could be proven to be BOUNDS(1, n^1).

(0) CpxRelTRS
(1) SInnermostTerminationProof [BOTH CONCRETE BOUNDS(ID, ID), 181 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)), 39 ms]
(10) CdtProblem
(11) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 8 ms]
(12) CdtProblem
(13) SIsEmptyProof [BOTH BOUNDS(ID, ID), 0 ms]
(14) BOUNDS(1, 1)


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

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


The TRS R consists of the following rules:

   ordered(Cons(x', Cons(x, xs))) -> ordered[Ite](<(x', x), Cons(x', Cons(x, xs)))
   ordered(Cons(x, Nil)) -> True
   ordered(Nil) -> True
   notEmpty(Cons(x, xs)) -> True
   notEmpty(Nil) -> False
   goal(xs) -> ordered(xs)

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

   <(S(x), S(y)) -> <(x, y)
   <(0, S(y)) -> True
   <(x, 0) -> False
   ordered[Ite](True, Cons(x, xs)) -> ordered(xs)
   ordered[Ite](False, xs) -> False

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

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

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


The TRS R consists of the following rules:

   ordered(Cons(x', Cons(x, xs))) -> ordered[Ite](<(x', x), Cons(x', Cons(x, xs)))
   ordered(Cons(x, Nil)) -> True
   ordered(Nil) -> True
   notEmpty(Cons(x, xs)) -> True
   notEmpty(Nil) -> False
   goal(xs) -> ordered(xs)

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

   <(S(x), S(y)) -> <(x, y)
   <(0, S(y)) -> True
   <(x, 0) -> False
   ordered[Ite](True, Cons(x, xs)) -> ordered(xs)
   ordered[Ite](False, xs) -> False

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, S(z0)) -> True
   <(z0, 0) -> False
   ordered[Ite](True, Cons(z0, z1)) -> ordered(z1)
   ordered[Ite](False, z0) -> False
   ordered(Cons(z0, Cons(z1, z2))) -> ordered[Ite](<(z0, z1), Cons(z0, Cons(z1, z2)))
   ordered(Cons(z0, Nil)) -> True
   ordered(Nil) -> True
   notEmpty(Cons(z0, z1)) -> True
   notEmpty(Nil) -> False
   goal(z0) -> ordered(z0)
Tuples:
   <'(S(z0), S(z1)) -> c(<'(z0, z1))
   <'(0, S(z0)) -> c1
   <'(z0, 0) -> c2
   ORDERED[ITE](True, Cons(z0, z1)) -> c3(ORDERED(z1))
   ORDERED[ITE](False, z0) -> c4
   ORDERED(Cons(z0, Cons(z1, z2))) -> c5(ORDERED[ITE](<(z0, z1), Cons(z0, Cons(z1, z2))), <'(z0, z1))
   ORDERED(Cons(z0, Nil)) -> c6
   ORDERED(Nil) -> c7
   NOTEMPTY(Cons(z0, z1)) -> c8
   NOTEMPTY(Nil) -> c9
   GOAL(z0) -> c10(ORDERED(z0))
S tuples:
   ORDERED(Cons(z0, Cons(z1, z2))) -> c5(ORDERED[ITE](<(z0, z1), Cons(z0, Cons(z1, z2))), <'(z0, z1))
   ORDERED(Cons(z0, Nil)) -> c6
   ORDERED(Nil) -> c7
   NOTEMPTY(Cons(z0, z1)) -> c8
   NOTEMPTY(Nil) -> c9
   GOAL(z0) -> c10(ORDERED(z0))
K tuples:none
Defined Rule Symbols:   ordered_1, notEmpty_1, goal_1, <_2, ordered[Ite]_2

Defined Pair Symbols:   <'_2, ORDERED[ITE]_2, ORDERED_1, NOTEMPTY_1, GOAL_1

Compound Symbols:   c_1, c1, c2, c3_1, c4, c5_2, c6, c7, c8, c9, c10_1


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

(5) CdtLeafRemovalProof (ComplexityIfPolyImplication)
Removed 1 leading nodes:
   GOAL(z0) -> c10(ORDERED(z0))
Removed 5 trailing nodes:
   NOTEMPTY(Nil) -> c9
   ORDERED[ITE](False, z0) -> c4
   <'(z0, 0) -> c2
   NOTEMPTY(Cons(z0, z1)) -> c8
   <'(0, S(z0)) -> c1

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

(6)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   <(S(z0), S(z1)) -> <(z0, z1)
   <(0, S(z0)) -> True
   <(z0, 0) -> False
   ordered[Ite](True, Cons(z0, z1)) -> ordered(z1)
   ordered[Ite](False, z0) -> False
   ordered(Cons(z0, Cons(z1, z2))) -> ordered[Ite](<(z0, z1), Cons(z0, Cons(z1, z2)))
   ordered(Cons(z0, Nil)) -> True
   ordered(Nil) -> True
   notEmpty(Cons(z0, z1)) -> True
   notEmpty(Nil) -> False
   goal(z0) -> ordered(z0)
Tuples:
   <'(S(z0), S(z1)) -> c(<'(z0, z1))
   ORDERED[ITE](True, Cons(z0, z1)) -> c3(ORDERED(z1))
   ORDERED(Cons(z0, Cons(z1, z2))) -> c5(ORDERED[ITE](<(z0, z1), Cons(z0, Cons(z1, z2))), <'(z0, z1))
   ORDERED(Cons(z0, Nil)) -> c6
   ORDERED(Nil) -> c7
S tuples:
   ORDERED(Cons(z0, Cons(z1, z2))) -> c5(ORDERED[ITE](<(z0, z1), Cons(z0, Cons(z1, z2))), <'(z0, z1))
   ORDERED(Cons(z0, Nil)) -> c6
   ORDERED(Nil) -> c7
K tuples:none
Defined Rule Symbols:   ordered_1, notEmpty_1, goal_1, <_2, ordered[Ite]_2

Defined Pair Symbols:   <'_2, ORDERED[ITE]_2, ORDERED_1

Compound Symbols:   c_1, c3_1, c5_2, c6, c7


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

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

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

(8)
Obligation:
Complexity Dependency Tuples Problem

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

Defined Pair Symbols:   <'_2, ORDERED[ITE]_2, ORDERED_1

Compound Symbols:   c_1, c3_1, c5_2, c6, 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.
   ORDERED(Cons(z0, Nil)) -> c6
   ORDERED(Nil) -> c7
We considered the (Usable) Rules:
   <(z0, 0) -> False
   <(0, S(z0)) -> True
   <(S(z0), S(z1)) -> <(z0, z1)
And the Tuples:
   <'(S(z0), S(z1)) -> c(<'(z0, z1))
   ORDERED[ITE](True, Cons(z0, z1)) -> c3(ORDERED(z1))
   ORDERED(Cons(z0, Cons(z1, z2))) -> c5(ORDERED[ITE](<(z0, z1), Cons(z0, Cons(z1, z2))), <'(z0, z1))
   ORDERED(Cons(z0, Nil)) -> c6
   ORDERED(Nil) -> 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)) = [1]
   POL(False) = 0
   POL(Nil) = [1]
   POL(ORDERED(x_1)) = [1]
   POL(ORDERED[ITE](x_1, x_2)) = x_1 + x_2
   POL(S(x_1)) = [1] + x_1
   POL(True) = 0
   POL(c(x_1)) = x_1
   POL(c3(x_1)) = x_1
   POL(c5(x_1, x_2)) = x_1 + x_2
   POL(c6) = 0
   POL(c7) = 0

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(10)
Obligation:
Complexity Dependency Tuples Problem

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

Defined Pair Symbols:   <'_2, ORDERED[ITE]_2, ORDERED_1

Compound Symbols:   c_1, c3_1, c5_2, c6, 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.
   ORDERED(Cons(z0, Cons(z1, z2))) -> c5(ORDERED[ITE](<(z0, z1), Cons(z0, Cons(z1, z2))), <'(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:
   <'(S(z0), S(z1)) -> c(<'(z0, z1))
   ORDERED[ITE](True, Cons(z0, z1)) -> c3(ORDERED(z1))
   ORDERED(Cons(z0, Cons(z1, z2))) -> c5(ORDERED[ITE](<(z0, z1), Cons(z0, Cons(z1, z2))), <'(z0, z1))
   ORDERED(Cons(z0, Nil)) -> c6
   ORDERED(Nil) -> 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)) = [1] + x_2
   POL(False) = [3]
   POL(Nil) = 0
   POL(ORDERED(x_1)) = [1] + x_1
   POL(ORDERED[ITE](x_1, x_2)) = x_2
   POL(S(x_1)) = [3] + x_1
   POL(True) = [3]
   POL(c(x_1)) = x_1
   POL(c3(x_1)) = x_1
   POL(c5(x_1, x_2)) = x_1 + x_2
   POL(c6) = 0
   POL(c7) = 0

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

(12)
Obligation:
Complexity Dependency Tuples Problem

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

Defined Pair Symbols:   <'_2, ORDERED[ITE]_2, ORDERED_1

Compound Symbols:   c_1, c3_1, c5_2, c6, c7


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

(14)
BOUNDS(1, 1)