WORST_CASE(?, O(n^1))
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


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

(0) DCpxTrs
(1) DerivationalComplexityToRuntimeComplexityProof [BOTH BOUNDS(ID, ID), 0 ms]
(2) CpxRelTRS
(3) SInnermostTerminationProof [BOTH CONCRETE BOUNDS(ID, ID), 166 ms]
(4) CpxRelTRS
(5) CpxTrsToCdtProof [UPPER BOUND(ID), 0 ms]
(6) CdtProblem
(7) CdtLeafRemovalProof [BOTH BOUNDS(ID, ID), 0 ms]
(8) CdtProblem
(9) CdtRhsSimplificationProcessorProof [BOTH BOUNDS(ID, ID), 0 ms]
(10) CdtProblem
(11) CdtGraphSplitRhsProof [BOTH BOUNDS(ID, ID), 0 ms]
(12) CdtProblem
(13) CdtLeafRemovalProof [ComplexityIfPolyImplication, 0 ms]
(14) CdtProblem
(15) CdtUsableRulesProof [BOTH BOUNDS(ID, ID), 0 ms]
(16) CdtProblem
(17) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 43 ms]
(18) CdtProblem
(19) SIsEmptyProof [BOTH BOUNDS(ID, ID), 0 ms]
(20) BOUNDS(1, 1)


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

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


The TRS R consists of the following rules:

   f(0, 1, x) -> f(x, x, x)

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

(1) DerivationalComplexityToRuntimeComplexityProof (BOTH BOUNDS(ID, ID))
The following rules have been added to S to convert the given derivational complexity problem to a runtime complexity problem:

   encArg(0) -> 0
   encArg(1) -> 1
   encArg(cons_f(x_1, x_2, x_3)) -> f(encArg(x_1), encArg(x_2), encArg(x_3))
   encode_f(x_1, x_2, x_3) -> f(encArg(x_1), encArg(x_2), encArg(x_3))
   encode_0 -> 0
   encode_1 -> 1


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

(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:

   f(0, 1, x) -> f(x, x, x)

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

   encArg(0) -> 0
   encArg(1) -> 1
   encArg(cons_f(x_1, x_2, x_3)) -> f(encArg(x_1), encArg(x_2), encArg(x_3))
   encode_f(x_1, x_2, x_3) -> f(encArg(x_1), encArg(x_2), encArg(x_3))
   encode_0 -> 0
   encode_1 -> 1

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

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

(4)
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:

   f(0, 1, x) -> f(x, x, x)

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

   encArg(0) -> 0
   encArg(1) -> 1
   encArg(cons_f(x_1, x_2, x_3)) -> f(encArg(x_1), encArg(x_2), encArg(x_3))
   encode_f(x_1, x_2, x_3) -> f(encArg(x_1), encArg(x_2), encArg(x_3))
   encode_0 -> 0
   encode_1 -> 1

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

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

(6)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   encArg(0) -> 0
   encArg(1) -> 1
   encArg(cons_f(z0, z1, z2)) -> f(encArg(z0), encArg(z1), encArg(z2))
   encode_f(z0, z1, z2) -> f(encArg(z0), encArg(z1), encArg(z2))
   encode_0 -> 0
   encode_1 -> 1
   f(0, 1, z0) -> f(z0, z0, z0)
Tuples:
   ENCARG(0) -> c
   ENCARG(1) -> c1
   ENCARG(cons_f(z0, z1, z2)) -> c2(F(encArg(z0), encArg(z1), encArg(z2)), ENCARG(z0), ENCARG(z1), ENCARG(z2))
   ENCODE_F(z0, z1, z2) -> c3(F(encArg(z0), encArg(z1), encArg(z2)), ENCARG(z0), ENCARG(z1), ENCARG(z2))
   ENCODE_0 -> c4
   ENCODE_1 -> c5
   F(0, 1, z0) -> c6(F(z0, z0, z0))
S tuples:
   F(0, 1, z0) -> c6(F(z0, z0, z0))
K tuples:none
Defined Rule Symbols:   f_3, encArg_1, encode_f_3, encode_0, encode_1

Defined Pair Symbols:   ENCARG_1, ENCODE_F_3, ENCODE_0, ENCODE_1, F_3

Compound Symbols:   c, c1, c2_4, c3_4, c4, c5, c6_1


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

(7) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
Removed 4 trailing nodes:
   ENCODE_1 -> c5
   ENCARG(1) -> c1
   ENCODE_0 -> c4
   ENCARG(0) -> c

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

(8)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   encArg(0) -> 0
   encArg(1) -> 1
   encArg(cons_f(z0, z1, z2)) -> f(encArg(z0), encArg(z1), encArg(z2))
   encode_f(z0, z1, z2) -> f(encArg(z0), encArg(z1), encArg(z2))
   encode_0 -> 0
   encode_1 -> 1
   f(0, 1, z0) -> f(z0, z0, z0)
Tuples:
   ENCARG(cons_f(z0, z1, z2)) -> c2(F(encArg(z0), encArg(z1), encArg(z2)), ENCARG(z0), ENCARG(z1), ENCARG(z2))
   ENCODE_F(z0, z1, z2) -> c3(F(encArg(z0), encArg(z1), encArg(z2)), ENCARG(z0), ENCARG(z1), ENCARG(z2))
   F(0, 1, z0) -> c6(F(z0, z0, z0))
S tuples:
   F(0, 1, z0) -> c6(F(z0, z0, z0))
K tuples:none
Defined Rule Symbols:   f_3, encArg_1, encode_f_3, encode_0, encode_1

Defined Pair Symbols:   ENCARG_1, ENCODE_F_3, F_3

Compound Symbols:   c2_4, c3_4, c6_1


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

(9) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID))
Removed 1 trailing tuple parts
----------------------------------------

(10)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   encArg(0) -> 0
   encArg(1) -> 1
   encArg(cons_f(z0, z1, z2)) -> f(encArg(z0), encArg(z1), encArg(z2))
   encode_f(z0, z1, z2) -> f(encArg(z0), encArg(z1), encArg(z2))
   encode_0 -> 0
   encode_1 -> 1
   f(0, 1, z0) -> f(z0, z0, z0)
Tuples:
   ENCARG(cons_f(z0, z1, z2)) -> c2(F(encArg(z0), encArg(z1), encArg(z2)), ENCARG(z0), ENCARG(z1), ENCARG(z2))
   ENCODE_F(z0, z1, z2) -> c3(F(encArg(z0), encArg(z1), encArg(z2)), ENCARG(z0), ENCARG(z1), ENCARG(z2))
   F(0, 1, z0) -> c6
S tuples:
   F(0, 1, z0) -> c6
K tuples:none
Defined Rule Symbols:   f_3, encArg_1, encode_f_3, encode_0, encode_1

Defined Pair Symbols:   ENCARG_1, ENCODE_F_3, F_3

Compound Symbols:   c2_4, c3_4, c6


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

(11) CdtGraphSplitRhsProof (BOTH BOUNDS(ID, ID))
Split RHS of tuples not part of any SCC
----------------------------------------

(12)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   encArg(0) -> 0
   encArg(1) -> 1
   encArg(cons_f(z0, z1, z2)) -> f(encArg(z0), encArg(z1), encArg(z2))
   encode_f(z0, z1, z2) -> f(encArg(z0), encArg(z1), encArg(z2))
   encode_0 -> 0
   encode_1 -> 1
   f(0, 1, z0) -> f(z0, z0, z0)
Tuples:
   ENCARG(cons_f(z0, z1, z2)) -> c2(F(encArg(z0), encArg(z1), encArg(z2)), ENCARG(z0), ENCARG(z1), ENCARG(z2))
   F(0, 1, z0) -> c6
   ENCODE_F(z0, z1, z2) -> c(F(encArg(z0), encArg(z1), encArg(z2)))
   ENCODE_F(z0, z1, z2) -> c(ENCARG(z0))
   ENCODE_F(z0, z1, z2) -> c(ENCARG(z1))
   ENCODE_F(z0, z1, z2) -> c(ENCARG(z2))
S tuples:
   F(0, 1, z0) -> c6
K tuples:none
Defined Rule Symbols:   f_3, encArg_1, encode_f_3, encode_0, encode_1

Defined Pair Symbols:   ENCARG_1, F_3, ENCODE_F_3

Compound Symbols:   c2_4, c6, c_1


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

(13) CdtLeafRemovalProof (ComplexityIfPolyImplication)
Removed 3 leading nodes:
   ENCODE_F(z0, z1, z2) -> c(ENCARG(z0))
   ENCODE_F(z0, z1, z2) -> c(ENCARG(z1))
   ENCODE_F(z0, z1, z2) -> c(ENCARG(z2))

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

(14)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   encArg(0) -> 0
   encArg(1) -> 1
   encArg(cons_f(z0, z1, z2)) -> f(encArg(z0), encArg(z1), encArg(z2))
   encode_f(z0, z1, z2) -> f(encArg(z0), encArg(z1), encArg(z2))
   encode_0 -> 0
   encode_1 -> 1
   f(0, 1, z0) -> f(z0, z0, z0)
Tuples:
   ENCARG(cons_f(z0, z1, z2)) -> c2(F(encArg(z0), encArg(z1), encArg(z2)), ENCARG(z0), ENCARG(z1), ENCARG(z2))
   F(0, 1, z0) -> c6
   ENCODE_F(z0, z1, z2) -> c(F(encArg(z0), encArg(z1), encArg(z2)))
S tuples:
   F(0, 1, z0) -> c6
K tuples:none
Defined Rule Symbols:   f_3, encArg_1, encode_f_3, encode_0, encode_1

Defined Pair Symbols:   ENCARG_1, F_3, ENCODE_F_3

Compound Symbols:   c2_4, c6, c_1


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(15) CdtUsableRulesProof (BOTH BOUNDS(ID, ID))
The following rules are not usable and were removed:
   encode_f(z0, z1, z2) -> f(encArg(z0), encArg(z1), encArg(z2))
   encode_0 -> 0
   encode_1 -> 1

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

(16)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   encArg(0) -> 0
   encArg(1) -> 1
   encArg(cons_f(z0, z1, z2)) -> f(encArg(z0), encArg(z1), encArg(z2))
   f(0, 1, z0) -> f(z0, z0, z0)
Tuples:
   ENCARG(cons_f(z0, z1, z2)) -> c2(F(encArg(z0), encArg(z1), encArg(z2)), ENCARG(z0), ENCARG(z1), ENCARG(z2))
   F(0, 1, z0) -> c6
   ENCODE_F(z0, z1, z2) -> c(F(encArg(z0), encArg(z1), encArg(z2)))
S tuples:
   F(0, 1, z0) -> c6
K tuples:none
Defined Rule Symbols:   encArg_1, f_3

Defined Pair Symbols:   ENCARG_1, F_3, ENCODE_F_3

Compound Symbols:   c2_4, c6, c_1


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(17) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   F(0, 1, z0) -> c6
We considered the (Usable) Rules:none
And the Tuples:
   ENCARG(cons_f(z0, z1, z2)) -> c2(F(encArg(z0), encArg(z1), encArg(z2)), ENCARG(z0), ENCARG(z1), ENCARG(z2))
   F(0, 1, z0) -> c6
   ENCODE_F(z0, z1, z2) -> c(F(encArg(z0), encArg(z1), encArg(z2)))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0) = [1]
   POL(1) = [1]
   POL(ENCARG(x_1)) = x_1
   POL(ENCODE_F(x_1, x_2, x_3)) = [1]
   POL(F(x_1, x_2, x_3)) = [1]
   POL(c(x_1)) = x_1
   POL(c2(x_1, x_2, x_3, x_4)) = x_1 + x_2 + x_3 + x_4
   POL(c6) = 0
   POL(cons_f(x_1, x_2, x_3)) = [1] + x_1 + x_2 + x_3
   POL(encArg(x_1)) = [1] + x_1
   POL(f(x_1, x_2, x_3)) = [1] + x_3

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

(18)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   encArg(0) -> 0
   encArg(1) -> 1
   encArg(cons_f(z0, z1, z2)) -> f(encArg(z0), encArg(z1), encArg(z2))
   f(0, 1, z0) -> f(z0, z0, z0)
Tuples:
   ENCARG(cons_f(z0, z1, z2)) -> c2(F(encArg(z0), encArg(z1), encArg(z2)), ENCARG(z0), ENCARG(z1), ENCARG(z2))
   F(0, 1, z0) -> c6
   ENCODE_F(z0, z1, z2) -> c(F(encArg(z0), encArg(z1), encArg(z2)))
S tuples:none
K tuples:
   F(0, 1, z0) -> c6
Defined Rule Symbols:   encArg_1, f_3

Defined Pair Symbols:   ENCARG_1, F_3, ENCODE_F_3

Compound Symbols:   c2_4, c6, c_1


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

(20)
BOUNDS(1, 1)