WORST_CASE(?,O(n^1))
proof of input_BcGXo4aero.trs
# AProVE Commit ID: aff8ecad908e01718a4c36e68d2e55d5e0f16e15 fuhs 20220216 unpublished


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

(0) CpxTRS
(1) CpxTrsToCdtProof [UPPER BOUND(ID), 0 ms]
(2) CdtProblem
(3) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 89 ms]
(4) CdtProblem
(5) CdtInstantiationProof [BOTH BOUNDS(ID, ID), 0 ms]
(6) CdtProblem
(7) CdtRhsSimplificationProcessorProof [BOTH BOUNDS(ID, ID), 0 ms]
(8) CdtProblem
(9) CdtUsableRulesProof [BOTH BOUNDS(ID, ID), 0 ms]
(10) CdtProblem
(11) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 7 ms]
(12) CdtProblem
(13) SIsEmptyProof [BOTH BOUNDS(ID, ID), 0 ms]
(14) BOUNDS(1, 1)


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


The TRS R consists of the following rules:

   f(x, c(y)) -> f(x, s(f(y, y)))
   f(s(x), y) -> f(x, s(c(y)))

S is empty.
Rewrite Strategy: PARALLEL_INNERMOST
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(1) CpxTrsToCdtProof (UPPER BOUND(ID))
Converted Cpx (relative) TRS with rewrite strategy PARALLEL_INNERMOST to CDT
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(2)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   f(z0, c(z1)) -> f(z0, s(f(z1, z1)))
   f(s(z0), z1) -> f(z0, s(c(z1)))
Tuples:
   F(z0, c(z1)) -> c1(F(z0, s(f(z1, z1))), F(z1, z1))
   F(s(z0), z1) -> c2(F(z0, s(c(z1))))
S tuples:
   F(z0, c(z1)) -> c1(F(z0, s(f(z1, z1))), F(z1, z1))
   F(s(z0), z1) -> c2(F(z0, s(c(z1))))
K tuples:none
Defined Rule Symbols:   f_2

Defined Pair Symbols:   F_2

Compound Symbols:   c1_2, c2_1


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(3) 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, c(z1)) -> c1(F(z0, s(f(z1, z1))), F(z1, z1))
We considered the (Usable) Rules:none
And the Tuples:
   F(z0, c(z1)) -> c1(F(z0, s(f(z1, z1))), F(z1, z1))
   F(s(z0), z1) -> c2(F(z0, s(c(z1))))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(F(x_1, x_2)) = x_2
   POL(c(x_1)) = [1] + x_1
   POL(c1(x_1, x_2)) = x_1 + x_2
   POL(c2(x_1)) = x_1
   POL(f(x_1, x_2)) = [1] + x_2
   POL(s(x_1)) = 0

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(4)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   f(z0, c(z1)) -> f(z0, s(f(z1, z1)))
   f(s(z0), z1) -> f(z0, s(c(z1)))
Tuples:
   F(z0, c(z1)) -> c1(F(z0, s(f(z1, z1))), F(z1, z1))
   F(s(z0), z1) -> c2(F(z0, s(c(z1))))
S tuples:
   F(s(z0), z1) -> c2(F(z0, s(c(z1))))
K tuples:
   F(z0, c(z1)) -> c1(F(z0, s(f(z1, z1))), F(z1, z1))
Defined Rule Symbols:   f_2

Defined Pair Symbols:   F_2

Compound Symbols:   c1_2, c2_1


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(5) CdtInstantiationProof (BOTH BOUNDS(ID, ID))
Use instantiation to replace F(z0, c(z1)) -> c1(F(z0, s(f(z1, z1))), F(z1, z1)) by 
   F(c(z1), c(z1)) -> c1(F(c(z1), s(f(z1, z1))), F(z1, z1))

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(6)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   f(z0, c(z1)) -> f(z0, s(f(z1, z1)))
   f(s(z0), z1) -> f(z0, s(c(z1)))
Tuples:
   F(s(z0), z1) -> c2(F(z0, s(c(z1))))
   F(c(z1), c(z1)) -> c1(F(c(z1), s(f(z1, z1))), F(z1, z1))
S tuples:
   F(s(z0), z1) -> c2(F(z0, s(c(z1))))
K tuples:
   F(c(z1), c(z1)) -> c1(F(c(z1), s(f(z1, z1))), F(z1, z1))
Defined Rule Symbols:   f_2

Defined Pair Symbols:   F_2

Compound Symbols:   c2_1, c1_2


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(7) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID))
Removed 1 trailing tuple parts
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(8)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   f(z0, c(z1)) -> f(z0, s(f(z1, z1)))
   f(s(z0), z1) -> f(z0, s(c(z1)))
Tuples:
   F(s(z0), z1) -> c2(F(z0, s(c(z1))))
   F(c(z1), c(z1)) -> c1(F(z1, z1))
S tuples:
   F(s(z0), z1) -> c2(F(z0, s(c(z1))))
K tuples:
   F(c(z1), c(z1)) -> c1(F(z1, z1))
Defined Rule Symbols:   f_2

Defined Pair Symbols:   F_2

Compound Symbols:   c2_1, c1_1


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(9) CdtUsableRulesProof (BOTH BOUNDS(ID, ID))
The following rules are not usable and were removed:
   f(z0, c(z1)) -> f(z0, s(f(z1, z1)))
   f(s(z0), z1) -> f(z0, s(c(z1)))

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(10)
Obligation:
Complexity Dependency Tuples Problem

Rules:none
Tuples:
   F(s(z0), z1) -> c2(F(z0, s(c(z1))))
   F(c(z1), c(z1)) -> c1(F(z1, z1))
S tuples:
   F(s(z0), z1) -> c2(F(z0, s(c(z1))))
K tuples:
   F(c(z1), c(z1)) -> c1(F(z1, z1))
Defined Rule Symbols:none

Defined Pair Symbols:   F_2

Compound Symbols:   c2_1, c1_1


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(11) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   F(s(z0), z1) -> c2(F(z0, s(c(z1))))
We considered the (Usable) Rules:none
And the Tuples:
   F(s(z0), z1) -> c2(F(z0, s(c(z1))))
   F(c(z1), c(z1)) -> c1(F(z1, z1))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(F(x_1, x_2)) = x_1
   POL(c(x_1)) = [1] + x_1
   POL(c1(x_1)) = x_1
   POL(c2(x_1)) = x_1
   POL(s(x_1)) = [1] + x_1

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(12)
Obligation:
Complexity Dependency Tuples Problem

Rules:none
Tuples:
   F(s(z0), z1) -> c2(F(z0, s(c(z1))))
   F(c(z1), c(z1)) -> c1(F(z1, z1))
S tuples:none
K tuples:
   F(c(z1), c(z1)) -> c1(F(z1, z1))
   F(s(z0), z1) -> c2(F(z0, s(c(z1))))
Defined Rule Symbols:none

Defined Pair Symbols:   F_2

Compound Symbols:   c2_1, c1_1


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(13) SIsEmptyProof (BOTH BOUNDS(ID, ID))
The set S is empty
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(14)
BOUNDS(1, 1)