WORST_CASE(?,O(n^1))
proof of input_lVxxkHQmOG.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# AProVE Commit ID: aff8ecad908e01718a4c36e68d2e55d5e0f16e15 fuhs 20220216 unpublished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) 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)), 43 ms]
(8) CdtProblem
(9) SIsEmptyProof [BOTH BOUNDS(ID, ID), 0 ms]
(10) 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:

   fib(0) -> 0
   fib(s(0)) -> s(0)
   fib(s(s(x))) -> +(fib(s(x)), fib(x))

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:
   fib(0) -> 0
   fib(s(0)) -> s(0)
   fib(s(s(z0))) -> +(fib(s(z0)), fib(z0))
Tuples:
   FIB(0) -> c
   FIB(s(0)) -> c1
   FIB(s(s(z0))) -> c2(FIB(s(z0)))
   FIB(s(s(z0))) -> c3(FIB(z0))
S tuples:
   FIB(0) -> c
   FIB(s(0)) -> c1
   FIB(s(s(z0))) -> c2(FIB(s(z0)))
   FIB(s(s(z0))) -> c3(FIB(z0))
K tuples:none
Defined Rule Symbols:   fib_1

Defined Pair Symbols:   FIB_1

Compound Symbols:   c, c1, c2_1, c3_1


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(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
Removed 2 trailing nodes:
   FIB(s(0)) -> c1
   FIB(0) -> c

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(4)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   fib(0) -> 0
   fib(s(0)) -> s(0)
   fib(s(s(z0))) -> +(fib(s(z0)), fib(z0))
Tuples:
   FIB(s(s(z0))) -> c2(FIB(s(z0)))
   FIB(s(s(z0))) -> c3(FIB(z0))
S tuples:
   FIB(s(s(z0))) -> c2(FIB(s(z0)))
   FIB(s(s(z0))) -> c3(FIB(z0))
K tuples:none
Defined Rule Symbols:   fib_1

Defined Pair Symbols:   FIB_1

Compound Symbols:   c2_1, c3_1


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

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(6)
Obligation:
Complexity Dependency Tuples Problem

Rules:none
Tuples:
   FIB(s(s(z0))) -> c2(FIB(s(z0)))
   FIB(s(s(z0))) -> c3(FIB(z0))
S tuples:
   FIB(s(s(z0))) -> c2(FIB(s(z0)))
   FIB(s(s(z0))) -> c3(FIB(z0))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:   FIB_1

Compound Symbols:   c2_1, c3_1


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(7) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   FIB(s(s(z0))) -> c2(FIB(s(z0)))
   FIB(s(s(z0))) -> c3(FIB(z0))
We considered the (Usable) Rules:none
And the Tuples:
   FIB(s(s(z0))) -> c2(FIB(s(z0)))
   FIB(s(s(z0))) -> c3(FIB(z0))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(FIB(x_1)) = x_1
   POL(c2(x_1)) = x_1
   POL(c3(x_1)) = x_1
   POL(s(x_1)) = [3] + x_1

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(8)
Obligation:
Complexity Dependency Tuples Problem

Rules:none
Tuples:
   FIB(s(s(z0))) -> c2(FIB(s(z0)))
   FIB(s(s(z0))) -> c3(FIB(z0))
S tuples:none
K tuples:
   FIB(s(s(z0))) -> c2(FIB(s(z0)))
   FIB(s(s(z0))) -> c3(FIB(z0))
Defined Rule Symbols:none

Defined Pair Symbols:   FIB_1

Compound Symbols:   c2_1, c3_1


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(9) SIsEmptyProof (BOTH BOUNDS(ID, ID))
The set S is empty
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(10)
BOUNDS(1, 1)