WORST_CASE(?,O(n^1))
proof of input_24nKBIUQN8.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# AProVE Commit ID: aff8ecad908e01718a4c36e68d2e55d5e0f16e15 fuhs 20220216 unpublished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)), 53 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:

   bin(x, 0) -> s(0)
   bin(0, s(y)) -> 0
   bin(s(x), s(y)) -> +(bin(x, s(y)), bin(x, 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:
   bin(z0, 0) -> s(0)
   bin(0, s(z0)) -> 0
   bin(s(z0), s(z1)) -> +(bin(z0, s(z1)), bin(z0, z1))
Tuples:
   BIN(z0, 0) -> c
   BIN(0, s(z0)) -> c1
   BIN(s(z0), s(z1)) -> c2(BIN(z0, s(z1)))
   BIN(s(z0), s(z1)) -> c3(BIN(z0, z1))
S tuples:
   BIN(z0, 0) -> c
   BIN(0, s(z0)) -> c1
   BIN(s(z0), s(z1)) -> c2(BIN(z0, s(z1)))
   BIN(s(z0), s(z1)) -> c3(BIN(z0, z1))
K tuples:none
Defined Rule Symbols:   bin_2

Defined Pair Symbols:   BIN_2

Compound Symbols:   c, c1, c2_1, c3_1


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(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
Removed 2 trailing nodes:
   BIN(0, s(z0)) -> c1
   BIN(z0, 0) -> c

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(4)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   bin(z0, 0) -> s(0)
   bin(0, s(z0)) -> 0
   bin(s(z0), s(z1)) -> +(bin(z0, s(z1)), bin(z0, z1))
Tuples:
   BIN(s(z0), s(z1)) -> c2(BIN(z0, s(z1)))
   BIN(s(z0), s(z1)) -> c3(BIN(z0, z1))
S tuples:
   BIN(s(z0), s(z1)) -> c2(BIN(z0, s(z1)))
   BIN(s(z0), s(z1)) -> c3(BIN(z0, z1))
K tuples:none
Defined Rule Symbols:   bin_2

Defined Pair Symbols:   BIN_2

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:
   bin(z0, 0) -> s(0)
   bin(0, s(z0)) -> 0
   bin(s(z0), s(z1)) -> +(bin(z0, s(z1)), bin(z0, z1))

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(6)
Obligation:
Complexity Dependency Tuples Problem

Rules:none
Tuples:
   BIN(s(z0), s(z1)) -> c2(BIN(z0, s(z1)))
   BIN(s(z0), s(z1)) -> c3(BIN(z0, z1))
S tuples:
   BIN(s(z0), s(z1)) -> c2(BIN(z0, s(z1)))
   BIN(s(z0), s(z1)) -> c3(BIN(z0, z1))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:   BIN_2

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.
   BIN(s(z0), s(z1)) -> c2(BIN(z0, s(z1)))
   BIN(s(z0), s(z1)) -> c3(BIN(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:
   BIN(s(z0), s(z1)) -> c2(BIN(z0, s(z1)))
   BIN(s(z0), s(z1)) -> c3(BIN(z0, z1))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(BIN(x_1, x_2)) = x_1
   POL(c2(x_1)) = x_1
   POL(c3(x_1)) = x_1
   POL(s(x_1)) = [1] + x_1

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(8)
Obligation:
Complexity Dependency Tuples Problem

Rules:none
Tuples:
   BIN(s(z0), s(z1)) -> c2(BIN(z0, s(z1)))
   BIN(s(z0), s(z1)) -> c3(BIN(z0, z1))
S tuples:none
K tuples:
   BIN(s(z0), s(z1)) -> c2(BIN(z0, s(z1)))
   BIN(s(z0), s(z1)) -> c3(BIN(z0, z1))
Defined Rule Symbols:none

Defined Pair Symbols:   BIN_2

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)