WORST_CASE(?,O(n^1))
proof of input_KaCUPZMQ9k.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# AProVE Commit ID: aff8ecad908e01718a4c36e68d2e55d5e0f16e15 fuhs 20220216 unpublished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)), 65 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:

   purge(nil) -> nil
   purge(.(x, y)) -> .(x, purge(remove(x, y)))
   remove(x, nil) -> nil
   remove(x, .(y, z)) -> if(=(x, y), remove(x, z), .(y, remove(x, z)))

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:
   purge(nil) -> nil
   purge(.(z0, z1)) -> .(z0, purge(remove(z0, z1)))
   remove(z0, nil) -> nil
   remove(z0, .(z1, z2)) -> if(=(z0, z1), remove(z0, z2), .(z1, remove(z0, z2)))
Tuples:
   PURGE(nil) -> c
   PURGE(.(z0, z1)) -> c1(PURGE(remove(z0, z1)), REMOVE(z0, z1))
   REMOVE(z0, nil) -> c2
   REMOVE(z0, .(z1, z2)) -> c3(REMOVE(z0, z2))
   REMOVE(z0, .(z1, z2)) -> c4(REMOVE(z0, z2))
S tuples:
   PURGE(nil) -> c
   PURGE(.(z0, z1)) -> c1(PURGE(remove(z0, z1)), REMOVE(z0, z1))
   REMOVE(z0, nil) -> c2
   REMOVE(z0, .(z1, z2)) -> c3(REMOVE(z0, z2))
   REMOVE(z0, .(z1, z2)) -> c4(REMOVE(z0, z2))
K tuples:none
Defined Rule Symbols:   purge_1, remove_2

Defined Pair Symbols:   PURGE_1, REMOVE_2

Compound Symbols:   c, c1_2, c2, c3_1, c4_1


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(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
Removed 2 trailing nodes:
   REMOVE(z0, nil) -> c2
   PURGE(nil) -> c

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(4)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   purge(nil) -> nil
   purge(.(z0, z1)) -> .(z0, purge(remove(z0, z1)))
   remove(z0, nil) -> nil
   remove(z0, .(z1, z2)) -> if(=(z0, z1), remove(z0, z2), .(z1, remove(z0, z2)))
Tuples:
   PURGE(.(z0, z1)) -> c1(PURGE(remove(z0, z1)), REMOVE(z0, z1))
   REMOVE(z0, .(z1, z2)) -> c3(REMOVE(z0, z2))
   REMOVE(z0, .(z1, z2)) -> c4(REMOVE(z0, z2))
S tuples:
   PURGE(.(z0, z1)) -> c1(PURGE(remove(z0, z1)), REMOVE(z0, z1))
   REMOVE(z0, .(z1, z2)) -> c3(REMOVE(z0, z2))
   REMOVE(z0, .(z1, z2)) -> c4(REMOVE(z0, z2))
K tuples:none
Defined Rule Symbols:   purge_1, remove_2

Defined Pair Symbols:   PURGE_1, REMOVE_2

Compound Symbols:   c1_2, c3_1, c4_1


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(5) CdtUsableRulesProof (BOTH BOUNDS(ID, ID))
The following rules are not usable and were removed:
   purge(nil) -> nil
   purge(.(z0, z1)) -> .(z0, purge(remove(z0, z1)))

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(6)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   remove(z0, nil) -> nil
   remove(z0, .(z1, z2)) -> if(=(z0, z1), remove(z0, z2), .(z1, remove(z0, z2)))
Tuples:
   PURGE(.(z0, z1)) -> c1(PURGE(remove(z0, z1)), REMOVE(z0, z1))
   REMOVE(z0, .(z1, z2)) -> c3(REMOVE(z0, z2))
   REMOVE(z0, .(z1, z2)) -> c4(REMOVE(z0, z2))
S tuples:
   PURGE(.(z0, z1)) -> c1(PURGE(remove(z0, z1)), REMOVE(z0, z1))
   REMOVE(z0, .(z1, z2)) -> c3(REMOVE(z0, z2))
   REMOVE(z0, .(z1, z2)) -> c4(REMOVE(z0, z2))
K tuples:none
Defined Rule Symbols:   remove_2

Defined Pair Symbols:   PURGE_1, REMOVE_2

Compound Symbols:   c1_2, c3_1, c4_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.
   PURGE(.(z0, z1)) -> c1(PURGE(remove(z0, z1)), REMOVE(z0, z1))
   REMOVE(z0, .(z1, z2)) -> c3(REMOVE(z0, z2))
   REMOVE(z0, .(z1, z2)) -> c4(REMOVE(z0, z2))
We considered the (Usable) Rules:
   remove(z0, nil) -> nil
   remove(z0, .(z1, z2)) -> if(=(z0, z1), remove(z0, z2), .(z1, remove(z0, z2)))
And the Tuples:
   PURGE(.(z0, z1)) -> c1(PURGE(remove(z0, z1)), REMOVE(z0, z1))
   REMOVE(z0, .(z1, z2)) -> c3(REMOVE(z0, z2))
   REMOVE(z0, .(z1, z2)) -> c4(REMOVE(z0, z2))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(.(x_1, x_2)) = [2] + x_1 + x_2
   POL(=(x_1, x_2)) = 0
   POL(PURGE(x_1)) = [2]x_1
   POL(REMOVE(x_1, x_2)) = x_1 + x_2
   POL(c1(x_1, x_2)) = x_1 + x_2
   POL(c3(x_1)) = x_1
   POL(c4(x_1)) = x_1
   POL(if(x_1, x_2, x_3)) = x_1
   POL(nil) = 0
   POL(remove(x_1, x_2)) = 0

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(8)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   remove(z0, nil) -> nil
   remove(z0, .(z1, z2)) -> if(=(z0, z1), remove(z0, z2), .(z1, remove(z0, z2)))
Tuples:
   PURGE(.(z0, z1)) -> c1(PURGE(remove(z0, z1)), REMOVE(z0, z1))
   REMOVE(z0, .(z1, z2)) -> c3(REMOVE(z0, z2))
   REMOVE(z0, .(z1, z2)) -> c4(REMOVE(z0, z2))
S tuples:none
K tuples:
   PURGE(.(z0, z1)) -> c1(PURGE(remove(z0, z1)), REMOVE(z0, z1))
   REMOVE(z0, .(z1, z2)) -> c3(REMOVE(z0, z2))
   REMOVE(z0, .(z1, z2)) -> c4(REMOVE(z0, z2))
Defined Rule Symbols:   remove_2

Defined Pair Symbols:   PURGE_1, REMOVE_2

Compound Symbols:   c1_2, c3_1, c4_1


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