WORST_CASE(?,O(n^1))
proof of input_iUUQYoeoAw.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# AProVE Commit ID: aff8ecad908e01718a4c36e68d2e55d5e0f16e15 fuhs 20220216 unpublished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) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 60 ms]
(6) CdtProblem
(7) SIsEmptyProof [BOTH BOUNDS(ID, ID), 0 ms]
(8) 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(0, y) -> 0
   f(s(x), y) -> f(f(x, y), 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(0, z0) -> 0
   f(s(z0), z1) -> f(f(z0, z1), z1)
Tuples:
   F(0, z0) -> c
   F(s(z0), z1) -> c1(F(f(z0, z1), z1), F(z0, z1))
S tuples:
   F(0, z0) -> c
   F(s(z0), z1) -> c1(F(f(z0, z1), z1), F(z0, z1))
K tuples:none
Defined Rule Symbols:   f_2

Defined Pair Symbols:   F_2

Compound Symbols:   c, c1_2


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(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
Removed 1 trailing nodes:
   F(0, z0) -> c

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(4)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   f(0, z0) -> 0
   f(s(z0), z1) -> f(f(z0, z1), z1)
Tuples:
   F(s(z0), z1) -> c1(F(f(z0, z1), z1), F(z0, z1))
S tuples:
   F(s(z0), z1) -> c1(F(f(z0, z1), z1), F(z0, z1))
K tuples:none
Defined Rule Symbols:   f_2

Defined Pair Symbols:   F_2

Compound Symbols:   c1_2


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(5) 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) -> c1(F(f(z0, z1), z1), F(z0, z1))
We considered the (Usable) Rules:
   f(0, z0) -> 0
   f(s(z0), z1) -> f(f(z0, z1), z1)
And the Tuples:
   F(s(z0), z1) -> c1(F(f(z0, z1), z1), F(z0, z1))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0) = 0
   POL(F(x_1, x_2)) = [1] + [3]x_1
   POL(c1(x_1, x_2)) = x_1 + x_2
   POL(f(x_1, x_2)) = 0
   POL(s(x_1)) = [3] + x_1

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(6)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   f(0, z0) -> 0
   f(s(z0), z1) -> f(f(z0, z1), z1)
Tuples:
   F(s(z0), z1) -> c1(F(f(z0, z1), z1), F(z0, z1))
S tuples:none
K tuples:
   F(s(z0), z1) -> c1(F(f(z0, z1), z1), F(z0, z1))
Defined Rule Symbols:   f_2

Defined Pair Symbols:   F_2

Compound Symbols:   c1_2


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(7) SIsEmptyProof (BOTH BOUNDS(ID, ID))
The set S is empty
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(8)
BOUNDS(1, 1)