WORST_CASE(?,O(n^1))
proof of input_mxICHJNz0i.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# AProVE Commit ID: aff8ecad908e01718a4c36e68d2e55d5e0f16e15 fuhs 20220216 unpublished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)), 66 ms]
(8) CdtProblem
(9) CdtNarrowingProof [BOTH BOUNDS(ID, ID), 0 ms]
(10) CdtProblem
(11) CdtLeafRemovalProof [BOTH BOUNDS(ID, ID), 0 ms]
(12) CdtProblem
(13) CdtRhsSimplificationProcessorProof [BOTH BOUNDS(ID, ID), 0 ms]
(14) CdtProblem
(15) CdtLeafRemovalProof [ComplexityIfPolyImplication, 0 ms]
(16) CdtProblem
(17) CdtUsableRulesProof [BOTH BOUNDS(ID, ID), 0 ms]
(18) CdtProblem
(19) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 0 ms]
(20) CdtProblem
(21) SIsEmptyProof [BOTH BOUNDS(ID, ID), 1 ms]
(22) 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, 0) -> s(0)
   f(s(x), s(y)) -> s(f(x, y))
   g(0, x) -> g(f(x, x), 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:
   f(z0, 0) -> s(0)
   f(s(z0), s(z1)) -> s(f(z0, z1))
   g(0, z0) -> g(f(z0, z0), z0)
Tuples:
   F(z0, 0) -> c
   F(s(z0), s(z1)) -> c1(F(z0, z1))
   G(0, z0) -> c2(G(f(z0, z0), z0), F(z0, z0))
S tuples:
   F(z0, 0) -> c
   F(s(z0), s(z1)) -> c1(F(z0, z1))
   G(0, z0) -> c2(G(f(z0, z0), z0), F(z0, z0))
K tuples:none
Defined Rule Symbols:   f_2, g_2

Defined Pair Symbols:   F_2, G_2

Compound Symbols:   c, c1_1, c2_2


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

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(4)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   f(z0, 0) -> s(0)
   f(s(z0), s(z1)) -> s(f(z0, z1))
   g(0, z0) -> g(f(z0, z0), z0)
Tuples:
   F(s(z0), s(z1)) -> c1(F(z0, z1))
   G(0, z0) -> c2(G(f(z0, z0), z0), F(z0, z0))
S tuples:
   F(s(z0), s(z1)) -> c1(F(z0, z1))
   G(0, z0) -> c2(G(f(z0, z0), z0), F(z0, z0))
K tuples:none
Defined Rule Symbols:   f_2, g_2

Defined Pair Symbols:   F_2, G_2

Compound Symbols:   c1_1, c2_2


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

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(6)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   f(z0, 0) -> s(0)
   f(s(z0), s(z1)) -> s(f(z0, z1))
Tuples:
   F(s(z0), s(z1)) -> c1(F(z0, z1))
   G(0, z0) -> c2(G(f(z0, z0), z0), F(z0, z0))
S tuples:
   F(s(z0), s(z1)) -> c1(F(z0, z1))
   G(0, z0) -> c2(G(f(z0, z0), z0), F(z0, z0))
K tuples:none
Defined Rule Symbols:   f_2

Defined Pair Symbols:   F_2, G_2

Compound Symbols:   c1_1, c2_2


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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.
   G(0, z0) -> c2(G(f(z0, z0), z0), F(z0, z0))
We considered the (Usable) Rules:
   f(s(z0), s(z1)) -> s(f(z0, z1))
   f(z0, 0) -> s(0)
And the Tuples:
   F(s(z0), s(z1)) -> c1(F(z0, z1))
   G(0, z0) -> c2(G(f(z0, z0), z0), F(z0, z0))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0) = [2]
   POL(F(x_1, x_2)) = [1]
   POL(G(x_1, x_2)) = [2]x_1
   POL(c1(x_1)) = x_1
   POL(c2(x_1, x_2)) = x_1 + x_2
   POL(f(x_1, x_2)) = 0
   POL(s(x_1)) = 0

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(8)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   f(z0, 0) -> s(0)
   f(s(z0), s(z1)) -> s(f(z0, z1))
Tuples:
   F(s(z0), s(z1)) -> c1(F(z0, z1))
   G(0, z0) -> c2(G(f(z0, z0), z0), F(z0, z0))
S tuples:
   F(s(z0), s(z1)) -> c1(F(z0, z1))
K tuples:
   G(0, z0) -> c2(G(f(z0, z0), z0), F(z0, z0))
Defined Rule Symbols:   f_2

Defined Pair Symbols:   F_2, G_2

Compound Symbols:   c1_1, c2_2


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(9) CdtNarrowingProof (BOTH BOUNDS(ID, ID))
Use narrowing to replace G(0, z0) -> c2(G(f(z0, z0), z0), F(z0, z0)) by 
   G(0, 0) -> c2(G(s(0), 0), F(0, 0))
   G(0, s(z0)) -> c2(G(s(f(z0, z0)), s(z0)), F(s(z0), s(z0)))

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(10)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   f(z0, 0) -> s(0)
   f(s(z0), s(z1)) -> s(f(z0, z1))
Tuples:
   F(s(z0), s(z1)) -> c1(F(z0, z1))
   G(0, 0) -> c2(G(s(0), 0), F(0, 0))
   G(0, s(z0)) -> c2(G(s(f(z0, z0)), s(z0)), F(s(z0), s(z0)))
S tuples:
   F(s(z0), s(z1)) -> c1(F(z0, z1))
K tuples:
   G(0, z0) -> c2(G(f(z0, z0), z0), F(z0, z0))
Defined Rule Symbols:   f_2

Defined Pair Symbols:   F_2, G_2

Compound Symbols:   c1_1, c2_2


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(11) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
Removed 1 trailing nodes:
   G(0, 0) -> c2(G(s(0), 0), F(0, 0))

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(12)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   f(z0, 0) -> s(0)
   f(s(z0), s(z1)) -> s(f(z0, z1))
Tuples:
   F(s(z0), s(z1)) -> c1(F(z0, z1))
   G(0, s(z0)) -> c2(G(s(f(z0, z0)), s(z0)), F(s(z0), s(z0)))
S tuples:
   F(s(z0), s(z1)) -> c1(F(z0, z1))
K tuples:none
Defined Rule Symbols:   f_2

Defined Pair Symbols:   F_2, G_2

Compound Symbols:   c1_1, c2_2


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(13) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID))
Removed 1 trailing tuple parts
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(14)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   f(z0, 0) -> s(0)
   f(s(z0), s(z1)) -> s(f(z0, z1))
Tuples:
   F(s(z0), s(z1)) -> c1(F(z0, z1))
   G(0, s(z0)) -> c2(F(s(z0), s(z0)))
S tuples:
   F(s(z0), s(z1)) -> c1(F(z0, z1))
K tuples:none
Defined Rule Symbols:   f_2

Defined Pair Symbols:   F_2, G_2

Compound Symbols:   c1_1, c2_1


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(15) CdtLeafRemovalProof (ComplexityIfPolyImplication)
Removed 1 leading nodes:
   G(0, s(z0)) -> c2(F(s(z0), s(z0)))

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(16)
Obligation:
Complexity Dependency Tuples Problem

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

Defined Pair Symbols:   F_2

Compound Symbols:   c1_1


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(17) CdtUsableRulesProof (BOTH BOUNDS(ID, ID))
The following rules are not usable and were removed:
   f(z0, 0) -> s(0)
   f(s(z0), s(z1)) -> s(f(z0, z1))

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(18)
Obligation:
Complexity Dependency Tuples Problem

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

Defined Pair Symbols:   F_2

Compound Symbols:   c1_1


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

Polynomial interpretation :

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

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(20)
Obligation:
Complexity Dependency Tuples Problem

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

Defined Pair Symbols:   F_2

Compound Symbols:   c1_1


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(21) SIsEmptyProof (BOTH BOUNDS(ID, ID))
The set S is empty
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(22)
BOUNDS(1, 1)