The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(1, n^1).

0 CpxTRS
    ↳1 CpxTrsMatchBoundsTAProof (⇔, 0 ms)
    ↳2 BOUNDS(1, n^1)
    ↳3 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)
    ↳4 CdtProblem
        ↳5 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳6 CdtProblem
        ↳7 CdtInstantiationProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳8 CdtProblem
        ↳9 CdtGraphSplitRhsProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳10 CdtProblem
        ↳11 CdtLeafRemovalProof (ComplexityIfPolyImplication, 0 ms)
        ↳12 CdtProblem
        ↳13 CdtKnowledgeProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳14 CdtProblem
        ↳15 CdtNarrowingProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳16 CdtProblem
        ↳17 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳18 CdtProblem
        ↳19 CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)), 355 ms)
        ↳20 CdtProblem
        ↳21 SIsEmptyProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳22 BOUNDS(1, 1)

(0) Obligation:

The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(1, n^1).


The TRS R consists of the following rules:

c(c(b(c(x)))) → b(a(0, c(x)))
c(c(x)) → b(c(b(c(x))))
a(0, x) → c(c(x))

Rewrite Strategy: INNERMOST

(1) CpxTrsMatchBoundsTAProof (EQUIVALENT transformation)

A linear upper bound on the runtime complexity of the TRS R could be shown with a Match-Bound[TAB_LEFTLINEAR,TAB_NONLEFTLINEAR] (for contructor-based start-terms) of 2.

The compatible tree automaton used to show the Match-Boundedness (for constructor-based start-terms) is represented by:
final states : [1, 2]
transitions:
b0(0) → 0
00() → 0
c0(0) → 1
a0(0, 0) → 2
c1(0) → 3
c1(3) → 2
c2(0) → 6
b2(6) → 5
c2(5) → 4
b2(4) → 2

(2) BOUNDS(1, n^1)

(3) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted Cpx (relative) TRS to CDT

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

c(c(b(c(z0)))) → b(a(0, c(z0)))
c(c(z0)) → b(c(b(c(z0))))
a(0, z0) → c(c(z0))
Tuples:

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(b(c(z0))), C(z0))
A(0, z0) → c3(C(c(z0)), C(z0))
S tuples:

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(b(c(z0))), C(z0))
A(0, z0) → c3(C(c(z0)), C(z0))
K tuples:none
Defined Rule Symbols:

c, a

Defined Pair Symbols:

C, A

Compound Symbols:

c1, c2, c3

(5) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing tuple parts

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

c(c(b(c(z0)))) → b(a(0, c(z0)))
c(c(z0)) → b(c(b(c(z0))))
a(0, z0) → c(c(z0))
Tuples:

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
A(0, z0) → c3(C(c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
S tuples:

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
A(0, z0) → c3(C(c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
K tuples:none
Defined Rule Symbols:

c, a

Defined Pair Symbols:

C, A

Compound Symbols:

c1, c3, c2

(7) CdtInstantiationProof (BOTH BOUNDS(ID, ID) transformation)

Use instantiation to replace A(0, z0) → c3(C(c(z0)), C(z0)) by

A(0, c(x0)) → c3(C(c(c(x0))), C(c(x0)))
A1(0, z0) → c3(C(c(z0)), C(z0))

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

c(c(b(c(z0)))) → b(a(0, c(z0)))
c(c(z0)) → b(c(b(c(z0))))
a(0, z0) → c(c(z0))
Tuples:

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
A(0, c(x0)) → c3(C(c(c(x0))), C(c(x0)))
A1(0, z0) → c3(C(c(z0)), C(z0))
S tuples:

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
A(0, c(x0)) → c3(C(c(c(x0))), C(c(x0)))
A1(0, z0) → c3(C(c(z0)), C(z0))
K tuples:none
Defined Rule Symbols:

c, a

Defined Pair Symbols:

C, A, A1

Compound Symbols:

c1, c2, c3

(9) CdtGraphSplitRhsProof (BOTH BOUNDS(ID, ID) transformation)

Split RHS of tuples not part of any SCC

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

c(c(b(c(z0)))) → b(a(0, c(z0)))
c(c(z0)) → b(c(b(c(z0))))
a(0, z0) → c(c(z0))
Tuples:

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
A(0, c(x0)) → c3(C(c(c(x0))), C(c(x0)))
A1(0, z0) → c4(C(c(z0)))
A1(0, z0) → c4(C(z0))
S tuples:

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
A(0, c(x0)) → c3(C(c(c(x0))), C(c(x0)))
A1(0, z0) → c4(C(c(z0)))
A1(0, z0) → c4(C(z0))
K tuples:none
Defined Rule Symbols:

c, a

Defined Pair Symbols:

C, A, A1

Compound Symbols:

c1, c2, c3, c4

(11) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

A1(0, z0) → c4(C(z0))

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

c(c(b(c(z0)))) → b(a(0, c(z0)))
c(c(z0)) → b(c(b(c(z0))))
a(0, z0) → c(c(z0))
Tuples:

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
A(0, c(x0)) → c3(C(c(c(x0))), C(c(x0)))
A1(0, z0) → c4(C(c(z0)))
S tuples:

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
A(0, c(x0)) → c3(C(c(c(x0))), C(c(x0)))
A1(0, z0) → c4(C(c(z0)))
K tuples:none
Defined Rule Symbols:

c, a

Defined Pair Symbols:

C, A, A1

Compound Symbols:

c1, c2, c3, c4

(13) CdtKnowledgeProof (BOTH BOUNDS(ID, ID) transformation)

The following tuples could be moved from S to K by knowledge propagation:

A1(0, z0) → c4(C(c(z0)))

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

c(c(b(c(z0)))) → b(a(0, c(z0)))
c(c(z0)) → b(c(b(c(z0))))
a(0, z0) → c(c(z0))
Tuples:

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
A(0, c(x0)) → c3(C(c(c(x0))), C(c(x0)))
A1(0, z0) → c4(C(c(z0)))
S tuples:

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
A(0, c(x0)) → c3(C(c(c(x0))), C(c(x0)))
K tuples:

A1(0, z0) → c4(C(c(z0)))
Defined Rule Symbols:

c, a

Defined Pair Symbols:

C, A, A1

Compound Symbols:

c1, c2, c3, c4

(15) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)

Use narrowing to replace A(0, c(x0)) → c3(C(c(c(x0))), C(c(x0))) by

A(0, c(b(c(z0)))) → c3(C(b(a(0, c(z0)))), C(c(b(c(z0)))))
A(0, c(z0)) → c3(C(b(c(b(c(z0))))), C(c(z0)))

(16) Obligation:

Complexity Dependency Tuples Problem
Rules:

c(c(b(c(z0)))) → b(a(0, c(z0)))
c(c(z0)) → b(c(b(c(z0))))
a(0, z0) → c(c(z0))
Tuples:

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
A1(0, z0) → c4(C(c(z0)))
A(0, c(b(c(z0)))) → c3(C(b(a(0, c(z0)))), C(c(b(c(z0)))))
A(0, c(z0)) → c3(C(b(c(b(c(z0))))), C(c(z0)))
S tuples:

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
A(0, c(b(c(z0)))) → c3(C(b(a(0, c(z0)))), C(c(b(c(z0)))))
A(0, c(z0)) → c3(C(b(c(b(c(z0))))), C(c(z0)))
K tuples:

A1(0, z0) → c4(C(c(z0)))
Defined Rule Symbols:

c, a

Defined Pair Symbols:

C, A1, A

Compound Symbols:

c1, c2, c4, c3

(17) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

Removed 2 trailing tuple parts

(18) Obligation:

Complexity Dependency Tuples Problem
Rules:

c(c(b(c(z0)))) → b(a(0, c(z0)))
c(c(z0)) → b(c(b(c(z0))))
a(0, z0) → c(c(z0))
Tuples:

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
A1(0, z0) → c4(C(c(z0)))
A(0, c(b(c(z0)))) → c3(C(c(b(c(z0)))))
A(0, c(z0)) → c3(C(c(z0)))
S tuples:

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
A(0, c(b(c(z0)))) → c3(C(c(b(c(z0)))))
A(0, c(z0)) → c3(C(c(z0)))
K tuples:

A1(0, z0) → c4(C(c(z0)))
Defined Rule Symbols:

c, a

Defined Pair Symbols:

C, A1, A

Compound Symbols:

c1, c2, c4, c3

(19) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
A(0, c(b(c(z0)))) → c3(C(c(b(c(z0)))))
A(0, c(z0)) → c3(C(c(z0)))
We considered the (Usable) Rules:

a(0, z0) → c(c(z0))
c(c(z0)) → b(c(b(c(z0))))
c(c(b(c(z0)))) → b(a(0, c(z0)))
And the Tuples:

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
A1(0, z0) → c4(C(c(z0)))
A(0, c(b(c(z0)))) → c3(C(c(b(c(z0)))))
A(0, c(z0)) → c3(C(c(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = [2]   
POL(A(x1, x2)) = [2]x2 + x12   
POL(A1(x1, x2)) = [2]x1·x2 + x12   
POL(C(x1)) = [2]x1   
POL(a(x1, x2)) = [2] + [2]x1 + [2]x2 + x1·x2   
POL(b(x1)) = x1   
POL(c(x1)) = [2] + [2]x1   
POL(c1(x1, x2)) = x1 + x2   
POL(c2(x1)) = x1   
POL(c3(x1)) = x1   
POL(c4(x1)) = x1   

(20) Obligation:

Complexity Dependency Tuples Problem
Rules:

c(c(b(c(z0)))) → b(a(0, c(z0)))
c(c(z0)) → b(c(b(c(z0))))
a(0, z0) → c(c(z0))
Tuples:

C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
A1(0, z0) → c4(C(c(z0)))
A(0, c(b(c(z0)))) → c3(C(c(b(c(z0)))))
A(0, c(z0)) → c3(C(c(z0)))
S tuples:none
K tuples:

A1(0, z0) → c4(C(c(z0)))
C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
A(0, c(b(c(z0)))) → c3(C(c(b(c(z0)))))
A(0, c(z0)) → c3(C(c(z0)))
Defined Rule Symbols:

c, a

Defined Pair Symbols:

C, A1, A

Compound Symbols:

c1, c2, c4, c3

(21) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty

(22) BOUNDS(1, 1)