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

0 CpxTRS
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)
↳2 CdtProblem
↳3 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID), 0 ms)
↳4 CdtProblem
↳5 CdtNarrowingProof (BOTH BOUNDS(ID, ID), 5 ms)
↳6 CdtProblem
↳7 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID), 0 ms)
↳8 CdtProblem
↳9 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))), 0 ms)
↳10 CdtProblem
↳11 CdtNarrowingProof (BOTH BOUNDS(ID, ID), 0 ms)
↳12 CdtProblem
↳13 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID), 0 ms)
↳14 CdtProblem
↳15 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))), 0 ms)
↳16 CdtProblem
↳17 SIsEmptyProof (⇔, 0 ms)
↳18 BOUNDS(O(1), O(1))

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

f(c(n, x)) → c(f(g(c(n, x))), f(h(c(n, x))))
g(c(0, x)) → x
h(c(1, x)) → x

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(c(z0, z1)) → c(f(g(c(z0, z1))), f(h(c(z0, z1))))
g(c(0, z0)) → z0
h(c(1, z0)) → z0
Tuples:

F(c(z0, z1)) → c1(F(g(c(z0, z1))), G(c(z0, z1)), F(h(c(z0, z1))), H(c(z0, z1)))
S tuples:

F(c(z0, z1)) → c1(F(g(c(z0, z1))), G(c(z0, z1)), F(h(c(z0, z1))), H(c(z0, z1)))
K tuples:none
Defined Rule Symbols:

f, g, h

Defined Pair Symbols:

F

Compound Symbols:

c1

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

Removed 2 trailing tuple parts

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(c(z0, z1)) → c(f(g(c(z0, z1))), f(h(c(z0, z1))))
g(c(0, z0)) → z0
h(c(1, z0)) → z0
Tuples:

F(c(z0, z1)) → c1(F(g(c(z0, z1))), F(h(c(z0, z1))))
S tuples:

F(c(z0, z1)) → c1(F(g(c(z0, z1))), F(h(c(z0, z1))))
K tuples:none
Defined Rule Symbols:

f, g, h

Defined Pair Symbols:

F

Compound Symbols:

c1

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

Use narrowing to replace F(c(z0, z1)) → c1(F(g(c(z0, z1))), F(h(c(z0, z1)))) by

F(c(0, z0)) → c1(F(z0), F(h(c(0, z0))))
F(c(x0, x1)) → c1(F(h(c(x0, x1))))

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(c(z0, z1)) → c(f(g(c(z0, z1))), f(h(c(z0, z1))))
g(c(0, z0)) → z0
h(c(1, z0)) → z0
Tuples:

F(c(0, z0)) → c1(F(z0), F(h(c(0, z0))))
F(c(x0, x1)) → c1(F(h(c(x0, x1))))
S tuples:

F(c(0, z0)) → c1(F(z0), F(h(c(0, z0))))
F(c(x0, x1)) → c1(F(h(c(x0, x1))))
K tuples:none
Defined Rule Symbols:

f, g, h

Defined Pair Symbols:

F

Compound Symbols:

c1, c1

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

Removed 1 trailing tuple parts

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(c(z0, z1)) → c(f(g(c(z0, z1))), f(h(c(z0, z1))))
g(c(0, z0)) → z0
h(c(1, z0)) → z0
Tuples:

F(c(x0, x1)) → c1(F(h(c(x0, x1))))
F(c(0, z0)) → c1(F(z0))
S tuples:

F(c(x0, x1)) → c1(F(h(c(x0, x1))))
F(c(0, z0)) → c1(F(z0))
K tuples:none
Defined Rule Symbols:

f, g, h

Defined Pair Symbols:

F

Compound Symbols:

c1

(9) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

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

F(c(0, z0)) → c1(F(z0))
We considered the (Usable) Rules:

h(c(1, z0)) → z0
And the Tuples:

F(c(x0, x1)) → c1(F(h(c(x0, x1))))
F(c(0, z0)) → c1(F(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = [5]   
POL(1) = 0   
POL(F(x1)) = [2]x1   
POL(c(x1, x2)) = [1] + x2   
POL(c1(x1)) = x1   
POL(h(x1)) = x1   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(c(z0, z1)) → c(f(g(c(z0, z1))), f(h(c(z0, z1))))
g(c(0, z0)) → z0
h(c(1, z0)) → z0
Tuples:

F(c(x0, x1)) → c1(F(h(c(x0, x1))))
F(c(0, z0)) → c1(F(z0))
S tuples:

F(c(x0, x1)) → c1(F(h(c(x0, x1))))
K tuples:

F(c(0, z0)) → c1(F(z0))
Defined Rule Symbols:

f, g, h

Defined Pair Symbols:

F

Compound Symbols:

c1

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

Use narrowing to replace F(c(x0, x1)) → c1(F(h(c(x0, x1)))) by

F(c(1, z0)) → c1(F(z0))
F(c(x0, x1)) → c1

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(c(z0, z1)) → c(f(g(c(z0, z1))), f(h(c(z0, z1))))
g(c(0, z0)) → z0
h(c(1, z0)) → z0
Tuples:

F(c(0, z0)) → c1(F(z0))
F(c(1, z0)) → c1(F(z0))
F(c(x0, x1)) → c1
S tuples:

F(c(1, z0)) → c1(F(z0))
F(c(x0, x1)) → c1
K tuples:

F(c(0, z0)) → c1(F(z0))
Defined Rule Symbols:

f, g, h

Defined Pair Symbols:

F

Compound Symbols:

c1, c1

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

Removed 1 trailing nodes:

F(c(x0, x1)) → c1

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(c(z0, z1)) → c(f(g(c(z0, z1))), f(h(c(z0, z1))))
g(c(0, z0)) → z0
h(c(1, z0)) → z0
Tuples:

F(c(0, z0)) → c1(F(z0))
F(c(1, z0)) → c1(F(z0))
S tuples:

F(c(1, z0)) → c1(F(z0))
K tuples:

F(c(0, z0)) → c1(F(z0))
Defined Rule Symbols:

f, g, h

Defined Pair Symbols:

F

Compound Symbols:

c1

(15) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

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

F(c(1, z0)) → c1(F(z0))
We considered the (Usable) Rules:none
And the Tuples:

F(c(0, z0)) → c1(F(z0))
F(c(1, z0)) → c1(F(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(1) = [3]   
POL(F(x1)) = [2]x1   
POL(c(x1, x2)) = [1] + x2   
POL(c1(x1)) = x1   

(16) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(c(z0, z1)) → c(f(g(c(z0, z1))), f(h(c(z0, z1))))
g(c(0, z0)) → z0
h(c(1, z0)) → z0
Tuples:

F(c(0, z0)) → c1(F(z0))
F(c(1, z0)) → c1(F(z0))
S tuples:none
K tuples:

F(c(0, z0)) → c1(F(z0))
F(c(1, z0)) → c1(F(z0))
Defined Rule Symbols:

f, g, h

Defined Pair Symbols:

F

Compound Symbols:

c1

(17) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(18) BOUNDS(O(1), O(1))