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

(0) Obligation:

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

dx(X) → one
dx(a) → zero
dx(plus(ALPHA, BETA)) → plus(dx(ALPHA), dx(BETA))
dx(times(ALPHA, BETA)) → plus(times(BETA, dx(ALPHA)), times(ALPHA, dx(BETA)))
dx(minus(ALPHA, BETA)) → minus(dx(ALPHA), dx(BETA))
dx(neg(ALPHA)) → neg(dx(ALPHA))
dx(div(ALPHA, BETA)) → minus(div(dx(ALPHA), BETA), times(ALPHA, div(dx(BETA), exp(BETA, two))))
dx(ln(ALPHA)) → div(dx(ALPHA), ALPHA)
dx(exp(ALPHA, BETA)) → plus(times(BETA, times(exp(ALPHA, minus(BETA, one)), dx(ALPHA))), times(exp(ALPHA, BETA), times(ln(ALPHA), dx(BETA))))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

dx(z0) → one
dx(a) → zero
dx(plus(z0, z1)) → plus(dx(z0), dx(z1))
dx(times(z0, z1)) → plus(times(z1, dx(z0)), times(z0, dx(z1)))
dx(minus(z0, z1)) → minus(dx(z0), dx(z1))
dx(neg(z0)) → neg(dx(z0))
dx(div(z0, z1)) → minus(div(dx(z0), z1), times(z0, div(dx(z1), exp(z1, two))))
dx(ln(z0)) → div(dx(z0), z0)
dx(exp(z0, z1)) → plus(times(z1, times(exp(z0, minus(z1, one)), dx(z0))), times(exp(z0, z1), times(ln(z0), dx(z1))))

Tuples:

DX(plus(z0, z1)) → c2(DX(z0), DX(z1))
DX(times(z0, z1)) → c3(DX(z0), DX(z1))
DX(minus(z0, z1)) → c4(DX(z0), DX(z1))
DX(neg(z0)) → c5(DX(z0))
DX(div(z0, z1)) → c6(DX(z0), DX(z1))
DX(ln(z0)) → c7(DX(z0))
DX(exp(z0, z1)) → c8(DX(z0), DX(z1))

S tuples:

DX(plus(z0, z1)) → c2(DX(z0), DX(z1))
DX(times(z0, z1)) → c3(DX(z0), DX(z1))
DX(minus(z0, z1)) → c4(DX(z0), DX(z1))
DX(neg(z0)) → c5(DX(z0))
DX(div(z0, z1)) → c6(DX(z0), DX(z1))
DX(ln(z0)) → c7(DX(z0))
DX(exp(z0, z1)) → c8(DX(z0), DX(z1))

K tuples:none
Defined Rule Symbols:

dx

Defined Pair Symbols:

DX

Compound Symbols:

c2, c3, c4, c5, c6, c7, c8

(3) 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.

DX(plus(z0, z1)) → c2(DX(z0), DX(z1))
DX(times(z0, z1)) → c3(DX(z0), DX(z1))
DX(minus(z0, z1)) → c4(DX(z0), DX(z1))
DX(neg(z0)) → c5(DX(z0))
DX(ln(z0)) → c7(DX(z0))

We considered the (Usable) Rules:none
And the Tuples:

DX(plus(z0, z1)) → c2(DX(z0), DX(z1))
DX(times(z0, z1)) → c3(DX(z0), DX(z1))
DX(minus(z0, z1)) → c4(DX(z0), DX(z1))
DX(neg(z0)) → c5(DX(z0))
DX(div(z0, z1)) → c6(DX(z0), DX(z1))
DX(ln(z0)) → c7(DX(z0))
DX(exp(z0, z1)) → c8(DX(z0), DX(z1))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(DX(x₁)) = [2]x₁   
POL(c2(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(c4(x₁, x₂)) = x₁ + x₂   
POL(c5(x₁)) = x₁   
POL(c6(x₁, x₂)) = x₁ + x₂   
POL(c7(x₁)) = x₁   
POL(c8(x₁, x₂)) = x₁ + x₂   
POL(div(x₁, x₂)) = x₁ + x₂   
POL(exp(x₁, x₂)) = x₁ + x₂   
POL(ln(x₁)) = [3] + x₁   
POL(minus(x₁, x₂)) = [2] + x₁ + x₂   
POL(neg(x₁)) = [5] + x₁   
POL(plus(x₁, x₂)) = [4] + x₁ + x₂   
POL(times(x₁, x₂)) = [3] + x₁ + x₂   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

dx(z0) → one
dx(a) → zero
dx(plus(z0, z1)) → plus(dx(z0), dx(z1))
dx(times(z0, z1)) → plus(times(z1, dx(z0)), times(z0, dx(z1)))
dx(minus(z0, z1)) → minus(dx(z0), dx(z1))
dx(neg(z0)) → neg(dx(z0))
dx(div(z0, z1)) → minus(div(dx(z0), z1), times(z0, div(dx(z1), exp(z1, two))))
dx(ln(z0)) → div(dx(z0), z0)
dx(exp(z0, z1)) → plus(times(z1, times(exp(z0, minus(z1, one)), dx(z0))), times(exp(z0, z1), times(ln(z0), dx(z1))))

Tuples:

DX(plus(z0, z1)) → c2(DX(z0), DX(z1))
DX(times(z0, z1)) → c3(DX(z0), DX(z1))
DX(minus(z0, z1)) → c4(DX(z0), DX(z1))
DX(neg(z0)) → c5(DX(z0))
DX(div(z0, z1)) → c6(DX(z0), DX(z1))
DX(ln(z0)) → c7(DX(z0))
DX(exp(z0, z1)) → c8(DX(z0), DX(z1))

S tuples:

DX(div(z0, z1)) → c6(DX(z0), DX(z1))
DX(exp(z0, z1)) → c8(DX(z0), DX(z1))

K tuples:

DX(plus(z0, z1)) → c2(DX(z0), DX(z1))
DX(times(z0, z1)) → c3(DX(z0), DX(z1))
DX(minus(z0, z1)) → c4(DX(z0), DX(z1))
DX(neg(z0)) → c5(DX(z0))
DX(ln(z0)) → c7(DX(z0))

Defined Rule Symbols:

dx

Defined Pair Symbols:

DX

Compound Symbols:

c2, c3, c4, c5, c6, c7, c8

(5) 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.

DX(div(z0, z1)) → c6(DX(z0), DX(z1))
DX(exp(z0, z1)) → c8(DX(z0), DX(z1))

We considered the (Usable) Rules:none
And the Tuples:

DX(plus(z0, z1)) → c2(DX(z0), DX(z1))
DX(times(z0, z1)) → c3(DX(z0), DX(z1))
DX(minus(z0, z1)) → c4(DX(z0), DX(z1))
DX(neg(z0)) → c5(DX(z0))
DX(div(z0, z1)) → c6(DX(z0), DX(z1))
DX(ln(z0)) → c7(DX(z0))
DX(exp(z0, z1)) → c8(DX(z0), DX(z1))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(DX(x₁)) = [2] + [4]x₁   
POL(c2(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(c4(x₁, x₂)) = x₁ + x₂   
POL(c5(x₁)) = x₁   
POL(c6(x₁, x₂)) = x₁ + x₂   
POL(c7(x₁)) = x₁   
POL(c8(x₁, x₂)) = x₁ + x₂   
POL(div(x₁, x₂)) = [5] + x₁ + x₂   
POL(exp(x₁, x₂)) = [4] + x₁ + x₂   
POL(ln(x₁)) = x₁   
POL(minus(x₁, x₂)) = [5] + x₁ + x₂   
POL(neg(x₁)) = x₁   
POL(plus(x₁, x₂)) = [4] + x₁ + x₂   
POL(times(x₁, x₂)) = [2] + x₁ + x₂   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

dx(z0) → one
dx(a) → zero
dx(plus(z0, z1)) → plus(dx(z0), dx(z1))
dx(times(z0, z1)) → plus(times(z1, dx(z0)), times(z0, dx(z1)))
dx(minus(z0, z1)) → minus(dx(z0), dx(z1))
dx(neg(z0)) → neg(dx(z0))
dx(div(z0, z1)) → minus(div(dx(z0), z1), times(z0, div(dx(z1), exp(z1, two))))
dx(ln(z0)) → div(dx(z0), z0)
dx(exp(z0, z1)) → plus(times(z1, times(exp(z0, minus(z1, one)), dx(z0))), times(exp(z0, z1), times(ln(z0), dx(z1))))

Tuples:

DX(plus(z0, z1)) → c2(DX(z0), DX(z1))
DX(times(z0, z1)) → c3(DX(z0), DX(z1))
DX(minus(z0, z1)) → c4(DX(z0), DX(z1))
DX(neg(z0)) → c5(DX(z0))
DX(div(z0, z1)) → c6(DX(z0), DX(z1))
DX(ln(z0)) → c7(DX(z0))
DX(exp(z0, z1)) → c8(DX(z0), DX(z1))

S tuples:none
K tuples:

DX(plus(z0, z1)) → c2(DX(z0), DX(z1))
DX(times(z0, z1)) → c3(DX(z0), DX(z1))
DX(minus(z0, z1)) → c4(DX(z0), DX(z1))
DX(neg(z0)) → c5(DX(z0))
DX(ln(z0)) → c7(DX(z0))
DX(div(z0, z1)) → c6(DX(z0), DX(z1))
DX(exp(z0, z1)) → c8(DX(z0), DX(z1))

Defined Rule Symbols:

dx

Defined Pair Symbols:

DX

Compound Symbols:

c2, c3, c4, c5, c6, c7, c8

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(8) BOUNDS(O(1), O(1))