### (0) Obligation:

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

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: FULL

### (1) RcToIrcProof (BOTH BOUNDS(ID, ID) transformation)

Converted rc-obligation to irc-obligation.

As the TRS does not nest defined symbols, we have rc = irc.

### (2) 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:

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

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

Converted Cpx (relative) TRS to CDT

### (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(z0) → c
DX(a) → c1
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(z0) → c
DX(a) → c1
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:

c, c1, c2, c3, c4, c5, c6, c7, c8

### (5) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)

Removed 2 trailing nodes:

DX(a) → c1
DX(z0) → c

### (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:

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

### (7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

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))))

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
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:none

Defined Pair Symbols:

DX

Compound Symbols:

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

### (9) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)

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

DX(times(z0, z1)) → c3(DX(z0), DX(z1))
DX(neg(z0)) → c5(DX(z0))
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)) = 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)) = [1] + x1 + x2
POL(ln(x1)) = x1
POL(minus(x1, x2)) = x1 + x2
POL(neg(x1)) = [1] + x1
POL(plus(x1, x2)) = x1 + x2
POL(times(x1, x2)) = [1] + x1 + x2

### (10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
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(minus(z0, z1)) → c4(DX(z0), DX(z1))
DX(div(z0, z1)) → c6(DX(z0), DX(z1))
DX(ln(z0)) → c7(DX(z0))
K tuples:

DX(times(z0, z1)) → c3(DX(z0), DX(z1))
DX(neg(z0)) → c5(DX(z0))
DX(exp(z0, z1)) → c8(DX(z0), DX(z1))
Defined Rule Symbols:none

Defined Pair Symbols:

DX

Compound Symbols:

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

### (11) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)

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

DX(minus(z0, z1)) → c4(DX(z0), DX(z1))
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)) = 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)) = [1] + x1
POL(minus(x1, x2)) = [1] + x1 + x2
POL(neg(x1)) = x1
POL(plus(x1, x2)) = x1 + x2
POL(times(x1, x2)) = x1 + x2

### (12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
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(div(z0, z1)) → c6(DX(z0), DX(z1))
K tuples:

DX(times(z0, z1)) → c3(DX(z0), DX(z1))
DX(neg(z0)) → c5(DX(z0))
DX(exp(z0, z1)) → c8(DX(z0), DX(z1))
DX(minus(z0, z1)) → c4(DX(z0), DX(z1))
DX(ln(z0)) → c7(DX(z0))
Defined Rule Symbols:none

Defined Pair Symbols:

DX

Compound Symbols:

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

### (13) CdtRuleRemovalProof (UPPER BOUND(ADD(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(div(z0, z1)) → c6(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]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)) = [1] + x1 + x2
POL(exp(x1, x2)) = x1 + x2
POL(ln(x1)) = x1
POL(minus(x1, x2)) = [3] + x1 + x2
POL(neg(x1)) = x1
POL(plus(x1, x2)) = [1] + x1 + x2
POL(times(x1, x2)) = x1 + x2

### (14) Obligation:

Complexity Dependency Tuples Problem
Rules:none
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(times(z0, z1)) → c3(DX(z0), DX(z1))
DX(neg(z0)) → c5(DX(z0))
DX(exp(z0, z1)) → c8(DX(z0), DX(z1))
DX(minus(z0, z1)) → c4(DX(z0), DX(z1))
DX(ln(z0)) → c7(DX(z0))
DX(plus(z0, z1)) → c2(DX(z0), DX(z1))
DX(div(z0, z1)) → c6(DX(z0), DX(z1))
Defined Rule Symbols:none

Defined Pair Symbols:

DX

Compound Symbols:

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

### (15) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty