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

D(t) → 1
D(constant) → 0
D(+(x, y)) → +(D(x), D(y))
D(*(x, y)) → +(*(y, D(x)), *(x, D(y)))
D(-(x, y)) → -(D(x), D(y))
D(minus(x)) → minus(D(x))
D(div(x, y)) → -(div(D(x), y), div(*(x, D(y)), pow(y, 2)))
D(ln(x)) → div(D(x), x)
D(pow(x, y)) → +(*(*(y, pow(x, -(y, 1))), D(x)), *(*(pow(x, y), ln(x)), D(y)))

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

D(t) → 1
D(constant) → 0
D(+(z0, z1)) → +(D(z0), D(z1))
D(*(z0, z1)) → +(*(z1, D(z0)), *(z0, D(z1)))
D(-(z0, z1)) → -(D(z0), D(z1))
D(minus(z0)) → minus(D(z0))
D(div(z0, z1)) → -(div(D(z0), z1), div(*(z0, D(z1)), pow(z1, 2)))
D(ln(z0)) → div(D(z0), z0)
D(pow(z0, z1)) → +(*(*(z1, pow(z0, -(z1, 1))), D(z0)), *(*(pow(z0, z1), ln(z0)), D(z1)))

Tuples:

D'(t) → c
D'(constant) → c1
D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
D'(minus(z0)) → c5(D'(z0))
D'(div(z0, z1)) → c6(D'(z0), D'(z1))
D'(ln(z0)) → c7(D'(z0))
D'(pow(z0, z1)) → c8(D'(z0), D'(z1))

S tuples:

D'(t) → c
D'(constant) → c1
D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
D'(minus(z0)) → c5(D'(z0))
D'(div(z0, z1)) → c6(D'(z0), D'(z1))
D'(ln(z0)) → c7(D'(z0))
D'(pow(z0, z1)) → c8(D'(z0), D'(z1))

K tuples:none
Defined Rule Symbols:

D

Defined Pair Symbols:

D'

Compound Symbols:

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

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

Removed 2 trailing nodes:

D'(constant) → c1
D'(t) → c

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

D(t) → 1
D(constant) → 0
D(+(z0, z1)) → +(D(z0), D(z1))
D(*(z0, z1)) → +(*(z1, D(z0)), *(z0, D(z1)))
D(-(z0, z1)) → -(D(z0), D(z1))
D(minus(z0)) → minus(D(z0))
D(div(z0, z1)) → -(div(D(z0), z1), div(*(z0, D(z1)), pow(z1, 2)))
D(ln(z0)) → div(D(z0), z0)
D(pow(z0, z1)) → +(*(*(z1, pow(z0, -(z1, 1))), D(z0)), *(*(pow(z0, z1), ln(z0)), D(z1)))

Tuples:

D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
D'(minus(z0)) → c5(D'(z0))
D'(div(z0, z1)) → c6(D'(z0), D'(z1))
D'(ln(z0)) → c7(D'(z0))
D'(pow(z0, z1)) → c8(D'(z0), D'(z1))

S tuples:

D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
D'(minus(z0)) → c5(D'(z0))
D'(div(z0, z1)) → c6(D'(z0), D'(z1))
D'(ln(z0)) → c7(D'(z0))
D'(pow(z0, z1)) → c8(D'(z0), D'(z1))

K tuples:none
Defined Rule Symbols:

D

Defined Pair Symbols:

D'

Compound Symbols:

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

(5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

D(t) → 1
D(constant) → 0
D(+(z0, z1)) → +(D(z0), D(z1))
D(*(z0, z1)) → +(*(z1, D(z0)), *(z0, D(z1)))
D(-(z0, z1)) → -(D(z0), D(z1))
D(minus(z0)) → minus(D(z0))
D(div(z0, z1)) → -(div(D(z0), z1), div(*(z0, D(z1)), pow(z1, 2)))
D(ln(z0)) → div(D(z0), z0)
D(pow(z0, z1)) → +(*(*(z1, pow(z0, -(z1, 1))), D(z0)), *(*(pow(z0, z1), ln(z0)), D(z1)))

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
D'(minus(z0)) → c5(D'(z0))
D'(div(z0, z1)) → c6(D'(z0), D'(z1))
D'(ln(z0)) → c7(D'(z0))
D'(pow(z0, z1)) → c8(D'(z0), D'(z1))

S tuples:

D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
D'(minus(z0)) → c5(D'(z0))
D'(div(z0, z1)) → c6(D'(z0), D'(z1))
D'(ln(z0)) → c7(D'(z0))
D'(pow(z0, z1)) → c8(D'(z0), D'(z1))

K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

D'

Compound Symbols:

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

(7) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)

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

D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
D'(minus(z0)) → c5(D'(z0))
D'(pow(z0, z1)) → c8(D'(z0), D'(z1))

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

D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
D'(minus(z0)) → c5(D'(z0))
D'(div(z0, z1)) → c6(D'(z0), D'(z1))
D'(ln(z0)) → c7(D'(z0))
D'(pow(z0, z1)) → c8(D'(z0), D'(z1))

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

POL(*(x₁, x₂)) = [1] + x₁ + x₂   
POL(+(x₁, x₂)) = x₁ + x₂   
POL(-(x₁, x₂)) = [1] + x₁ + x₂   
POL(D'(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(ln(x₁)) = x₁   
POL(minus(x₁)) = [1] + x₁   
POL(pow(x₁, x₂)) = [1] + x₁ + x₂   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
D'(minus(z0)) → c5(D'(z0))
D'(div(z0, z1)) → c6(D'(z0), D'(z1))
D'(ln(z0)) → c7(D'(z0))
D'(pow(z0, z1)) → c8(D'(z0), D'(z1))

S tuples:

D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(div(z0, z1)) → c6(D'(z0), D'(z1))
D'(ln(z0)) → c7(D'(z0))

K tuples:

D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
D'(minus(z0)) → c5(D'(z0))
D'(pow(z0, z1)) → c8(D'(z0), D'(z1))

Defined Rule Symbols:none

Defined Pair Symbols:

D'

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.

D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(div(z0, z1)) → c6(D'(z0), D'(z1))

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

D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
D'(minus(z0)) → c5(D'(z0))
D'(div(z0, z1)) → c6(D'(z0), D'(z1))
D'(ln(z0)) → c7(D'(z0))
D'(pow(z0, z1)) → c8(D'(z0), D'(z1))

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

POL(*(x₁, x₂)) = x₁ + x₂   
POL(+(x₁, x₂)) = [1] + x₁ + x₂   
POL(-(x₁, x₂)) = x₁ + x₂   
POL(D'(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)) = [1] + x₁ + x₂   
POL(ln(x₁)) = x₁   
POL(minus(x₁)) = x₁   
POL(pow(x₁, x₂)) = x₁ + x₂   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
D'(minus(z0)) → c5(D'(z0))
D'(div(z0, z1)) → c6(D'(z0), D'(z1))
D'(ln(z0)) → c7(D'(z0))
D'(pow(z0, z1)) → c8(D'(z0), D'(z1))

S tuples:

D'(ln(z0)) → c7(D'(z0))

K tuples:

D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
D'(minus(z0)) → c5(D'(z0))
D'(pow(z0, z1)) → c8(D'(z0), D'(z1))
D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(div(z0, z1)) → c6(D'(z0), D'(z1))

Defined Rule Symbols:none

Defined Pair Symbols:

D'

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.

D'(ln(z0)) → c7(D'(z0))

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

D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
D'(minus(z0)) → c5(D'(z0))
D'(div(z0, z1)) → c6(D'(z0), D'(z1))
D'(ln(z0)) → c7(D'(z0))
D'(pow(z0, z1)) → c8(D'(z0), D'(z1))

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

POL(*(x₁, x₂)) = [2] + x₁ + x₂   
POL(+(x₁, x₂)) = [2] + x₁ + x₂   
POL(-(x₁, x₂)) = [2] + x₁ + x₂   
POL(D'(x₁)) = [1] + 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)) = [2] + x₁ + x₂   
POL(ln(x₁)) = [1] + x₁   
POL(minus(x₁)) = x₁   
POL(pow(x₁, x₂)) = [2] + x₁ + x₂   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
D'(minus(z0)) → c5(D'(z0))
D'(div(z0, z1)) → c6(D'(z0), D'(z1))
D'(ln(z0)) → c7(D'(z0))
D'(pow(z0, z1)) → c8(D'(z0), D'(z1))

S tuples:none
K tuples:

D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
D'(minus(z0)) → c5(D'(z0))
D'(pow(z0, z1)) → c8(D'(z0), D'(z1))
D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(div(z0, z1)) → c6(D'(z0), D'(z1))
D'(ln(z0)) → c7(D'(z0))

Defined Rule Symbols:none

Defined Pair Symbols:

D'

Compound Symbols:

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

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

The set S is empty

(14) BOUNDS(1, 1)